Question 513 Marks
Prove that $\text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2$ is the general solution of differential equation $(\text{x}^3-3\text{x y}^2) \text{dx}=(\text{y}^3-3\text{x}^2\text{y})\text{dy, where c}$ is a parameter.
Answer$\text{Here},\ \ \text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2\ \ ...{(1)}$
Differentiating w.r.t.x, we get,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]$
$\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big]\ \ ...(2)$
Dividing (2) by (1), we get.
$\frac{\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}}{\text{x}^2-\text{y}^2}=\frac{2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)}{\text{x}^2+\text{y}^2}$
$\therefore\ \ \text{x}(\text{x}^2+\text{y}^2)-\text{y}(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}$ $=2\text{x}(\text{x}^2-\text{y}^2)+2\text{y}(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}$
$\therefore\ \ \big[2\text{y}(\text{x}^2-\text{y}^2)+\text{y}(\text{x}^2+\text{y}^2)\big]\frac{\text{dy}}{\text{dx}}$ $=\text{x}(\text{x}^2+\text{y}^2)-2\text{x}(\text{x}^2-\text{y}^2)$
$\therefore\ \ (2\text{x}^2\text{y}-2\text{y}^3+\text{x}^2\text{y}+\text{y}^3)\frac{\text{dy}}{\text{dx}}$ $=\text{x}^3+\text{xy}^2-2\text{x}^3+2\text{xy}^2$
$\therefore\ \ (3\text{x}^2\text{y}-\text{y}^3)\frac{\text{dy}}{\text{dx}}=3\text{xy}^2-\text{x}^3$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^3-3\text{xy}^2}{\text{y}^3-3\text{x}^2\text{y}}$ which is the required equation.
Hence the result.
View full question & answer→Question 523 Marks
Solve the following differential equations:
$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
Answer$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
$\text{dy}=-(\text{x}+1)(\text{y}+1)\text{dx}$
$\int\frac{\text{dy}}{\text{y}+1}=-\int(\text{x}+1)\text{dx}$
$\log|\text{y}+1|=-\frac{\text{x}^2}{2}-\text{x}+\text{C}$
$\log|\text{y}+1|+\frac{\text{x}^2}{2}+\text{x = C}$
View full question & answer→Question 533 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{y}}(\text{e}^{\text{x}}+\text{e}^{-\text{x}})$
$\Rightarrow\text{e}^{-\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
$\Rightarrow-\text{e}^{-\text{y}}=\text{e}^{\text{x}}-\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$
Hence, $\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$ is the required solution.
View full question & answer→Question 543 Marks
In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b. y = e^{2x} (a + bx)$
Answer$y = e^{2x} (a + bx) ....(1)$
Differentiating both sides with respect to $x,$
we get: $\text{y}\ '=2\text{e}^{2\text{x}}(\text{a+bx}) + \text{e}^{2\text{x}}\cdot\text{b}$
$\Rightarrow \text{y}\ '=\text{e}^{2\text{x}} (2\text{a}+2\text{bx}+\text{b}) \ ....(2)$
MUltiplying equation $(1)$ by $(2)$ and than subtracting it equation $(2),$
we get:$\text{y}\ '-2\text{y = e}^{2\text{x}} (2\text{a + 2bx + b)}-\text{e}^{2\text{x}}(2\text{a + 2bx})$
$\Rightarrow \text{y}'-2\text{y}=\text{be}^{2\text{x}} \ ...(3)$
Differentiating both sides with respect to $x,$
we get: $\text{y}\ ''-2\text{y}\ '=2\text{be}^{2\text{x}} \ ...(4)$
Dividing equation $(4)$ by equation $(3),$
we get: $\frac{\text{y}\ ''-2\text{y}\ '}{\text{y}\ '-2\text{y}}=2$
$\Rightarrow \text{y}\ ''-2\text{y}\ '=2\text{y}\ '-4\text{y}$
$\Rightarrow \text{y}\ ''-4\text{y}\ '+4\text{y}=0$
This is the required differential equation of the given curve.
View full question & answer→Question 553 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$
AnswerWe have $\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$ $\Rightarrow\text{dx}=\frac{1}{\sin^2\text{y}}$ $\Rightarrow\text{dx}=\text{cosec}^2\text{y dy}$ Integrating both sides, we get $\int\text{dx}=\text{cosec}^2\text{y dy}$ $\Rightarrow\text{x}=-\cot\text{y}+\text{C}$$\Rightarrow\text{x}+\cot\text{y}=\text{C}$
Hence, $\Rightarrow\text{x}+\cot\text{y}=\text{C}$ is the required solution.
