Question 1515 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\Rightarrow\text{dy}=(\tan^{-1}\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\tan^{-1}\text{x})\text{dx}$
$\Rightarrow\text{y}=\int1\times\tan^{-1}\text{x}\text{ dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x }-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$
So, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is the solution o the given differential equation.
View full question & answer→Question 1525 Marks
Form the differential equation by eliminating A and B in $Ax^2 + By^2 = 1.$
AnswerGiven equation is $Ax^2 + By^2 = 1$
On differentiating both sides w.r.t.x, we get
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2\text{By}\frac{\text{dy}}{\text{dx}}=-2\text{Ax}$
$\Rightarrow\text{By}\frac{\text{dy}}{\text{dx}}=-\text{Ax}$
$\Rightarrow\frac{\text{y}}{\text{x}}.\frac{\text{dy}}{\text{dx}}=-\frac{\text{A}}{\text{B}}$
Again, differentiating w.r.t.x, we get
$\frac{\text{y}}{\text{x}}.\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\frac{\text{dy}}{\text{dx}}.\bigg(\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}}{\text{x}^2}\bigg)=0$
$\Rightarrow\frac{\text{y}}{\text{x}}.\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\frac{\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\text{x}^2}=0$
$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2-\text{y}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)=0$
$\Rightarrow\text{xy}\text{y}''+\text{x}(\text{y}')^2-\text{y}\text{y}'=0$
View full question & answer→Question 1535 Marks
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0,\text{y}(2)=\text{x}$
Answer$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0,\text{y}(2)=\text{x}$
It is a homogeneous equation. put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So, $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}-\sin\Big(\frac{\text{vx}}{\text{x}}\Big)$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\sin\text{v}$
$\frac{\text{dv}}{\sin\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\text{cosec(v)dv}=-\frac{\text{dx}}{\text{x}}$
integrating both sides we get,
$\log(\text{cosec(v)}-\cot(\text{v}))=-\log\text{x}+\log\text{c}$
$\log\Big(\text{cosec}\Big(\frac{\text{y}}{\text{x}}\Big)-\cot\Big(\frac{\text{y}}{\text{x}}\Big)\Big)=-\log\text{x}+\log\text{c}$
Putting the values $\text{x}=2$ and $\text{y}=\pi$
$\log\Big(\text{cosec}\Big(\frac{\pi}{2}\Big)-\cot\Big(\frac{\pi}{2}\Big)\Big)=-\log2+\log\text{C}$
$\text{C}=0$
$\log\Big(\text{cosec}\Big(\frac{\text{y}}{\text{x}}\Big)-\cot\Big(\frac{\text{y}}{\text{x}}\Big)\Big)=-\log\text{x}$
View full question & answer→Question 1545 Marks
Solve the following differential equations $\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}(\log\text{x}+1)}{\sin\text{y+y}\cos\text{y}},$ given that $\text{y}=0,$ when $\text{x}=1.$
Answer$\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}(\log\text{x}+1)}{\sin\text{y+y}\cos\text{y}},\text{y}=0$ at $\text{x}=1$
$\int(\sin\text{y+y}\cos\text{y})\text{dy}=\int2\text{x}(\log\text{x}+1)\text{dx}$
$\Rightarrow\int\sin\text{y dy}+\int\text{y}\cos\text{y dy}=\int2\text{x}\log\text{x dx}+2\int\text{x dx}$
$\Rightarrow-\cos\text{y}+\big[\text{y}\times\int\cos\text{y dy}-\int(1\times\int\cos\text{y dy})\text{dy}\big]\\=2\Big[\log\text{x}\int\text{x dx}-\int\Big(\frac{1}{\text{x}}\int\text{x dx}\Big)\text{dx}\Big]+\text{x}^2+\text{C}$
$\Rightarrow-\cos\text{y + y}\sin\text{y}-\int\sin\text{y dy}=2\frac{\text{x}^2}{2}\log\text{x}-2\int\frac{\text{x}}{2}\text{dx}+\text{x}^2+\text{C}$
$\Rightarrow-\cos\text{y + y}\sin\text{y}+\cos\text{y}=\text{x}^2\log\text{x}-\frac{\text{x}^2}{2}+\text{x}^2+\text{C}$
$\text{y}\sin\text{y}=\text{x}^2\log\text{x}+\frac{\text{x}^2}{2}+\text{C}$
Put $\text{y}=0,\text{x}=1$
$0=0+\frac{1}{2}+\text{C}$
$\text{C}=-\frac{1}{2}$
Put $\text{C}=-\frac{1}{2}$ in equation (1),
$\text{y}\sin\text{y = x}^2\log\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}$
$2\text{y}\sin\text{y}=2\text{x}^2\log\text{x + x}^2-1$
View full question & answer→Question 1555 Marks
Show that $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
AnswerWe have,
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}+2\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\big(2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}\big)+2\big(\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}\big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 1565 Marks
Solve the following differential equation:
$\big(\text{y}^2-2\text{xy}\big)\text{dx}=\big(\text{x}^2-2\text{xy}\big)\text{dy}$
AnswerHere, $\big(\text{y}^2-2\text{xy}\big)\text{dx}=\big(\text{x}^2-2\text{xy}\big)\text{dy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-2\text{xy}}{\text{x}^2-2\text{xy}}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-2\text{xvx}}{\text{x}^2-2\text{xvx}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-2\text{v}}{1-2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-2\text{v}}{1-2\text{v}}-\text{v}$
$=\frac{\text{v}^2-2\text{v}-\text{v}+2\text{v}^2}{1-2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{3\text{v}^2-3\text{v}}{1-2\text{v}}$
$\frac{1-2\text{v}}{3(\text{v}^2-\text{v})}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{-(2\text{v}-1)}{3(\text{v}^2-\text{v})}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\int\frac{2\text{v}-1}{\text{v}^2-\text{v}}\text{dv}=-3\int\frac{\text{dx}}{\text{x}}$
$\log\big|\text{v}^2-\text{v}\big|-3\log|\text{x}|+\log\text{C}$
$\text{v}^2-\text{v}=\frac{\text{C}}{\text{x}^3}$
$\frac{\text{y}^2}{\text{x}^2}-\frac{\text{y}}{\text{x}}=\frac{\text{C}}{\text{x}^3}$
$\text{y}^2-\text{xy}=\frac{\text{C}}{\text{x}}$
$\text{x}\big(\text{y}^2-\text{xy}\big)=\text{C}$
View full question & answer→Question 1575 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})}{\text{y}(2\log\text{y}+1)}$
Answer$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})}{\text{y}(2\log\text{y}+1)}$
$\Rightarrow\text{y}(2\log\text{y}+1)\text{dy}=\text{e}^{\text{x}}(\sin^2\text{x}+\sin2\text{x})\text{dx}$
$\Rightarrow(2\text{y}\log\text{y+y})\text{dy}=(\text{e}^{\text{x}}\sin^2\text{x + e}^{\text{x}}\sin2\text{x})\text{dx}$
$\Rightarrow2\text{y}\log\text{y}\text{ dy}+\text{y dy}=\text{e}^{\text{x}}\sin^2\text{x dx}+\text{e}^{\text{x}}\sin2\text{x}\text{ dx}$
Integrating both sides, we get
$2\int\text{y}\log\text{y dy}+\int\text{y dy}=\int\text{e}^{\text{x}}\sin^2\text{x dx}+\int\text{e}^{\text{x}}\sin2\text{x dx}$
$\Rightarrow2\Big[\log\text{y}\int\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}(\log\text{ y})\int\text{y dy}\Big\}\Big]\text{dy}+\int\text{y dy}\\=\sin^2\text{x}\int\text{e}^{\text{x}}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\sin^2\text{x})\int\text{e}^{\text{x}}\text{dx}\Big]\text{dx}+\int\text{e}^{\text{x}}\sin2\text{x dx} $
$\Rightarrow2\Big[\log\text{y}\Big(\frac{\text{y}^2}{2}\Big)-\int\Big(\frac{1}{\text{y}}\Big)\frac{\text{y}^2}{2}\text{dy}\Big]+\int\text{y dy}\\=\sin^2\text{x }\text{e}^{\text{x}}-\int\big[2\sin\text{x}\cos\text{x}\text{ e}^{\text{x}}\big]\text{dx}+\int\text{e}^{\text{x}}\sin2\text{x dx + C}$
$\Rightarrow\text{y}^2\log\text{ y}-\int\text{y dy}+\int\text{y dy}\\=\text{e}^{\text{x}}\sin^2\text{x}-\int\text{e}^{\text{x}}\sin2\text{x dx}+\int\text{e}^{\text{x}}\sin2\text{x dx + C}$
$\Rightarrow\text{y}^2\log\text{y}=\text{e}^{\text{x}}\sin^2\text{x + C}$
View full question & answer→Question 1585 Marks
Solve the following initial value problems:
$\text{dy}=\cos\text{x}(2-\text{y cosecx})\text{dx}$
AnswerWe have,
$\text{dy}=\cos\text{x}(2-\text{y cosecx})\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\cos\text{x}-\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=2\cos\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{x dx}}$
$=\text{e}^{\log\sin\text{x}}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=2\sin\text{x }\cos\text{x}$
$\Rightarrow\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=\int\sin2\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}$
Hence, $\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}$ is the required solution.
