Question 13 Marks
Form the differential equation of the family of curves y = a cos(x + b), where a and b are arbitrary constants.
Answery = a cos (x + b) $\Rightarrow\frac{\text{dy}}{\text{dx}}$ = - a sin (x + b)
$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ = - a cos (x + b)
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y OR }\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$.
View full question & answer→Question 23 Marks
Solve the following differential equation:
$4\frac{\text{dy}}{\text{dx}}+\text{8y}=\text{5e}^{-3x}$.
Answer$\frac{\text{dy}}{\text{dy}}+\text{2y}=\frac{5}{4}\text{e}^{-3x}\Rightarrow\text{P(x) = 2, Q(x)}=\frac{5}{4}\text{e}^{-3x}$
I.F. = $\text{e}^{\int\text{p(x)dx}}=\text{e}^{\text{2x}}$
$\therefore \text{We have y}\cdot\text{e}^{\text{2x}}=\int\frac{5}{4}\text{e}^{\text{-3x}}\cdot\text{ e}^{\text{2x}}\text{dx}$
$\Rightarrow\text{y}\cdot\text{e}^{\text{2x}}=\frac{-5}{4}\text{e}^{\text{-x}}+\text{c }\text{ OR }\text{y}=\frac{-5}{4}\text{e}^{\text{-3x}}+\text{c }\text{ e}^{\text{-2x}}.$
View full question & answer→Question 33 Marks
Solve the following differential equation:
$y(1 - x^2) \frac{\text{dy}}{\text{dx}} = x(1 + y^2)$.
AnswerWriting $y(1 - x^2) \frac{\text{dy}}{\text{dx}} = x(1 + y^2)$ as $\int\frac{\text{ydy}}{\text{1 + y}^{2}}=\int\frac{\text{x}}{\text{1 - x}^{2}}\text{dx}$
$\Rightarrow \log|1 + y^2| = - \log|1 - x^2| + \log C_1$
$\Rightarrow (1 + y^2)(1 - x^2) = C$.
View full question & answer→Question 43 Marks
Verify that y = A cos x - b sin x is a solution of the differential equation.
$\frac{\text{d}^{2} \text{y}}{\text{dx}^{2}}+\text{y}=0.$
Answer$\frac{\text{dy}}{\text{dx}}=-\text{A sin x - B cos x}$
$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =-\text{A cos x + B sin x}$
$\therefore\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{(A cos x - B sin x})=-\text{y }\text{ }\therefore\text{ }\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
$\therefore$ y = A cos x - B sin x is the solution of given diffential equation.
View full question & answer→Question 53 Marks
Solve the following differential equation:
$(y^2 - x^2) dy = 3xy\ dx$.
AnswerWriting $\frac{\text{dy}}{\text{dx}}=\frac{\text{3xy}}{\text{y}^{2}-\text{x}^{2}}=\frac{\text{3y/x}}{\text{y}^{2}/\text{x}^{2}-1}$
Putting $\frac{\text{y}}{\text{x}}=\text{v }\Rightarrow\text{v + x }\frac{\text{dv}}{\text{dx}}=\frac{\text{3v}}{\text{v}^{2}-1}\Rightarrow\text{ x }\frac{\text{dv}}{\text{dx}}=-\frac{\text{v}^{3}-\text{4v}}{\text{v}^{2}-1}$
$\therefore \int\frac{\text{v}^{2}-1}{\text{v}^{3}-\text{4v}}\text{ dv}=-\int\frac{\text{dx}}{\text{x}}\Rightarrow\frac{1}{8}\int\Bigg(\frac{2}{\text{v}}+\frac{3}{\text{v - 2}}+\frac{3}{\text{v + 2}}\Bigg)\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\therefore 2 log\ v + 3 log\ (v - 2) + 3 log\ (v + 2) + 8 log\ x = log\ c$
$\Rightarrow v^2 (v^2 - 4)^3 x^8 = c$
$\Rightarrow y^2 (y^2 - 4x^2)^3 = c$.