View full question & answer→Question 563 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$ |
: |
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$ |
AnswerThe given differential equation is
$\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$
$\Rightarrow\ \text{e}^{-\text{x}}\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}\ \ ...(1)$
Differentiating (1) twice w.r.t. .x. we get
$\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}(-1)=-\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{and}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\text{e}^{-\text{x}}(-1)-\Big\{-\text{y e}^{-\text{x}}+\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}\Big\}$ $=-\text{a}\cos\text{x}-\text{b}\sin\text{x}$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}=-\text{y x}^{-\text{x}}\ \ [\because \text{of}\ (1)]$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+2\text{y e}^{-\text{x}}=0$ $\ \text{or}\ \text{e}^{-\text{x}}\Big\{\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}\Big\}=0$
$\ \text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence the result.
View full question & answer→Question 573 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{xy} = \text{log}\ \text{y} + \text{C} \ \ : \ \text{y}'=\frac{\text{y}^2}{1-\text{xy}}\ (\text{xy} \neq1)$
AnswerGiven: xy = log y + C ....(i) To prove: y given by eq. (i) is a solution of differential equation $\text{y}'=\frac{\text{y}^2}{1-\text{xy}} \ ...(\text{ii})$ Proof: Differentiating both sides of eq. (i) w.r.t x, we have $\text{xy}'+\text{y}(1)=\frac{1}{\text{y}}\text{y}'+0 \ $ $\Rightarrow \ \text{xy}'-\frac{\text{y}'}{\text{y}}=-\text{y} $ $ \Rightarrow \ \text{y}'\bigg(\text{x}-\frac{\text{1}}{\text{y}}\bigg)=-\text{y}$ $\Rightarrow \ \text{y}'\bigg(\frac{\text{xy}-1}{\text{y}}\bigg)=-\text{y}\ $ $\Rightarrow \ \text{y}'(\text{xy}-1)=-\text{y}^2$ $ \Rightarrow \ \text{y}'=\frac{-\text{y}^2}{\text{xy}-1}$$\Rightarrow \ \ \text{y}'=\frac{-\text{y}^2}{-(1-\text{xy})}=\frac{\text{y}^2}{1-\text{xy}}$
Hence, Function (implicit) given by eq. (i) is a solution of ${\text{y}}'=\frac{\text{y}^2}{1-\text{xy}}.$
View full question & answer→Question 583 Marks
For each of the differential equations in find the general solution:
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5$
AnswerThe given differential equation is
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5 \ \ \text{or} \ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^5}{\text{x}^5}$
Separating the variables, we get,
$\frac{1}{\text{y}^5}\text{dy}=-\frac{1}{\text{x}^5}\text{dx}$
Integrating, $\int \frac{\text{1}}{\text{y}^5}\text{dy}=-\int \frac{1}{\text{x}^5}\ \text{dx} \ \text{or}$ $ \int \text{y}^{-5}\text{dy}=-\int\text{x}^{-5}\text{dx}$
$\therefore\ \ \frac{\text{y}^{-4}}{-4}=-\frac{\text{x}^{-4}}{-4}+\text{c}'\ \text{or}$ $\frac{1}{-4\text{y}^4}=\frac{1}{4\text{x}^4}+\text{c}'$
$\text{or} \ \frac{1}{\text{x}^4}+\frac{1}{\text{y}^4}=-4\text{c}'$
$\text{or}\ \text{x}^{-4}+\text{y}^{-4}=\text{c}, \ \text{where c}=-4\text{c}'.$
View full question & answer→Question 593 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$
AnswerThe given differential equation is$\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
Separating the variables and integrating,$\int \text{dy}=\int\text{sin}^{-1}\text{x dx}$
$\therefore\ \int \text{1 dy}= \int \sin^{-1}\text{x}\cdot 1\ \text{dx}$ $\therefore \ \text{y}=(\text{sin}^{-1}\text{x}).\text{x}-\int\frac{1}{\sqrt{1-\text{x}^2}}.\text{x dx}$ $\text{or}\ \text{y = x sin}^{-1}\text{x}+\frac{1}{2}\int(1-\text{x}^2)^{-\frac{1}{2}}(-2\ \text{x}) \ \text{dx}$ $\therefore\ \text{y}=\text{x sin}^{-1}\text{x}+\frac{1}{2}\frac{(1-\text{x}^2)\frac{1}{2}}{\frac{1}{2}}+\text{c}$ $\therefore\ \text{y}= \text{x sin}^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$ which is the required solution.