View full question & answer→Question 1595 Marks
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Answer$\text{V}=\frac{4}{3}\pi\text{r}^3$
Given:
$\frac{\text{dv}}{\text{dt}}=-\text{k}$ (where k > 0)
$\Rightarrow\frac{\text{d}}{\text{dt}}\Big(\frac{4}{3}\pi\text{r}^3\Big)=-\text{k}$
$\Rightarrow4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}=-\text{k}$
$\Rightarrow4\pi\text{r}^{2}\text{dr}=-\text{kdt}$
Integrating both sides, we get
$\int4\pi\text{r}^2\text{dr}=-\int\text{kdt}$
$\frac{4}{3}\pi\text{r}^3=-\text{kt + C}...(1)$
It is given that at $\text{t}=0,\text{r}=3.$
$\text{C}=36\pi$
putting $\text{C}=36\pi$ in (1), we get
$\frac{4}{3}\pi\text{r}^3=-\text{kt}+36\pi...(2)$
It is also given that at $\text{t}=3,\text{r}=6.$
Putting $\text{t}=3$ and $\text{r}=6$ in (1), we get
$288\pi=-3\text{k}+36\pi$
$\Rightarrow\text{k}=-84\pi$
Putting $\text{k}=-84\pi$ in (2), we get
$\frac{4}{3}\pi\text{r}^3=84\pi\text{t}+36\pi$
$\Rightarrow\text{r}^3=63\text{t}+27$
$\Rightarrow\text{r}=(63\text{t}+27)^{\frac{1}{3}}$
View full question & answer→Question 1605 Marks
Solve the following differential equations:$\tan\text{y dx}+\sec^2\text{y}\tan\text{x dy}=0$
AnswerWe have,
$\tan\text{y dx}+\sec^2\text{y}\tan\text{x dy}=0$
$\Rightarrow\sec^2\text{y}\tan\text{x dy}=-\tan\text{y dx}$
$\Rightarrow\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}=-\frac{1}{\tan\text{x}}\text{dx}$
$\Rightarrow\frac{1}{\cos^2\text{y}}\times\frac{\cos\text{y}}{\sin\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow\frac{1}{\sin\text{y}\cos\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow\frac{2}{\sin2\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow2\text{ cosec }2\text{y dy}=-\cot\text{x dx}$
Integrating both sides, we get
$2\int\text{cosec}\text{ 2y dy}=-\int\cot\text{x dx}$
$\Rightarrow\log\tan\text{x}=-\log\sin\text{x}=\log\text{C}$
$\Rightarrow\log \tan\text{x}+\log\sin\text{x}=\log\text{C}$
$\Rightarrow\log(\tan\text{ x}\times\sin\text{x})=\log\text{C}$
$\Rightarrow\tan\text{x}\times\sin\text{x}=\text{C}$
View full question & answer→Question 1615 Marks
In a bank principal increases at the rate of 5% per year. An amount of Rs $1000$ is deposited with this bank, how much will it worth after $10$ years $(e^{0.5_=}1.648).$
AnswerLet $p$ and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of $5\%$ per year.
$\Rightarrow\frac{\text{dp}}{\text{dt}}=\Big(\frac{5}{100}\Big)\text{p}$
$\Rightarrow\frac{\text{dp}}{\text{dt}}=\frac{\text{P}}{20}$
$\Rightarrow\frac{\text{dp}}{\text{p}}=\frac{\text{dt}}{20}$
Integrating both sides, we get:
$\int\frac{\text{dp}}{\text{p}}=\frac{1}{20}\int\text{dt}$
$\Rightarrow\log\text{p}=\frac{\text{t}}{20}+\text{C}$
$\Rightarrow\text{p}=\text{e}^{\frac{\text{t}}{20}}+\text{C}...(1)$
Now, when $\text{t}=0,\text{P}=1000.$
$1000=\text{e}^{\text{C}}...(2)$
View full question & answer→Question 1625 Marks
The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?
AnswerLet the population at any time t be P.
Given: $\frac{\text{dP}}{\text{dt}}\propto\text{P}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=\beta\text{P}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\beta\text{dt}$
$\Rightarrow\log|\text{P}|=\beta\text{t}+\log\text{C}\ ...(\text{ii})$
Now,
At t = 1990, P = 200000 and at t = 2000, P = 250000
$\therefore \log 200000=1990\beta+\log\text{C}\ ...(\text{ii})$
$ \log 250000=2000\beta+\log\text{C}\ ...(\text{iii})$
Subtracting (iii) from (ii), we get
$\log 200000-\log25000=10\beta$
$\Rightarrow\beta=\frac{1}{10}\log(\frac{5}{4})$
Putting $\beta=\frac{1}{10}\log(\frac{5}{4})$ in (ii), we get
$\log200000=1990\times\frac{1}{10}\log(\frac{5}{4})+\log\text{C}$
$\Rightarrow\log200000=199\log(\frac{5}{4})+\log\text{C}$
$\Rightarrow\log\text{C}=\log200000-199\log(\frac{5}{4})$
Putting $\beta=\frac{1}{10}\log(\frac{5}{4}), \log\text{C}=\log200000-199\log(\frac{5}{4})$
$\log|\text{P}|=\frac{1}{10}\times2010\log(\frac{5}{4})+\log200000-199\log(\frac{5}{4})$
$\Rightarrow\log|\text{P}|=201\log(\frac{5}{4})+\log200000-199\log(\frac{5}{4})$
$\Rightarrow\log|\text{P}|=\log(\frac{5}{4})^{201}-\log(\frac{5}{4})^{199}+\log200000$
$\Rightarrow\log|\text{P}|=\log\left\{(\frac{5}{4})^{201}\log(\frac{5}{4})^{199}\right\}+\log200000$
$\Rightarrow\log|\text{P}|=\log\left\{(\frac{5}{4})^{2}\right\}+\log200000$
$\Rightarrow\log|\text{P}|=\log\big(\frac{25}{16}\times200000\big)$
$\Rightarrow\log|\text{P}|=\log312500$
$\Rightarrow \text{P}=312500$
View full question & answer→Question 1635 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0$
AnswerWe have, $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$ This is a homogeneous differential equation. Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v +x}\frac{\text{dv}}{\text{dx}},$ we get$\text{v +x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\sin\text{v}}{\text{x}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\sin\text{v}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-\sin\text{v}$
$\Rightarrow\ \text{cosec v dv}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get$\int\text{cosec v dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\log|\text{cosec v}-\cot\text{v}|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\frac{1}{\text{cosec v}-\cot\text{v}}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \log|\text{cosec v}+\cot\text{v}|=\log|\text{Cx}|$
$\Rightarrow\ \log\Big|\frac{1+\cos\text{v}}{\sin\text{v}}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \frac{1+\cos\text{v}}{\sin\text{v}}=\text{Cx}$
$\Rightarrow\ \text{x}\sin\text{v}=\frac{1}{\text{C}}(1+\cos\text{v})$
$\Rightarrow\ \text{x}\sin\text{v}=\text{K}(1+\cos\text{v})$ $\Big($where, $\text{K}=\frac{1}{\text{C}}\Big)$
Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get
$\Rightarrow\ \text{x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=\text{K}\Big[1+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]$
Hence, $\text{x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=\text{K}\Big[1+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]$ is the required solution.