View full question & answer→Question 63 Marks
Solve the following differential equation:
$\sin x \frac{dy}{dx} + \cos x y = \cos x \sin ^{2} x$
AnswerGiven equation can be written as
$\frac{dy}{dx} + \cot x . y- \cos x . \sin x \therefore \text{Integrating factor} = e^{\log\sin x} = \text{\sin x} $
$\therefore \text{y} . \sin \text{x} = \int \sin ^{2}\text{ x.} \cos \text{x. dx}$
$\Rightarrow \text{y}. \sin\text{x} = \frac{\sin^{3}\text{x}}{3} \text{+ c or } \text{y} = \frac{1}{3} \sin^{2}\text{x + c}. \text{cosec x}$
View full question & answer→Question 73 Marks
Form the differential equation of the family of curves $y = A\cos 2 x + B\sin2x,$where A and B are constants.
Answer$y = A\cos 2 x + B\sin2x,$$\therefore \frac{dy}{dx} = -\text{2 A} \sin 2x + \text{2 B}\cos 2x$
$\frac{d^{2} y}{dx^{2}} = -\text{4 A} \cos 2x - \text{4 B}\sin 2x$
$\frac{d^{2} y}{dx^{2}} = -\text{4 . y}$
or $\frac{d^{2}y}{dx^{2}} + 4y = 0$ is the required differential equation.
View full question & answer→Question 83 Marks
Solve the following differential equation:$\frac{dy}{dx} + 2y = \text6 {e^{x}}$
Answer$\text{I.F.} = e^{\int 2dx} = e^{2x}$$\therefore \text{the solution is} $
$y . e^{2{x}} = 6 \int e^{3x} \text{dx + c} $
$y . e^{2x} ={\text{2 e}}^{3x}+ c$
or $y = \text{2 e}^{x} + {\text{c e}^{-2x}}$
View full question & answer→Question 93 Marks
Solve the following differential equation:$2 xydx + (x^{2} +2y^2) dy = 0$
Answer$\frac{dy}{dx} = \frac{- 2xy}{x^{2}+ 2y^{2}} ; \text{Let} y = vx\Rightarrow \frac{dy}{dx} = v + x \frac{dy}{dx}$$\therefore v + x \frac{dv}{dx} = \frac{- 2vx^2}{x + 2v^2x^2} = \frac{- 2v}{1 + 2v^2}$
$\therefore x \frac{dv}{dx} = \frac{- 2v - v - 2v^{3}}{1 + 2v^{2}} = -\frac{2v^{3} + 3v}{1 + 2v^{2}}$
$\therefore \frac{dx}{x} = \frac{1 + 2v^{2}}{(2v^{2} + 3)v}$
$= -\bigg[ \frac{2v^{2} + 3 - 2}{v (2v^{2} +3}\bigg] dv = \frac{-dv}{v} + \frac{2dv}{v (2v^{2} + 3)}$
$\therefore \frac{dx}{v} = \frac{-dv}{v} + \bigg[ \frac{A}{v} + \frac{Bv + C}{2v^{2} +3}\bigg]$
$A = \frac{2}{3}, B = -\frac{4}{3}, C = 0$
$\therefore \frac{dx}{x} = -\frac{dv}{v} + \frac{2}{3} \frac{dv}{v} -\frac{1}{3} \frac{4vdv}{2v^{2} + 3}$
Intergrating, we get
$\log x = -\log v + \frac{2}{3} \log v -\frac{1}{3} \log(2v^{2} + 3) + \log \text{c}$
$= -\frac{1}{3} \log v -\frac{1}{3} \log (2v^{2} + 3) + \log c$
$\log x^-3 = \log cv(2v^{2} + 3)$ or
$\frac{1}{x^{3}} = c \frac{y}{x} \bigg(\frac{2y^{2}}{x^{2}} + 3 \bigg)$ or
$= cy \bigg(\frac{2y^{2}}{x^{3}} + \frac{3x^{2}}{x^{3}}\bigg)$ or
$cy (2y^{2} + 3x^{2} = 1$
View full question & answer→Question 103 Marks
Form the differential equation of the family of curves y = a sin (x + b), where a and b are arbitrary constants.