View full question & answer→Question 603 Marks
Find the general solution of $\frac{\text{dy}}{\text{dx}}+\text{ay}=\text{e}^{\text{mx}}.$
AnswerWe have, $\frac{\text{dy}}{\text{dx}}+\text{ay}=\text{e}^{\text{mx}}$
which is a linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\text{a},\text{Q}=\text{e}^{\text{mx}}$
$\text{I.F.}=\text{e}^{\int\text{P}\text{dx}}$
$\text{I.F.}=\text{e}^{\int\text{a}\text{dx}}$
$\text{I.F.}=\text{e}^{\text{ax}}$
The general solution is,
$\text{y.}\text{e}^{\text{ax}}=\int\text{e}^{\text{mx}}.\text{e}^{\text{ax}}\text{dx}+\text{C}$
$\Rightarrow\text{y.}\text{e}^{\text{ax}}=\int\text{e}^{(\text{m}+\text{a})\text{x}}\text{ dx}+\text{C}$
$\Rightarrow\text{y}.\text{e}^\text{ax}=\frac{\text{e}^{(\text{m}+\text{a})\text{x}}}{(\text{m}+\text{a})}+\text{C}$
View full question & answer→Question 613 Marks
Solve the differential equation $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}.$
AnswerWe have $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}\ ....(\text{i})$
Take $\text{x}+\text{y}=\text{t}$
$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=\frac{\text{dt}}{\text{dx}}$
Substituting $\text{x}+\text{y}=\text{t}$ in equation (i) we get,
$\frac{\text{dt}}{\text{dx}}=\text{e}^\text{t}$
$\Rightarrow\text{e}^{-\text{t}\text{dt}}=\text{dx}$
$\Rightarrow-\text{e}^{-\text{t}}=\text{x}+\text{C}$
$\Rightarrow\frac{-1}{\text{e}^\text{x+y}}=\text{x}+\text{C}$
$\Rightarrow-1=(\text{x}+\text{C})\text{e}^\text{x+y}$
$\Rightarrow(\text{x}+\text{C})\text{ e}^\text{x+y}1=0$
View full question & answer→Question 623 Marks
Form the differential equation corresponding to $\text{y}^2=\text{a}(\text{b}-\text{x}^2)$ bt eliminating a and b.
Answer$\text{y}^2=\text{a}(\text{b}-\text{x}^2)$
Differential it with respect to x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}(-2\text{x}) ...(1)$
Again, differential it with respect to x,
$2\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\times\frac{\text{dy}}{\text{dx}}\Big]=-2\text{a}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\Big(\frac{2\text{y}}{-2\text{x}}\frac{\text{dy}}{\text{dx}}\Big)$
using equation (1)
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
View full question & answer→Question 633 Marks
In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if $Rs. 100$ double itself in $10$ years $(log_e2 = 0.6931).$
AnswerLet $P$ denote the principal at any time t and $r\%$ per annum be the interest rate..
From the given condition,$\frac{\text{dP}}{\text{dt}}=(0.0\ \text{r})\text{P}$
Separatng the variables and integrating,
we get,$\int\frac{1}{\text{P}}\text{dP}=\int0.0\ \text{r}\ \text{dt}$
$\therefore\ \log\ \text{P}=0.0\ \text{r}\ \text{t}+\log\ \text{c}\ \ \text{or}\ \ \log\frac{\text{P}}{\text{c}}=0.0\ \text{r t}$
$\therefore\ \text{P}=\text{c e}^{0.00\ \text{r t}}\ ...(1)$
Now $P = 100$ when $t = 0$
$\therefore\ 100=\text{c e}^0$
$\Rightarrow\ \ \text{c}=100$
$\therefore\ \text{from (1), P}=100\text{e}^{0.0\ \text{r t}}$
Now $Rs.100$ become $Rs.200$ in $10$ years.