View full question & answer→Question 1645 Marks
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Answerwe know that the equation of said family of ellopsis is
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
Differentiating (1) w..r.t.x, we get
$\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}.\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{y}}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-\text{b}^2}{\text{a}^2}\ ...(2)$
Differentiating (2) w..r.t.x, we get
$\frac{\text{y}}{\text{x}}\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)+\bigg(\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}}{\text{x}^2}\bigg)\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\frac{\text{dy}}{\text{dx}}$
Which is the required difeerential equation.
View full question & answer→Question 1655 Marks
Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}-{\text{y}}=\cos\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}-{\text{y}}=\cos\text{x}$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Q}$
Where P = -1 and $\text{Q}=\cos\text{x}$
$\therefore\text{ I}.\text{F}.=\text{e}^{\int{\text{P}\text{dx}}}$
$=\text{e}^{-\int\text{dx}}$
$=\text{e}^{-\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{-\text{x}},$ we get
$\text{e}^{-\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=\text{e}^{-\text{x}}\cos\text{x}$
$\Rightarrow\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{-\text{x}}\text{y}=\text{e}^{-\text{x}}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{-\text{x}}=\int \ \text{e}^{-\text{x}} \cos\text{x}\text{ dx} \ + \ \text{C}$
$\Rightarrow\text{ye}^{-\text{x}}=\text{I}+\text{C} \ ....(2)$
Here,
$\text{I}=\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}\ ..(3)$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin{\text{x}}-\int\big(-\text{e}^{-\text{x}}\sin\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}+\int {\text{e}^{-\text{x}}}\sin\text{x}\text{ dx}$
$\Rightarrow\text{I}= \text{e}^{-\text{x}}\sin \text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int[(-\text{e}^{-\text{x}})\times(-\cos\text{x})]\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x} - \text{e}^{-\text{x}}\cos\text{x} - \text{I}$ [From (3)]
$ \Rightarrow2\text{I}=\text{e}^{-\text{x}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x}-\cos{\text{x}})\ ...(4)$
From (2) and (4) we get
$\Rightarrow\text{y}\text{e}^{-\text{x}}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x} - \cos\text{x})+\text{C}$
$\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$
Hence, $\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$ is the requires solution.
View full question & answer→Question 1665 Marks
For each of the differential equations given in find the general solution: $\cos^2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
AnswerGiven: Differential equation $\cos^2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos^2\text{x}}=\frac{\tan\text{x}}{\cos^2\text{x}}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}+(\sec^2\text{x})\text{y}=\sec^2\text{x}\tan\text{x}$
$\text{Comparing with}\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{we have P}=\sec^2\text{x and Q}=\sec^2\text{x}\tan\text{x}.$
$\therefore\ \ \int\text{P dx}=\int\sec^2\text{x dx}=\tan\text{x}\ \ \text{I.F}=\text{e}^{\int\text{P dx}}=\text{e}^{\tan\text{x}}$
$\text{Solution is y (I.F)}=\int\text{Q (I.F.) dx}+\text{c}$ $\Rightarrow\ \ \text{ye}^{\tan\text{x}}=\int\sec^2\text{x}\tan\text{xe}^{\tan\text{x}}\text{dx}+\text{c}\ \ ...\text{(i)}$
$\text{Putting}\tan\text{x}=\text{t and differentiating}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\ \ \int\sec^2\text{x}\ \tan\text{xe}^{\tan\text{x}}\text{ dx}=\int\text{te}^\text{t}\ \text{dt}$
Applying product rule,
$\Rightarrow\ \ \int\sec^2\text{x}\tan\text{x e}^{\tan\text{x}}$ $\text{dx}=\text{t.e}^\text{t}-\int1.\text{e}^\text{t}\ \text{dt}=\text{t.}\text{e}^\text{t}-\text{e}^\text{t}=(\text{t}-1)\text{e}^\text{t}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}$
Putting this value in eq. (i),
$\text{ye}^{\tan\text{x}}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{C}$ $\ \ \Rightarrow\ \ \text{y}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{ce}^{\tan\text{x}}$
View full question & answer→Question 1675 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x},\text{ y}=2,\text{ when x}=\frac{\pi}{2}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x}\ ....(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-3\cot\text{x}$ and $\text{Q}=\sin2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-3\int\cot\text{x dx}}$
$=\text{e}^{-3\log|\sin\text{x}|}$
$=\text{cosec}^3\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\text{cosec}^3\text{x},$ we get
$\text{cosec}^3\text{x}\Big(\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}\Big)=\sin2\text{x}(\text{cosec}^3\text{x})$
$\Rightarrow\text{cosec}^3\text{x}\Big(\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}\Big)=2\cot\text{x cosec x}$
Integrating both sides with respect to x, we get
$\text{y }\text{cosec}^3\text{x}=2\int\cot\text{x}\text{ cosec}\text{ x dx}+\text{C}$
$\Rightarrow\text{y }\text{cosec}^3\text{x}=-2\text{cosec}\text{ x}+\text{C}$
$\Rightarrow\text{y}=-2\sin^2\text{x}+\text{C}\sin^3\text{x}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=2$
$\therefore\ 2=-2\sin^2\frac{\pi}{2}+\text{C}\sin^3\frac{\pi}{2}$
$\Rightarrow\text{C}=4$
Putting the value of C in (2), we get
$\text{y}=-2\sin^2\text{x}+4\sin^3\text{x}$
$\Rightarrow\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$
Hence, $\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$ is the required solution.
View full question & answer→Question 1685 Marks
If the interest is compounded continuously at 6% per annum, how much worth Rs $100$ will be after 10 years? How long will it take to double Rs $1000?$
AnswerLet $P_0$ be the intial amount and $P$ be the amount at any time $t.$ Then,
$\frac{\text{dP}}{\text{dt}}=\frac{6\text{P}}{100}$
$\Rightarrow \frac{\text{dP}}{\text{dt}}=0.06\text{P}$
$\Rightarrow \frac{\text{dP}}{\text{P}}=0.06\text{dt}$
Integrating both sides with respect to t, we get
$\log \text{P}=0.06 \text{t}+\text{C}$
Now,
$\therefore \log\text{P}_{0}=0+\text{C}$
$\Rightarrow \text{C}=\log\text{P}_{0}$
$\log \text{P}=0.06\text{t}+\log\text{P}_{0}$
$\Rightarrow\log\frac{\text{P}}{\text{P}_{0}}=0.06\text{t}$
$\Rightarrow \text{e}^{0.06\text{t}}=\frac{\text{P}}{\text{P}_{0}}$
To find the amount $10$ years, we get
$\Rightarrow \text{e}^{0.06\text{t}\times10}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{e}^{0.6}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow 1.822=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{P}=1.822\ \text{P}_{0}$
To find the time after which the amount will doble, we have
$\text{P}=2\text{P}_{0}$
$\therefore \log\frac{2\text{P}_{0}}{\text{P}_{0}}=0.06\text{t}$
$\Rightarrow \log2=0.06\text{t}$
$\Rightarrow \text{t}=\frac{0.6931}{0.06}=11.55 \ \text{years}$
View full question & answer→Question 1695 Marks
Solve the following differential equations:$\cos\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}\sin\text{y}$
AnswerWe have,
$\cos\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}\sin\text{y}$
$\Rightarrow\frac{\cos\text{y}}{\sin\text{y}}\text{dy}=\frac{-\sin\text{x}}{\cos\text{x}}\text{dx}$
$\Rightarrow\cot\text{y dy}=-\tan\text{x dx}$
Integrating both sides, we get
$\int\cot\text{y dy}=-\int\tan\text{x dx}$
$\Rightarrow\log|\sin\text{y}|=-\log|\sec\text{x}|+\log\text{C}$
$\Rightarrow\log |\sin\text{y}|=\log|\cos\text{x}|+\log\text{C}$
$\Rightarrow\sin\text{y}=\text{C}\cos\text{x}$
Hence, $\sin\text{y =C}\cos\text{x}$ is the reguired solution.