Answer$y = \text {a}\sin (x + b)$$\therefore \frac {dy}{dx} = \text {a}\cos (x + b)$
$\frac{d^2y}{dx^2} = -\text {a}\sin (x+ b) =-y$
$\therefore \frac{d^{2}y}{dx^{2}} + y = 0$
View full question & answer→Question 113 Marks
Solve the following differential equation:$\frac{dy}{dx} -\frac{y}{x} = 2x^{2}$
AnswerIntegrating factor $= e^{-\int\frac{dx}{x}}$$= e^{-\int\frac{- logx}{}} = \frac{1}{x} $
The solution of different equation is
$ y^\frac{1} {x} = {\int\frac{2x^{2}}{x}} dx + c$
$y^{{}{}} \frac{1}{x} = x^{2} + c$ or
$y = x^{3} + cx$
View full question & answer→Question 123 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 133 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 $\left(\log _{\mathrm{e}} 2=0.6931\right)$.
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 143 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 153 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 163 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 173 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 183 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 193 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 203 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}})$
View full question & answer→Question 213 Marks
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.
View full question & answer→Question 223 Marks
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.
View full question & answer→Question 233 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
Answer$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
$\frac{\text{dy}}{\text{dx}}=-\cos\text{x}\tan\text{y}$
$\frac{\text{dy}}{\tan\text{y}}=-\cos\text{x dx}$
$\int\cot\text{ y dy}=-\int\cos\text{x dx}$
$\log|\sin\text{y}|=-\sin\text{x + C}$
$\sin\text{x}+\log|\sin\text{y}|=\text{C}$
View full question & answer→Question 243 Marks
For each of the differential equations given in find the general solution:$\text{y dx}+(\text{x}-\text{y}^{2})\ \text{dy}=0$
Answer$\text{y dx}+(\text{x}-\text{y}^{2})\text{dy}=0$$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{y}^2-\text{x}}{\text{y}}=\text{y}-\frac{\text{x}}{\text{y}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{\text{y}}=\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{px}=\text{Q}\ (\text{where p}=\frac{1}{\text{y}\ }\ \text{and}\ \text{Q}=\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{1}{\text{y}}\text{dy}}=\text{e}^{\log\text{y}}=\text{y}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\text{xy}=\int(\text{y}\cdot\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\int\text{y}^2\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\frac{\text{y}^3}{3}+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{3}+\frac{\text{C}}{\text{y}}$
View full question & answer→Question 253 Marks
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.
View full question & answer→Question 263 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 273 Marks
For each of the differential equations in find the general solution:
$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
AnswerThe given differential equation is
$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
or $\sec ^2 y \tan x d y=-\sec ^2 x \tan y\ d x$
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.
View full question & answer→Question 283 Marks
Solve the following differential equations:$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
Answer$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
$\int\frac{\text{dr}}{\text{r}}=-\int\text{tdt}$
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\text{C}...(1)$
Put $\text{t = 0, r = r}_{0}$ inequation (1),
$\log|\text{r}_0|=0+\text{C}$
$\log|\text{r}_0|=\text{C}$
Now,
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\log|\text{r}_0|$
$\frac{\text{r}}{\text{r}_0}=\text{e}^{-\frac{\text{t}^2}{2}}$
$\text{r}=\text{r}_0\text{e}^{-\frac{\text{t}^2}{2}}$
View full question & answer→Question 293 Marks
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.
View full question & answer→Question 303 Marks
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.