$\therefore\ 200=100\text{e}^{(0.0\ \text{r})\ 10}$
$\Rightarrow\ \ 2=\text{e}^\frac{\text{r}}{10}$
$\therefore\ \log_{\text{e}}2=\frac{\text{r}}{10}$
$\Rightarrow\ \ 0.6931=\frac{\text{r}}{10}$
$\Rightarrow\ \ \text{r}=6.931$
$\therefore$ required rate of interest $= 6.9\%$
View full question & answer→Question 643 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{y}=\text{x}\sin3\text{x}$ |
: |
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}-6\cos3\text{x}=0$ |
AnswerThe given differential equation is $\text{y}=\text{x}\sin3\text{x}\ \ ...(1)$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{x}\cos3\text{x}. 3+\sin3\text{x}$ $\text{and}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=3\{\text{x}(-\sin3\text{x}. 3)+\cos3\text{x}\}+\cos3\text{x}. 3$ $\text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{x}\sin3\text{x}+6\cos3\text{x}$ $\text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{y}+6\cos3\text{x}\ \ [\because\text{of (1)}]$ $\text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{y}+6\cos3\text{x}=0.$Hence the result.
View full question & answer→Question 653 Marks
For each of the differential equations in find the general solution: $(e^x + e^{–x}) dy – (e^x – e^{–x}) dx = 0$
AnswerGiven: Differential equation $(e^x + e^{–x}) dy – (e^x – e^{–x}) dx = 0$
$\Rightarrow \ (\text{e}^\text{x} + \text{e}^{-\text{x}}) \ \text{dy} = (\text{e}^\text{x} – \text{e}^{–\text{x}}) \ \text{dx} \ $ $\Rightarrow \ \text{dy}=\Bigg(\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{{\text{e}^\text{x}+\text{e}^{-\text{x}}}}\Bigg) \text{dx}$
Integrating both sides, $\ \int\text{dy}=\int\Bigg(\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}\Bigg)\text{dx}$
$\Rightarrow\ \text{y}=\text{log}\Big|\text{e}^\text{x}+\text{e}^{-\text{x}}\Big|+\text{c} \ \Bigg[\because\int\frac{f'(\text{x})}{f(\text{x})}\text{dx}=\text{log}\big|f(\text{x})\big|\Bigg]$
View full question & answer→Question 663 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
AnswerWe have
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
$=\frac{2\sin^2\text{y}}{2\cos^2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\tan^2\text{y}$
$\frac{\text{dy}}{\tan^2\text{y}}=\text{dx}$
$\int\cot^2\text{y dy}=\int\text{dx}$
$\int(\text{cosec}^2\text{y}-1)\text{dy}=\int\text{dx}$
$-\cot\text{y}-\text{y}+\text{C}=\text{x}$
$\text{C}=\text{x}+\text{y}+\cot\text{y}$
View full question & answer→Question 673 Marks
For each of the differential equation given in find the general solution: $\text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{2}{\text{x}}\log\text{x}$
AnswerThe given differential equation is: $\text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{2}{\text{x}}\log\text{x}$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{2}{\text{x}^2}$ This equation is the form of a linear differential equation as: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ \big(\text{where p}=\frac{1}{\text{x}\log\text{x}}\ \text{and}\ \text{Q}=\frac{2}{\text{x}^2}\big)$ $\text{Now, I.F}=\text{e}^{\int{\text{pdx}}}=\text{e}^{\int{\frac{1}{\text{x}\log\text{x}}\text{dx}}}=\text{e}^{\log(\log\text{x})}=\log\text{x}$The general solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F})\text{dx}+\text{C}$ $\Rightarrow\ \text{y}\log\text{x}=\int\Big(\frac{2}{\text{x}^2}\log\text{x}\Big)\text{dx}+\text{C}\ \ ..\text{(i)}$ $\text{Now},\ \int\Big(\frac{2}{\text{x}^2}\log\text{x}\Big)\text{dx}=2\int\Big(\log\text{x}\cdot\frac{1}{\text{x}^2}\Big)\text{dx}.$ $=2\bigg[\log\text{x}\cdot\int\frac{1}{\text{x}^2}\text{dx}-\int{\bigg\{\frac{\text{d}}{\text{dx}}(\log\text{x})\cdot\int\frac{1}{\text{x}^2}\text{dx}\bigg\}}\bigg]$ $=2\bigg[\log\text{x}\bigg(-\frac{1}{\text{x}}\bigg)-\int\bigg(\frac{1}{\text{x}}\cdot\bigg(-\frac{1}{\text{x}}\bigg)\bigg)\text{dx}\bigg]$ $=2\bigg[-\frac{\log\text{x}}{\text{x}}+\int\frac{1}{\text{x}^2}\text{dx}\bigg]$ $=2\bigg[-\frac{\log\text{x}}{\text{x}}-\frac{1}{\text{x}}\bigg]$ $=-\frac{2}{\text{x}}(1+\log\text{x})$ Substituting the value of $\int\bigg(\frac{2}{\text{x}}\log\text{x}\bigg)\text{dx}$ in equation (1), we get: $\text{y}\log\text{x}=-\frac{2}{\text{x}}(1+\log\text{x})+\text{C}$ This is the required general solution of the given differential equation.