View full question & answer→Question 1705 Marks
Solve the following differential equations:$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$
AnswerWe have,
$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$
$\Rightarrow\text{y}(1+\text{x})\text{dx = x}(\text{y}^2-1)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{\text{x}}\text{dx}=\frac{\text{y}^2-1}{\text{y}}\text{dy}$
Integrating both sides, we get
$\int\frac{1+\text{x}}{\text{x}}\text{dx}=\int\frac{\text{y}^2-1}{\text{y}}\text{dy}$
$\Rightarrow\int\frac{1}{\text{x}}\text{dx}+\int\text{dx}=\int\text{y dy}-\int\frac{1}{\text{y}}\text{dy}$
$\Rightarrow\log|\text{x}|+\text{x}=\frac{\text{y}^2}{2}-\log|\text{y}|+\text{C}$
$\Rightarrow\log|\text{x}|+\text{x}-\frac{\text{y}^2}{2}+\log|\text{y}|=\text{C}$
Hence, $\log|\text{x}|+\text{x}-\frac{\text{y}^2}{2}+\log|\text{y}|=\text{C}$ is the required solution.
View full question & answer→Question 1715 Marks
Find the particular solution of the differential equation$(1-\text{y}^2)(1+\log\text{x})\text{dx}+2\text{xy dy}=0,$ given that $\text{y}=0$ when $\text{x}=1.$
AnswerGiven:
$(1-\text{y}^2)(1+\log\text{x})\text{dx}+2\text{xy dy}=0$
$\Rightarrow(1-\text{y}^2)(1+\log\text{x})\text{dx}=-2\text{x y dy}$
$\Rightarrow\Big(\frac{1+\log\text{x}}{2\text{x}}\Big)\text{dx}=-\Big(\frac{\text{y}}{1-\text{y}^2}\Big)\text{dy}...(1)$
Let:
$1+\log\text{x = t}$
and
$(1-\text{y}^2)=\text{p}$
$\Rightarrow\frac{1}{\text{x}}\text{dx dt}$ and $-2\text{y dy = dp}$
Therefore, (1) becomes
$\int\frac{\text{t}}{2}\text{dt}=\int\frac{1}{2\text{p}}\text{dp}$
$\Rightarrow\frac{\text{t}^2}{4}=\frac{\log\text{p}}{2}+\text{C}...(2)$
Substituting the values of t and p in (2) we get
$\frac{(1+\log\text{x})^2}{4}=\frac{\log(1-\text{y}^2)}{2}+\text{C}...(3)$
At $\text{x}=1$ and $\text{y}=0,$ (3) becomes
$\text{C}=\frac{1}{4}$
Substituting the value of C in (3), we get
$\frac{(1+\log\text{x})^2}{4}=\frac{\log(1-\text{y}^2)}{2}+\frac{1}{4}$
$\Rightarrow(1+\log\text{x})^2=2\log(1-\text{y}^2)+1$
Or
$(\log\text{x})^2+\log\text{x}^2=\log(1-\text{y}^2)^2$
It is the required particular solution.
View full question & answer→Question 1725 Marks
Solve the following differential equations:$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{y}+2,\text{y}(2)=0$
AnswerWe have,
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{y}+2,\text{y}(2)=0$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\frac{\text{y}+2-2}{\text{y}+2}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\text{dy}-2\int\frac{1}{\text{y}+2}\text{dy}=\log\text{x + C}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log|\text{x}|+\text{C}\dots(1)$
It is given that at $\text{x}=2,\text{y}=0.$
Substituting the valuse of x and y in (1), we get
$-2\log2-\log2=\text{C}$
$\Rightarrow-\log(2^2\times2)=\text{C}$
$\Rightarrow\text{C}=-\log8$
Substituting the value of C in (1), we get
$\text{y}-2\log|\text{y}+2|=\log|\text{x}|-\log8$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$
Hence, $\text{y}-2\log|\text{y}+2|=\log\Big|\frac{\text{x}}{8}\Big|$ is the required solution.
View full question & answer→Question 1735 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}-\text{x}\text{e}^\text{x}-\frac{5}{2}+\cos^2\text{x}$
Answer$\frac{\text{dy}}{\text{dx}}-\text{x}\text{e}^\text{x}-\frac{5}{2}+\cos^2\text{x}$
$\text{dy}=\Big(\text{xe}^\text{x}-\frac{5}{2}+\cos^2\text{x}\Big)\text{dx}$
$\int\text{dy}=\int\text{xe}^\text{x}\text{dx}-\frac{5}{2}\int\text{dx}+\cos^2\text{x dx}$
$\int\text{dy}=\int\text{xe}^\text{x}\text{dx}-\frac{5}{2}\int\text{dx}+\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\text{xe}^\text{x}-\frac{5}{2}\int\text{dx}+\frac{1}{2}\int\text{dx}+\frac{1}{2}\int\cos2\text{x dx}$
$\int\text{dy}=\int\text{xe}^\text{x}-2\int\text{dx}+\frac{1}{2}\int\cos2\text{x dx}$
$\text{y}=[\text{x}\times\int\text{e}^\text{x}\text{dx}-\int(1\times\int\text{e}^\text{x}\text{dx})\text{dx}]-2\text{x}+\frac{1}{2}\frac{\sin2\text{x}}{2}+\text{C}$
Using integration by parts
$\text{y}=\text{xe}^\text{x}-\text{e}^\text{x}-2\text{x}+\frac{1}{4}\sin2\text{x}+\text{C}$
$\text{y}=\text{xe}^\text{x}-\text{e}^\text{x}-2\text{x}+\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 1745 Marks
The slope of the tangent at a point P(x, y) on a curve is $\frac{-\text{x}}{\text{y}}$. If the curve passs es through the point (3, -4). Find the equation of the curve.
AnswerAccording to the question,
$\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}$
$\text{y}\ \text{dy}=-\text{x}\ \text{dx}$
Integrating both sides with respect to x, we get
$\int\text{y}\ \text{dy}=-\int \text{x}\ \text{dx} $
$\Rightarrow \frac{\text{y}^{2}}{2}=-\frac{\text{x}^{2}}{2}+\text{C}$
Since the curve passes through (3, -4), it satisfies the above equation.
$\therefore \frac{(-4)^{2}}{2}=-\frac{3^{2}}{2}+\text{C}$
$\Rightarrow 8 = -\frac{9}{2}+\text{C}$
$\Rightarrow \text{C}=\frac{25}{2}$
Putting the value of C, we get
$\frac{\text{y}^{2}}{2}=-\frac{\text{x}^{2}}{2}+\frac{25}{2}$
$\Rightarrow \text{x}^{2}+\text{y}^{2}=25$
View full question & answer→Question 1755 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x + y}}{\text{x}-\text{y}}$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{x + y}}{\text{x}-\text{y}}$
This is a homogeneous differential equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{{\text{x}}+\text{vx}}{\text{x}-\text{vx}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{1-\text{v}}$
$\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\tan^{-1}\text{v}-\frac{1}2\log\big|1+\text{v}^2\big|=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)-\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|=\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\frac{\text{x}^2+\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\frac{1}2\log|\text{x}^2|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\log|\text{x}|+\log|\text{x}|+\text{C}$
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$
Hence, $\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$ is the required solution.
View full question & answer→Question 1765 Marks
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis twice abscissa of the pont of contact.