View full question & answer→Question 313 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\ \text{dy}=\big(1+\text{x}^2\big) \big(1+\text{y}^2\big)\text{dx}\ \Rightarrow \ \frac{\text{dy}}{1+\text{y}^2}$ $=\big(1+\text{x}^2\big)\text{dx}\ \ [\text{Separating variables}]$
Integrating both sides, $\int \frac{1}{\text{y}^2+1}\text{dy}=\int\big(\text{x}^2+1\big)\text{dx}$
$\Rightarrow \ \ \ \ \tan^{-1}\text{y}=\frac{\text{x}^3}{3}+\text{x+c}$
View full question & answer→Question 323 Marks
In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
$y^2 = a (b^2 – x^2)$
Answer$y^2 = a (b^2 – x^2)$
Differentiating both sides with respect to x, we get:
$2\text{y}\frac{\text{dy}}{\text{dx}}=a(-2\text{x})$
$\Rightarrow 2\text{yy}'=-2\text{ax}$
$\Rightarrow \text{yy}'=-\text{ax}\ \ ...(1)$
Again, differentiating both sides with respect to x, we get:
$\text{y}'.\text{y}'+\text{yy}''=-\text{a}$
$\Rightarrow (\text{y}')^2+\text{yy}''=-\text{a} \ \ ...(2)$
Dividing equation (2) by equation (1), we get:
$\frac{(\text{y}')^2+\text{yy}''}{\text{yy}'}=\frac{-\text{a}}{-\text{ax}}$
$\Rightarrow \text{xyy}''+\text{x(y}')^2-4\text{y}' = 0$
This is the required differential equation of the given curve.
View full question & answer→Question 333 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.
View full question & answer→Question 343 Marks
For each of the differential equations given in find the general solution:
$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}(\text{y}>0)$
Answer$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}+3\text{y}^2}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{x}+3\text{y}^2}{\text{y}}=\frac{\text{x}}{\text{y}}+3\text{y}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}-\frac{\text{x}}{\text{y}}=3\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dx}}{\text{dy}}+\text{px}=\text{Q}\ (\text{where p}=-\frac{1}{\text{y}}\ \text{and}\ \text{Q}=3\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{-\int\frac{\text{dy}}{\text{y}}}=\text{e}^{-\log\text{y}}=\text{e}^{\log\Big(\frac{1}{\text{y}}\Big)}=\frac{1}{\text{y}}.$
The general solution of the given differential equation is given by the relation,
$\text{x}\text{(I.F.)}=\int(\text{Q}\times\text{I.F.})\text{dy}+\text{C}$
$\Rightarrow\text{x}\times\frac{1}{\text{y}}=\int\Big(3\text{y}\times\frac{1}{\text{y}}\Big)\text{dy}+\text{C}$
$\Rightarrow\frac{\text{x}}{\text{y}}=3\text{y}+\text{C}$
$\Rightarrow\text{x}=3\text{y}^2+\text{Cy}$
View full question & answer→Question 353 Marks
For each of the differential equations in Exercises find the general solution:
$e^x \tan y ~d x+\left(1-e^x\right) \sec ^2 y ~d y=0$
AnswerGiven: Differential equation $e^x \tan y ~d x+\left(1-e^x\right) \sec ^2 y ~d y=0$
Dividing each term by $(1-e^x)$ tan y, we get
$\frac{\text{e}^\text{x}}{1-\text{e}^\text{x}} \text{dx}+\frac{\text{sec}^2\text{y}}{\tan \text{y}} \text{dy = 0}\ \text{[Separating variables]}$
Integrating both sides, $\ \int \frac{\text{e}^\text{x}}{1-\text{e}^\text{x}} \text{dx}+\int\frac{\text{sec}^2\text{y}}{\tan \text{y}} \text{dy}=\text{c}$
$\Rightarrow \ -\int\frac{-\text{e}^\text{x}}{1-\text{e}^\text{x}}\text{dx + log|tan y| = c} $ $ \Rightarrow \ -\text{log|1-}\text{e}^\text{x}|+\text{log}|\text{tan y}|=\text{c}$
$\Rightarrow \ \text{log}\frac{|\tan \text{y|}}{|1-\text{e}^\text{x}|}=\text{log c}'$ $\Rightarrow \ \frac{|\tan \text{y|}}{|1-\text{e}^\text{x}|} = \text{c}'$