View full question & answer→Question 683 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\text{y}=1(\text{y}\neq1)$
AnswerThe given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}=1\ \text{or} \ \frac{\text{dy}}{\text{dx}}=1-\text{y}$
Separating the variables, we get,
$\frac{1}{1-\text{y}}\text{dy}=\text{dx}$
Integrating, $\int\frac{1}{1-\text{y}}\text{dy}=\int1\ \text{dx}$
$\therefore \ \frac{\text{log|1-y|}}{-1}= \text{x+c}'$
$\therefore \ \text{log|1-y|}=-\text{x}-\text{c}'$
$\therefore\ |1-\text{y}|=\text{e}^{-\text{x}-\text{c}'}\ \text{or}\ |1-\text{y}|=\text{e}^{-\text{x}}.\text{e}^{-\text{c}}$
$\therefore \ 1-\text{y}=\pm\text{e}^{-\text{e}'}\ \text{e}^{-\text{x}}$
$\therefore \ 1-\text{y}=-\text{ce}^{-\text{x}} \ \ \text{where}-\text{c}=\pm\text{e}^{-\text{x}}$
$\therefore\ \text{y}=1+\text{ce}^{-\text{x}}$ is the required solution.
View full question & answer→Question 693 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=2\text{xy, y}(0)=1$
Answer$\frac{\text{dy}}{\text{dx}}=2\text{xy, y}(0)=1$
$\int\frac{\text{dy}}{\text{y}}=\int2\text{x dx}$
$\log|\text{y}|=2\frac{\text{x}^2}{2}+\text{C}$
$\log|\text{y}|=\text{x}^2+\text{C}...(1)$
Put $\text{x}=0,\text{y}=1$
$\log(1)=0+\text{C}$
$0=0+\text{C}$
$\text{C}=0$
Put $\text{C}=0$ in equation (1),
$\log\text{y = x}^2$
$\text{y = e}^{\text{x}^{2}}$
View full question & answer→Question 703 Marks
Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.
AnswerAccording to the quation,
$\frac{\text{dy}}{\text{dx}}=\text{x}+3\text{y}-1$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-3\text{y}=\text{x}-1$
Comparing with we get,
$\text{P}=-3, \text{Q}=\text{x}-1$
Now,
$\text{I.F}=\text{e}^-{\int3\text{dx}}$
$=\text{e}^{-3\text{x}}$
So, the solution is given by
$\text{y}\times\text{I.F}=\int\text{Q}\times\text{I.F}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\int(\text{x}-1)\text{e}^{-3\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\int\text{x}\text{e}^{-3\text{x}}\ \text{dx}-\int\text{e}^{-3\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\text{x}\int\text{e}^{-3\text{x}}\ \text{dx}-\int\big[ \frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{-3\text{x}}\text{dx}\big]\text{dx}-\text{e}^{-3\text{dx}}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\text{x}\int\text{e}^{-3\text{x}}\ \text{dx}+\frac{1}{3} \int\text{e}^{-3\text{x}}\text{dx}-\int\text{e}^{-3\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=-\frac{1}{3}\text{xe}^{-3\text{x}}+\frac{1}{3}\int\text{e}^{-3\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{1}{9}+\frac{1}{3}+\text{Ce}^{-3\text{x}}$
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{2}{9}+\text{Ce}^{-3\text{x}}$
Since the curve passes throught this origin it equation,
$\Rightarrow 0=-0+\frac{2}{9}+\text{Ce}^{0}$
$\Rightarrow \text{C}=-\frac{2}{9}$
Putting the value of C in the equation of the curve, we get
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{2}{9}(1-\text{e}^{-3\text{x}})$
$\Rightarrow \text{y} +\frac{1}{3}\text{x}=\frac{2}{9}(1-\text{e}^{-3\text{x}})$
$\Rightarrow 3(3\text{y}+\text{x})=2(1-\text{e}^{-3\text{x}})$