Answer
It is given that the distance between the foot of ordinate of point of contanct (A) and the point of intersection of tangent with x-axis (T) = 2x
Coordinate of $\text{T}=\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\big)$
$\text{AT}=\Big[\text{x}-\Big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\Big)\Big]=2\text{x}$
Equation of tangent,
$\text{y}-\text{y}=\frac{\text{dy}}{\text{dx}}(\text{x}-\text{x})$
$\Rightarrow\text{y}-\text{0}=\frac{\text{dy}}{\text{dx}}\Big(\text{x}-\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}\big)\Big)$
$\Rightarrow \text{y}\frac{\text{dx}}{\text{dy}}=2\text{x}$
$\Rightarrow \int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$
$\Rightarrow \log\text{x}=\log\text{y}^{2}+\log\text{C}$
$\text{x}=\text{Cy}^{2}$
As the circle passes through (1, 2)
$1=\text{C}\times2^{2}$
$\Rightarrow \text{C}=\frac{1}{4}$
$\Rightarrow 4\text{x}=\text{y}^{2}$ View full question & answer→Question 1775 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=2\cos\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{x dx}}$
$=\text{e}^{\log\sin\text{x}}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=2\sin\text{x }\cos\text{x}$
$\Rightarrow\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=\int\sin2\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ 0\times\sin\frac{\pi}{2}=-\frac{\cos\pi}{2}+\text{C}$
$\Rightarrow\text{C}=-\frac{1}{2}$
Putting the value of C in (2), we get
$\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}-\frac{1}{2}$
$\Rightarrow2\text{y}\sin\text{x}=-(1+\cos2\text{x})$
$\Rightarrow2\text{y}\sin\text{x}=-2\cos^2\text{x}$
$\Rightarrow\text{y}=-\cot\text{x}\cos\text{x}$
Hence, $\text{y}=-\cot\text{x}\cos\text{x}$ is the required solution.
View full question & answer→Question 1785 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}\cos\text{x}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=\cos\text{x}$ It is a linear differential equation. Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{2}{\text{x}},\text{Q}=\cos\text{x}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{2\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{2\log|\text{x}|}$ $=\text{x}^2$Solution of the equation is given by,
$\text{y}\times(\text{I.F.}=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}(\text{x}^2)=\int\cos\text{x}(\text{x}^2)\text{dx + C}$ $\text{yx}^2=\int\text{x}^2\cos\text{xdx + C}$ $=\text{x}^2\int\cos\text{x}-\int(2\text{x}\times\int\cos\text{xdx})\text{dx + C}$ Usind integration by parts $\text{yx}^2=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx + C}$ $=\text{x}^2\sin\text{x}-2\big[\text{x}\times\int\sin\text{xdx}-\int(1\times\int\sin\text{xdx})\text{dx}\big]+\text{C}$ $=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int\cos\text{xdx + C}$ $\text{yx}^2=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x + C}$ $\text{y}=\sin\text{x}+\frac{2}{\text{x}}\cos\text{x}-\frac{2}{\text{x}^2}\sin\text{x}+\frac{\text{C}}{\text{x}^2}$
View full question & answer→Question 1795 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
AnswerHere, $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
It is a linear differential equation. comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{1}{\text{x}},\text{Q}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log|\text{x}|}=\frac{1}{\text{x}},\text{x}>0$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}}\Big)=\int\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Since $\int[\text{f(x)}+\text{f}'(\text{x})]\text{e}^{\text{x}}\text{dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C}$
$\text{y}=\text{e}^{\text{x}}+\text{Cx},\text{x}>0$
View full question & answer→Question 1805 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{x}^2\cot\text{x}+2\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{x}^2\cot\text{x}+2\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\cot\text{x}$
$\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{xdx}}$
$=\text{e}^{\log|\sin\text{x}|}=\sin\text{x}$
Multiplying both sides of (1) by $\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}(\text{x}^2\cot\text{x}+2\text{x})$
$\Rightarrow\ \sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=\int\text{x}^2\cos\text{xdx}+\int2\text{x}\sin\text{xdx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\int\cos\text{xdx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x}^2)\int\cos\text{xdx}\Big]\text{dx}+\int2\text{x}\sin\text{x dx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx}+\int2\text{x}\sin\text{xdx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x +C}$
Hence, $\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x +C}$ is the required solution.
View full question & answer→Question 1815 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
$\Rightarrow\text{dy}=(\log\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\log\text{x})\text{dx}$
$\Rightarrow\text{dy}=\int1\times\log\text{x}\text{ dx}$
$\Rightarrow\text{dy}=\log\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x }-\int\frac{\text{x}}{\text{x}}\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\int1\text{dx}$
$\Rightarrow\text{y}=\text{x}\log\text{x}-\text{x}$
$\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$
So, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\text{x}(\log\text{x}-1)+\text{C}$ where $\text{x}\in\text{R}-\{0\}$ is the solution o the given differential equation.
View full question & answer→Question 1825 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}\cos(\text{x}-\text{y})=1$
Answer$\frac{\text{dy}}{\text{dx}}\times\cos(\text{x}-\text{y})=1$
Let $\text{x}-\text{y}=\text{v}$
$1-\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dv}}{\text{dx}}$
So,
$\Big(1-\frac{\text{dv}}{\text{dx}}\Big)\cos\text{v}=1$
$1-\frac{\text{dv}}{\text{dx}}=\sec\text{v}$
$1-\sec\text{v}=\frac{\text{dv}}{\text{dx}}$
$\text{dx}=\frac{\text{dv}}{1-\sec\text{v}}$
$\text{dx}=\frac{\cos\text{v}}{1-\cos\text{v}}\text{dv}$
$\int\text{dx}=\int\frac{\cos^{2}\frac{\text{v}}{2}-\sin^{2}\frac{\text{v}}{2}}{2\sin^{2}\frac{\text{v}}{2}}\text{dv}$
$\int\text{dx}=\int\frac{1}{2}\cot\big(\frac{\text{v}}{2}\big)\text{dv}-\frac{1}{2}\text{dv}$
$2\int\text{dx}=\int\cot^{2}\big(\frac{\text{v}}{2}\big)-\int\text{dv}$
$2\int\text{dx}=\int\Big(\text{cosec}^{2}\frac{\text{v}}{2}-1\Big)\text{dv}-\int\text{dv}$
$2\text{x}=-2\cot\big(\frac{\text{v}}{2}\big)\text{dv}-\text{v}-\text{v}+\text{C}_{1}$
$2(\text{x}+\text{v})=-2\cot\frac{\text{v}}{2}+\text{C}_{1}$
$\text{x}+\text{x}-\text{y}=-\cot\Big(\frac{\text{x}-\text{y}}{2}\Big)+\text{C}$
$\text{C}+\text{y}=\cot\Big(\frac{\text{x}-\text{y}}{2}\Big)$
View full question & answer→Question 1835 Marks
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
AnswerLet A be the surface area of balloon, So
$\frac{\text{dA}}{\text{dt}}\propto\text{t}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=\lambda\text{t}$
$\Rightarrow\frac{\text{d}}{\text{dt}}(4\pi\text{r}^{2})=\lambda\text{t}$
$\Rightarrow 8\pi\text{r}\frac{\text{dr}}{\text{dt}}=\lambda\text{t}$
$\Rightarrow8\pi\text{r}\ \text{dr}=\lambda\text{t}$
$\Rightarrow 8\pi\int\limits_{}{}\text{r}\ \text{dr}=\lambda\int_{}^{}\text{t}\ \text{dt} $
$\Rightarrow 8\pi\frac{\text{r}^{2}}{2}=\frac{\lambda\text{t}^{2}}{2}+\text{c}$
$\Rightarrow 4\pi\text{r}^{2}=\frac{\lambda\text{t}^{2}}{2}+\text{c}\ ...(\text{i})$
Given r = 1 units When t = 0, so
$4\pi(1)^{2}=0+\text{c}$
$4\pi=\text{c}$
Using it is equation (i),
$\Rightarrow 4\pi\text{r}^{2}=\frac{\lambda\text{t}^{2}}{2}+4\pi\ ...(\text{ii})$
Also, given r = 2 units when t = 3 Sec.
$4\pi\text{(2)}^{2}=\frac{\lambda\text{(3)}^{2}}{2}+4\pi$
$ \Rightarrow16\pi=\frac{9}{2}\lambda+4\pi$
$\Rightarrow\frac{9}{2}\lambda=12\pi$
$\Rightarrow\lambda=\frac{24}{9}\pi$
$\Rightarrow\lambda=\frac{8}{3}\pi$
Now, equation (ii) becomes
$ 4\pi\text{r}^{2}=\frac{8\pi}{6}\text{t}^{2}+4\pi$
$\Rightarrow 4\pi(\text{r}^{2}-1)=\frac{4}{3}\pi\text{t}^{2}$
$\Rightarrow\text{r}^{2}-1=\frac{1}{3}\text{t}^{2}$
$\Rightarrow\text{r}^{2}=1+\frac{1}{3}\text{t}^{2}$
$\therefore\ \text{r}=\sqrt{(1+\frac{1}{3}\text{t}^{2}})$
View full question & answer→Question 1845 Marks
Solve the following differential equation:
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2-\text{y}^2$
AnswerWe have,
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2-\text{y}^2$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-\text{y}^2}{\text{xy}}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-\text{v}^2\text{x}^2}{\text{vx}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}^2}{\text{v}}$
$\Rightarrow\ \frac{\text{v}}{1-2\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{v}}{1-2\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{-1}4\log\big|1-2\text{v}^2\big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\big|1-2\text{v}^2\big|=-4\log|\text{x}|-4\log\text{C}$
$\Rightarrow\ \log\big|\big(1-2\text{v}^2\big)\big(\text{x}^4\big)\big|=\log\frac{1}{\text{C}^4}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \log\big|\text{x}^2\big(\text{x}^2-2\text{y}^2\big)\big|=\log\frac{1}{\text{C}^4}$
$\Rightarrow\ \text{x}^2\big(\text{x}^2-2\text{y}^2\big)=\text{C}_1$
where
$\text{C}_1=\frac{1}{\text{C}^4}$
Hence, $\text{x}^2\big(\text{x}^2-2\text{y}^2\big)=\text{C}_1$ is the required solution.
View full question & answer→Question 1855 Marks
Given that $\frac{\text{dy}}{\text{dx}}=\text{e}^\text{-2y}$ and y = 0 when x = 5. Find the value of x when y = 3.
AnswerGiven that, $\frac{\text{dy}}{\text{dx}}=\text{e}^\text{-2y}$
$\Rightarrow\frac{\text{dy}}{\text{e}^\text{-2y}}=\text{dx}$
$\Rightarrow\int\text{e}^\text{2y}\text{dy}=\int\text{dx}$
$\Rightarrow\frac{\text{e}^\text{2y}}{2}=\text{x+C }....(\text{i})$
When x = 5 and y = 0, then substituting these values in equation (i), we get
$\frac{\text{e}^0}{2}=5+\text{C}$
$\Rightarrow\frac{1}{2}=5+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}-5$
$\Rightarrow\text{C}=-\frac{9}{2}$
Thus equation (i) becomes
$\text{e}^\text{2y}=2\text{x}-9$
When y = 3, then $\text{e}^6=2\text{x}-9$
$\Rightarrow2\text{x}=\text{e}^6+9$
$\Rightarrow\text{x}=\frac{(\text{e}^6+9)}{2}$
View full question & answer→Question 1865 Marks
Solve the following differential equations:
$2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0,\text{y}(1)=-2$
Answer$2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2(\text{y}+3)=\text{xy}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{2}{\text{x}}\text{dx}=\frac{\text{y}}{\text{y}+3}\text{dy}$
$\Rightarrow\frac{2}{\text{x}}\text{dx}=\frac{\text{y}+3-3}{\text{y}+3}\text{dy}$
$\Rightarrow\frac{2}{\text{x}}\text{dx}=\Big(1-\frac{3}{\text{y}+3}\Big)\text{dy}$
$\Rightarrow\int\frac{2}{\text{x}}\text{dx}=\int\Big(1-\frac{3}{\text{y}+3}\Big)\text{dy}$
$\Rightarrow2\log\text{x = y}-3\log|\text{y}+3|+\text{C}$
$\Rightarrow\log\text{x}^2+\log|(\text{y}+3)^3|=\text{y + C}$
$\Rightarrow\log|(\text{x}^2)(\text{y}+3)^3|=\text{y + C}...(1)$
$\Rightarrow\log|(1)^2(-2+3)^3|=-2+\text{C}$
$\Rightarrow\text{C}=2$
Substituting the value of C in (1), we get
$\log|(\text{x}^2)(\text{y}+3)^3|=\text{y}+2$
$\Rightarrow(\text{x}^2)(\text{y}+3)^3=\text{e}^{\text{y}+2}$
View full question & answer→Question 1875 Marks
Show that $\text{y}=\frac{\text{a}}{\text{x}}+\text{b}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=0$
AnswerWe have,
$\text{y}=\frac{\text{a}}{\text{x}}+\text{b}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{a}}{\text{x}^2}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{a}}{\text{x}^3}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(-\frac{\text{a}}{\text{x}^2}\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 1885 Marks
Solve the following differential equation:
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$
AnswerWe have,
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow 2\frac{\text{dy}}{\text{dx}} = 1 - \cos^2(\text{x}-2\text{y} )$
Let $\text{x}-2\text{y}=\text{v}$
$\Rightarrow1-2\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}$
$\Rightarrow 2\frac{\text{dy}}{\text{dx}} = 1 -\frac{\text{dv}}{\text{dx}}$
$\therefore 1 - \frac{\text{dv}}{\text{dx}} = 1 - \cos^2\text{v}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \cos^2\text{v}$
$\Rightarrow \sec^2 \text{v}\text{ dv} = \text{dx}$
Integrating both sides, we get
$\int\sec^2\text{v}\text{ dv} = \int \text{dx}$
$\Rightarrow \tan \text{v} = \text{x} - \text{C}$
$\Rightarrow \tan (\text{x}-2\text{y}) = \text{x}-\text{C}$
$\Rightarrow \text{x} = \tan (\text{x}-2\text{y})+\text{C}$
View full question & answer→Question 1895 Marks
In each of the show that the given differential equation is homogeneous and solve each of them.
$\text{y}'=\frac{\text{x}+\text{y}}{\text{x}}$
AnswerGiven: Differential equation $\text{y}'=\frac{\text{x}+\text{y}}{\text{x}}\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{x}+\frac{\text{y}}{\text{x}}$$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=1+\frac{\text{y}}{x}=f\Big(\frac{\text{y}}{\text{x}}\Big)\ \ ....(\text{i})$
Therefore, eq. (i) is homogeneous.