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For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{cos} \ \text{x}}{1+\text{cos} \ \text{x}}$
AnswerThe given differential equation is
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{cosx}}{1+\text{cosx}}$
$\text{or}\ \ \ \ \text{dy} = \frac{1-\text{cosx}}{1+\text{cosx}}\text{dx}\ \ \ \text{or} \ \ \int\text{dy}=\int\frac{1-\text{cosx}}{1+\text{cosx}}\text{dx}$
$\therefore \ \ \ \int\text{dy} = \int \frac{2\ \text{sin}^2\frac{\text{x}}{2}}{2\ \text{cos}^2\frac{\text{x}}{2}}\text{dx} \ \text{or} \ \int1.\text{dy}=\int\text{tan}^2\frac{\text{x}}{2}\text{dx}$
$\text{or} \ \ \int 1.\text{dy}=\int\Big[\text{sec}^2\frac{\text{x}}{2}-1\Big] \text{dx}, \ \text{or}\ \text{y}=\frac{\text{tan}\frac{\text{x}}{2}}{\frac{1}{2}}-\text{x+c}$
$\text{or} \ \ \text{y}=2\text{tan}\frac{\text{x}}{2}-\text{x}+\text{c}$ which is the required solution.
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Find the equation of a curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
AnswerLet y = f(x) be equation of curve $\text{Now}\frac{\text{dy}}{\text{dx}}$ is slope of tangent to the curve at point (x, y). From the given condition,$\frac{\text{dy}}{\text{dx}}\text{xy}=\text{x}\ \ \text{or}\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
Separating the variables and integrating,$\int\text{y dx}=\int\text{x dx}\ \ \text{or}\ \ \frac{\text{y}^2}{2}=\frac{\text{x}^2}{2}+\text{c}\ ...(1)$
Since it passes through (0, - 2) $\therefore\ \frac{(-2)^2}{2}=\frac{(0)^2}{2}+\text{c}\ \Rightarrow\ 2=\text{c}$ $\therefore\ \text{from}(1),\ \frac{\text{y}^2}{2}=\frac{\text{x}^2}{2}+2\ \text{or}\ \text{y}^2=\text{x}^2+4$ which is required equation of curve.
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For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ |
$\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ |
AnswerWe have $\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ $\Rightarrow\text{y}^2=\text{a}^2-\text{x}^2\ ...(1)$ Given differential equation $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ Differentiating both sides of (1) with respect to x, we get $2\text{y}\frac{\text{dy}}{\text{dx}}=-2\text{x}$ $\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}={\text{x}}$ $\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$Hence, the given function is the solution to the given differential equation.
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For each of the differential equations in find the general solution: $\sec^2\ x \tan\ y dx + sec^2\ y \tan x\ dy = 0$
AnswerThe given differential equation is
$\sec^2\ x \tan\ y dx + \sec^2\ y \tan\ x\ dy = 0$
or $\sec^2\ y \tan\ x$ dy $= -\sec^2\ x \tan\ y\ dx$
Separating the variables, we get,
$\frac{\text{sec}^2}{\text{tan}\ \text{y}}\text{dy}=-\frac{\text{sec}^2\text{x}}{\text{tan}\ \text{x}}\text{dx}$
$\text{Integrating},\ \text{log}|\text{tan}\ \text{y}|=-\text{log}|\text{tan}\ \text{x}|+\text{log}\ \text{A}$
$\text{or} \ \ \text{log} |\text{tan y}|+\text{log|tan x|}$
$=\text{log A$ or $log |tan x tan y|=log A}$
$\text{or}\ |\text{tan x tan y| = A}\ $
$\Rightarrow \ \text{tan x tan y =} \ \pm\text{A}\ \text{or}\ \text{tan x tan y}=\text{c} $
where $c$ is an arbitrary constant.