$\text{Putting}\ \frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}$ $\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}.1+\text{x}\ \frac{\text{dv}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
$\text{Putting value of y and}\ \frac{\text{dy}}{\text{dx}}\ \text{in eq. (i)}$
$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}\ \ \Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=1$ $\ \ \Rightarrow\ \ \text{x dv = dx}$
$\Rightarrow\ \ \text{dv}=\frac{\text{dx}}{\text{x}}\ \ [\text{Separating variables}]$
$\text{Interating both sides},\ \ \int1\ \text{dv}=\int\frac{\text{dv}}{\text{x}}$ $\ \ \Rightarrow\ \ \text{v}=\log|\text{x}|+\text{c}$
$\text{Putting}\ \frac{\text{y}}{\text{x}}=\text{v,}\ \ \frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$ $\ \ \Rightarrow\ \ \text{y}=\text{x}\log|\text{x}|+\text{xc}$
View full question & answer→Question 1905 Marks
Solve the following differential equation:
$\text{y}^2\frac{\text{dx}}{\text{dy}}+\text{x}-\frac{1}{\text{y}}=0$
AnswerHere, $\text{y}^2\frac{\text{dx}}{\text{dy}}+\text{x}-\frac{1}{\text{y}}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}^2}=\frac{1}{\text{y}^3}$
It is a linear differential equation. Comparing the equation with,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{\text{y}^2},\text{Q}=\frac{1}{\text{y}^3}$
I.F. $=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{1}{\text{y}^2}\text{dy}}$
$=\text{e}^{-\frac{1}{\text{y}}}$
Solution of the equation is given by,
$\text{x}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dy + C}$
$\text{x}\Big(\text{e}^{-\frac{1}{\text{y}}}\Big)=\int\frac{1}{\text{y}^3}\Big(\text{e}^{-\frac{1}{\text{y}}}\Big)\text{dy + C}$
Let $\text{e}^{-\frac{1}{\text{y}}}=\text{t}$
$\Rightarrow\frac{1}{\text{y}}=-\log\text{t}$
$\text{e}^{-\frac{1}{\text{y}}}\times\frac{1}{\text{y}^2}\text{dy = dt}$
$\text{x (t)}=\int\frac{1}{\text{y}}\text{dt + C}$
$=-\log+\text{dt + C}$
$=-\Big[\log\text{t}\times\int1\times\text{dt}-\int\Big(\frac{1}{\text{t}}\int1\times\text{dt}\Big)\text{dt}\Big]+\text{C}$
$=-\Big[\text{t}\log\text{t}-\int\frac{\text{t}}{\text{t}}\text{dt}\Big]+\text{C}$
$\text{x (t)}=-\text{t}\log\text{t + t + C}$
$\text{x (t)}=-\text{t}[\log\text{t}-1]+\text{C}$
$\text{x}=-\Big[-\frac{1}{\text{y}}-1\Big]\text{Ce}^{\frac{1}{\text{y}}}$
$\text{x}=\frac{1}{\text{y}}+1+\text{Ce}^{\frac{1}{\text{y}}}$
$\text{x}=\Big(\frac{1+\text{y}}{\text{y}}\Big)+\text{Ce}^{\frac{1}{\text{y}}}$
View full question & answer→Question 1915 Marks
For each of the differential equations given in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin 2\text{x};\ \text{y}=2\ \text{when x}=\frac{\pi}{2}$
AnswerThe given differential equation is $\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin\ 2\text{x}.$This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p}=-3\cot\text{x}\ \text{and}\ \text{Q}=\sin2\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{-3\int\cot\text{x}\ \text{dx}}=\text{e}^{-3\log|\sin\text{x}|}=\text{e}^{\log|\frac{1}{\sin^3\text{x}}|}=\frac{1}{\sin^3\text{x}}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\text{y}\cdot\frac{1}{\sin^3\text{x}}=\int\Big[\sin2\text{x}.\frac{1}{\sin^3\text{x}}\text{dx}+\text{C}\Big]$
$\Rightarrow\text{y cosec}^3\ \text{x}=2\int(\cot\text{x}\ \text{cosec}\ \text{x})\text{dx}+\text{C}$
$\Rightarrow\text{y cosec}^3\ \text{x}=-2\text{cosec}\ \text{x}+\text{C}$
$\Rightarrow\text{y}=-\frac{2}{\text{cosec}^2\text{x}}+\frac{\text{C}}{\text{cosec}^3\text{x}}$
$\Rightarrow\text{y}=-2\sin^2\text{x}+\text{C}\sin^3\text{x}\ \ ...(1)$
$\text{Now,}\ \text{y}=2\ \text{at}\ \text{x}=\frac{\pi}{2}.$
Therefore, we get:
$2=-2+\text{C}$
$\Rightarrow\text{C}=4$
Substituting C = 4 in equation (1), we get:
$\text{y}=-2\sin^2\text{x}+4\sin^3\text{x}$
$\Rightarrow\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$
This is the required particular solution of the given differential equation.
View full question & answer→Question 1925 Marks
For each of the differential equations in find a particular solution satisfying the given condition:
$(\text{x}^3+\text{x}^2+\text{x}+1) \frac{\text{dy}}{\text{dx}} = 2\text{x}^2+\text{x; y} =1 \ \text{when x = 0}$
AnswerThe given differential equation is
$(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
$\text{or} \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^3+\text{x}^2+\text{x}+1}$
$\text{or} \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^2(\text{x}+1)+1(\text{x}+1)}$
$\text{or} \ \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}$
Separting the variables, we get,
$\text{dy}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\text{dx}$
Integrating, $\int\text{dy=} \ \int\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\text{dx} \ \ ...(1)$
Put $\frac{2\text{x}^2+\text{x}}{\text{(x}-1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1} \ \ ...(2 )$
$\Rightarrow \ 2\text{x}^2 +\text{x}=\text{Ax}^2 + \text{A} +\text{Bx}^2 +\text{Cx + Bx + C}$
$\Rightarrow \ \ 2\text{x}^2 +\text{x} = (\text{A + B)x}^2+\ (\text{B + C)x + A + C} \ \ \ \ ......(3)$
Now comparing the coefficients of $x^2$ and $x$
$\Rightarrow \text{A + B} =2$
$\Rightarrow \text{B + C}=1$
$\Rightarrow \text{A + C}=0{}$
Solving tham we will get the values of $A, B, C$
$\text{A}= \frac{1}{2}, \text{B} =\frac{3}{2}-\frac{1}{2}$
Putting the values of $A,B,C$
$\therefore \text{from}(2), \frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)} = \frac{\frac{1}{2}}{\text{x}+1}+\frac{\frac{3}{2}\text{x}-\frac{1}{2}}{\text{x}^2+1}$
$\text{or}\ \ \frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)} \equiv \frac{1}{2(\text{x}+1)}+\frac{3}{2}\bigg(\frac{x}{x^2+1}\bigg)-\frac{1}{2}\bigg(\frac{1}{x^2+1}\bigg)$
$\therefore \text{from}(1), \ \int\text{dy}=\frac{1}{2} \int\frac{1}{x+1}\text{dx}+\frac{3}{2}\int\frac{\text{x}}{\text{x}^2+1}\text{dx}-\frac{1}{2}\int \frac{1}{\text{x}^2+1}\text{dx}$
$\therefore \ \int 1\ \text{dy}= \frac{1}{2}\int\frac{1}{\text{x}+1}\text{dx}+\frac{3}{4}\int\frac{2\text{x}}{\text{x}^2+1} \text{dx}-\frac{1}{2}\int\frac{1}{\text{x}^2+1}\text{dx}$
$\therefore \ \text{y}= \frac{1}{2}\text{log}|\text{x}+1|+ \frac{3}{4}\text{log}(\text{x}^2+1)-\frac{1}{2}\text{tan}^{-1}\text{x}+\text{c} \ \ \ \ \ .....(4)$
Now $y = 1$ when $x = 0$
$\therefore 1=\frac{1}{2}\text{log}(1)+\frac{3}{4}\text{log} \ 1-\frac{1}{2}\text{tan}^{-1} \text{0+c}$
$\therefore 1 =\frac{1}{2}(0)+\frac{3}{4}(0)-\frac{1}{2}(0)+\text{c} \ \ \Rightarrow\ \ \text{c}=1$
$\therefore \text{from}\ (4),\ \text{y}= \frac{1}{2}\text{log}|\text{x}+1|+\frac{3}{4}\text{log}(\text{x}^2+1)-\frac{1}{2}\text{tan}^{-1}\text{x}+1$
View full question & answer→Question 1935 Marks
Solve $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2.$
AnswerGiven that, $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$
Dividing both sides by $x^2,$ we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}\ ....(\text{i})$
Let $\text{f}(\text{x, y})=1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}$
$\text{f}(\lambda\text{x},\lambda\text{y})=1+\frac{\lambda\text{y}}{\lambda\text{x}}+\frac{\lambda^2\text{y}^2}{\lambda^2\text{x}^2}$
$\text{f}(\lambda\text{x},\lambda\text{y})=\lambda^0\Big(1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}\Big)$
$\text{f}(\lambda\text{x},\lambda\text{y})=\lambda^0\text{f}(\text{x},\text{ y})$
Which is homogeneous expression of degree 0.
Put $\text{y}=\text{vx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dy}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\Big(\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}\Big)=1+\text{v}+\text{v}^2$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}+\text{v}^2-\text{v}$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2$
$\Rightarrow\frac{\text{dv}}{1+\text{v}^2}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\int\frac{\text{dv}}{1+\text{v}^2}=\int\frac{\text{dx}}{\text{x}}$
$\tan^{-1}\text{v}=\log|\text{x}|+\text{C}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{x}|+\text{c}$
View full question & answer→Question 1945 Marks
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\log\text{x}$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\log\text{x}$
Dividing both sides by x, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\log{\text{x}}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\log\text{x}$
Now,
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{\log|\text{x}|}=\text{x}$
So, the solution is given by
$\text{y}\times\text{I.F.}=\int\text{Q}\times\text{I.F. dx + C}$
$\Rightarrow\ \text{xy}=\int\text{x}\log\text{x dx + C}$
$\Rightarrow\ \text{xy}=\log\text{x}\int\text{xdx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x dx}\Big]\text{ dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^2\log\text{x}}{2}-\int\frac{\text{x}}2\text{dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^2\log\text{x}}{2}-\frac{\text{x}^2}4+\text{C}$
$\Rightarrow\ 4\text{xy}=2\text{x}^2\log\text{x}-\text{x}^2+\text{K}$ (where, K = 2C)
View full question & answer→Question 1955 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$
AnswerHere, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$It is homogeneous equation
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-1}{2\text{v}}-\frac{\text{v}}1$
$=\frac{\text{v}^2-1-2\text{v}^2}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-\text{v}^2}{2\text{v}}$
$\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|1+\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$
$1+\text{v}^2=\frac{\text{C}}{\text{x}}$
$1+\frac{\text{y}^2}{\text{x}^2}=\frac{\text{C}}{\text{x}}$
$\text{x}^2+\text{y}^2=\text{Cx}$
View full question & answer→Question 1965 Marks
If $\vec{\text{a}}=\vec{\text{i}}+\vec{\text{j}}+2\vec{\text{k}}$ and $\vec{\text{b}}=2\vec{\text{i}}+\vec{\text{j}}-2\vec{\text{k}},$ find the unit vector in the direction of:
- $6\vec{\text{b}}$
- $2\vec{\text{a}}-\vec{\text{b}}$
AnswerHere, $\vec{\text{a}}=\vec{\text{i}}+\vec{\text{j}}+2\vec{\text{k}}$ and $\vec{\text{b}}=2\vec{\text{i}}+\vec{\text{j}}-2\vec{\text{k}}$
- $6\vec{\text{b}}=12\hat{\text{i}}+6\hat{\text{j}}-12\hat{\text{k}}$
$\therefore$ Unit vectors in the direction of $6\vec{\text{b}}=\frac{6\vec{\text{b}}}{|6\vec{\text{b}}|}$
$=\frac{12\hat{\text{i}}+6\hat{\text{j}}-12\hat{\text{k}}}{\sqrt{12^2+6^2+12^2}}$
$=\frac{6(12\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})}{\sqrt{324}}$
$=\frac{6(2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})}{18}$
$=\frac{2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}}{3}$
- $2\vec{\text{a}}-\vec{\text{b}}=2(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$
$=\hat{\text{j}}+6\hat{\text{k}}$
$\therefore$ Unit vectors in the direction of $2\vec{\text{a}}-\vec{\text{b}}$
$=\frac{2\vec{\text{a}}-\vec{\text{b}}}{|2\vec{\text{a}}-\vec{\text{b}}|}=\frac{\hat{\text{j}}+6\hat{\text{k}}}{\sqrt{1^2+6^2}}$
$=\frac{\hat{\text{j}}+6\hat{\text{k}}}{\sqrt{37}}$ View full question & answer→Question 1975 Marks
verify that $\text{y}=\text{e}^{\text{m}\cos^{-1}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{m}^2\text{y}=0$
AnswerWe have,
$\text{y}=\text{e}^{\text{m}\cos^{-1}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{me}^{\text{m}^{\cos^{-1}}}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{me}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^3}}\ ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{me}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-\text{m})\Bigg[\frac{\sqrt{1-\text{x}^2}\text{me}^{\text{m}^{\cos^{-1}}}\Big(-\frac{1}{\sqrt{1-\text{x}}}\Big)-\text{e}^{\text{m}^{\cos^{-1}}}\text{x}\frac{1}{2}\Big(-\frac{2\text{x}}{\sqrt{1-\text{x}^2}}\Big)}{(1-\text{x}^2)}\Bigg]$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-\text{m})\Big[-\text{me}^{\text{m}^{\cos^{-1}}}\text{x}+\frac{\text{xe}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{m}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{m}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{m}^2\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 1985 Marks
Solve the following differential equation
$\cos\text{x }\frac{\text{dy}}{\text{dx}}-\cos2\text{x}=\cos3\text{x}$
AnswerWe have,
$\cos\text{x }\frac{\text{dy}}{\text{dx}}-\cos2\text{x}=\cos3\text{x}$
$\Rightarrow\text{dy}=\frac{\cos3\text{x}+\cos2\text{x}}{\cos\text{x}}\ \text{dx}$
$\Rightarrow\text{dy}=\frac{4\cos^2\text{x}-3\cos\text{x}+2\cos^2\text{x}-1}{\cos\text{x}}\ \text{dx}$
$\Rightarrow\text{dy}=(4\cos^2\text{x}-3+2\cos\text{x}-\sec\text{x})\text{dx}$
$\Rightarrow\text{dy}[2(2\cos^2\text{x}-1)-1+2\cos\text{x}-\sec\text{x}]\text{dx}$
$\Rightarrow\text{dy}(2\cos2\text{x}-1+2\cos\text{x}-\sec\text{x})\text{ dx}$
Integrating both sides, we get
$\int\text{dy}=\int(2\cos2\text{x}-1+2\cos\text{x}-\sec\text{x})\text{dx}$
$\Rightarrow\text{y}=\sin2\text{x}-\text{x}+2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
hence, $\text{y}=\sin2\text{x}-\text{x}+2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$ is the solution to the given differential equation.
View full question & answer→Question 1995 Marks
For each of the differential equations given in find a particular solution satisfying the given condition:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{1+\text{x}^2};\text{y}=0\ \text{when x}=1$
Answer$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{1+\text{x}^2}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ This is a linear differential equation of the form: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ \Big(\text{where p}=\frac{2\text{x}}{1+\text{x}^2}\ \text{and}\ \text{Q}=\frac{1}{(1+\text{x}^2)^2}\Big)$ $\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{2\text{x dx}}{1+\text{x}^2}}=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2.$ The general solution of the given differential equation is given by the relation, $\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+\text{x}^2)=\int\bigg[\frac{1}{(1+\text{x}^2)^2}\cdot(1+\text{x}^2)\bigg]\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{1+\text{x}^2}\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}\ \ ...(1)$ Now, y = 0 at x =1. Therefore, $0=\tan^{-1}1+\text{C}$$\Rightarrow\text{C}=-\frac{\pi}{4}$
$\text{Substituting C}=-\frac{\pi}{4}\ \text{in equation (1), we get:}$$\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}-\frac{\pi}{4}$
This is the required general solution of the given differential equation.
View full question & answer→Question 2005 Marks
For each of the differential equation given in find the general solution:
$(1+\text{x}^2)\ \text{dy}+2\text{xy}\ \text{dx}=\cot\text{x}\ \text{dx}\ (\text{x}\neq0)$
Answer$(1+\text{x}^2)\ \text{dx}+2\text{xy}\ \text{dx}=\cot\text{xdx}$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{\cot\text{xdx}}{1+\text{x}^2}$This equation is a linear dyfferential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ \big(\text{where p}=\frac{2\text{x}}{1+\text{x}^2}\ \text{and}\ \text{Q}=\frac{\cot\text{x}}{1+\text{x}^2}\big)$ $\text{Now, I.F}=\text{e}^{\int{\text{pdx}}}=\text{e}^{\int{\frac{2\text{x}}{1+\text{x}^2}\text{dx}}}=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2.$ The general solution of the given differential equation is given by the relation, $\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F})\ \text{dx}+\text{C}$ $\Rightarrow\ \text{y}(1+\text{x}^2)=\int\Big[\frac{\cot\text{x}}{1+\text{x}}\times(1+\text{x}^2)\Big]\text{dx}+\text{C}$ $\Rightarrow\ \text{y}(1+\text{x}^2)=\int\cot\text{x dx}+\text{C}$ $\Rightarrow\ \text{y}(1+\text{x}^2)=\log|\sin\text{x}|+\text{C}$
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