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For each of the differential equations in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}=\text{y}$ tan x; y = 1 when x = 0
AnswerThe given differential equation is$\frac{\text{dy}}{\text{dx}}=\text{y tan x}$
Separating the variables, we get,$\frac{1}{\text{y}}\text{dy}= \text{tanx}\ \text{dx}$
$\text{Integrating,}\ \int\frac{1}{\text{y}}\text{dy}=\int\text{tanx dx}$ $\therefore\ \log|\text{y}|=-\log|\cos{\text{x}}|+\text{c'}\ \ ... (1)$ $\therefore\ \log|\text{y}|+\log|\cos{\text{x}}|=\text{c'}$ $\therefore\ \log|\text{y}\cos{\text{x}}|=\text{c'}$ $\therefore|\text{y}\cos{\text{x}}|=\text{e}^{\text{c'}}$ $\therefore\ \text{y}\cos{\text{x}}=\pm\text{e}^{\text{c}}\ \text{or}\ \text{y cosx}=\text{c}$ Now y = 1 when x = 0 $\therefore1\cos0=\text{c}\ \Rightarrow\ 1\times1=\text{c}\Rightarrow\text{c}=1$ $\therefore\text{form (1),}\ \text{y}\cos\text{x}=1,$ which is required solution.
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Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\frac{1}{(1+\text{y}^2)}\text{dy}=(1+\text{x}^2)\text{dx}$
Integrating both sides, we get
$\int\frac{1}{(1+\text{y}^2)}\text{dy}=\int(1+\text{x}^2)\text{dx}$
$\Rightarrow\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}$
Hence, $\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}$ is the required solution.
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Solve the following equation:
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
Answer$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
View full question & answer→Question 783 Marks
Verify that $\text{y}=4\sin3\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0.$
AnswerWe have,
$\text{y}=4\sin3\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=12\cos3\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-36\sin3\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9(4\sin3\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 793 Marks
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
AnswerWe have,$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}-\cot\text{y}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\frac{1}{\cot\text{y}}\ \text{dy}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\tan\text{y dy}$
Integrating both sides, we get
$\int\frac{1}{\text{x}}\ \text{dx}=-\int\tan\text{y dy}$
$\Rightarrow\int|\text{x}|=-\int|\sec\text{y}|+\int\text{C}$
$\Rightarrow\int|\text{x}|=-\int|\cos\text{y}|+\int\text{C}$
$\Rightarrow\text{x}=\text{C}\cos\text{y}$
Hence, $\text{x}=\text{C}\cos\text{y}$ is the required solution.
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Show that $\text{y}=\text{ax}^3+\text{bx}^2+\text{c}$ is a solution of the differential equation $\frac{\text{d}^3\text{y}}{\text{dx}^3}=6\text{a}$
AnswerWe have,
$\text{y}=\text{ax}^3+\text{bx}^2+\text{c}\ ...(1)$
Differentiating both sides of (1) with respect in x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{ax}^2+2\text{bx}\ ...(2)$
Differentiating both sides of (2) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{ax}+2\text{b}\ ...(3)$
Differentiating both sides of (3) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{a}$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 813 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=4\text{ax}$
AnswerThe equation of the family of curves is
$\text{y}=4\text{ax}\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dt}}{\text{dx}}=4\text{a}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}-2\text{x}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
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show that $\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}$ is a solution of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}\ ...(1)$
Differentiating both sides with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{be}^\text{x}+2\text{ce}^{2\text{x}}\ ...(2)$
Differentiating both sides with respect to x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{be}^\text{x}+4\text{ce}^{2\text{x}}\ ...(3)$
now,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3(\text{be}^\text{x}+2\text{ce}^{2\text{x}})+2(\text{be}^\text{x}+\text{ce}^{2\text{x}})$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}+2\text{be}^\text{x}+2\text{ce}^{2\text{x}}$
$=3\text{be}^\text{x}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}-6\text{ce}^{2\text{x}}$
$=0$
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Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$
AnswerWe have,
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
View full question & answer→Question 843 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
$\Rightarrow\text{dy}=\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
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Form the differential equation having $\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$ where A and B are aribitrary constants, as its general solution.
Answer$\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$
$\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}\times\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\text{Ax}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)=0$
$\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)(-2\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)-0$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
View full question & answer→Question 863 Marks
Represent the followinf families of curves by forming the corresponding differential equation:
$\text{y}=\text{e}^{\text{ax}}$
AnswerThe equation of the family of curves is
$\text{y}=\text{e}^{\text{ax}}\ ...(1)$
$\Rightarrow\log\text{y}=\text{ax}$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{y}}{\text{x}}$
$\Rightarrow\text{x}\frac{\text{dx}}{\text{dx}}=\text{y}\log\text{y}$
It is the required differential equation.
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