$\Rightarrow \ \tan \text{y} = \text{C}\big(1-\text{e}^\text{x}\big) \ \big[\because|\text{t}|=\text{c}' =\text{C (say)}\big]$
View full question & answer→Question 363 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\sec\text{xy}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
AnswerThe given differential equation is:$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q (where p}=\sec\text{x and Q}=\tan\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\sec\text{x dx}}=\text{e}^{\log(\sec\text{x}+\tan\text{x})}=\sec.\text{x}+\tan\text{x}.$
The general solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\tan\text{x}(\sec\text{x}+\tan\text{x})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\sec\text{x}\tan\text{x}\text{dx}+\int\tan^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\int(\sec^2 \text{x}-1)\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\tan\text{x}-\text{x}+\text{C}$
This is the required general solution of the given differential equation.
View full question & answer→Question 373 Marks
Solve $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}.$
AnswerGiven that, $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}$
$\Rightarrow\frac{1}{\text{x}^2}-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}=1$ [dividing throughout by $x^2ydx$]
$\Rightarrow-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}^2}-1=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{xy}}{\text{x}^2}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\Big(\text{x}-\frac{1}{\text{x}}\Big)\text{y}=0$
Which is linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\Big(\text{x}-\frac{1}{\text{x}}\Big),\text{Q}=0$
$\text{I.f}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\big(\text{x}-\frac{1}{\text{x}}\big)}\text{dx}$
$=\text{e}^{\frac{\text{x}^2}{2}-\log\text{x}}$
$=\text{e}^{\frac{\text{x}^2}{2}}.\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}$
The general solution is,
$\text{y}.\frac{1}{\text{x}}\text{e}^\frac{\text{x}^2}{2}=\int0.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}=\text{C}$
$\Rightarrow\text{y}=\text{C}\text{x}\text{e}^{\frac{-\text{x}^2}{2}}$
View full question & answer→Question 383 Marks
Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y' = e^x$ sin x.
AnswerThe given differential equation is
$\text{y'}=\text{e}^{\text{x}}\ \text{sin}\ \text{x} \ \text{or}\ \frac{\text{dy}}{\text{dx}}= \text{e}^{\text{x}}\ \text{sin}\ \text{x}$
Separting the variables, we get,
dy = $e^x$ sinx dx
Integrating, $\int\ \text{dy}=\int\text{e}^\text{x}\ \text{sinx}\ \text{dx}$
$\therefore\ \text{y}=\frac{1}{1+1}\text{e}^\text{x}(\text{sinx - cosx})+\text{c}$
$\Bigg[\because\ \int\text{e}^\text{ax}\ \text{sin b x dx}=\frac{1}{\text{a}^2+\text{b}^2}\text{e}^\text{ax}(\text{a sin bx - b cos bx})\Bigg]$
$\therefore\ \text{y}=\frac{1}{2}\text{e}^\text{x}(\text{sinx - cosx})+\text{c}\ ....(1)$
Now curve passes through (0, 0)
$\therefore\ 0=\frac{1}{2}\text{e}^0(\text{sin}0-\text{cos}0)+\text{c}\ \Rightarrow\ 0=\frac{1}{2}\text{x}(0 - 1)+\text{c}$
$\therefore\ \text{c}=\frac{1}{2}$
$\therefore\ \text{from}(1),\ \text{y}=\frac{1}{2}\text{e}^{\text{x}}(\text{sinx - cosx})+\frac{1}{2}$ which is required solution.
View full question & answer→Question 393 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{x}^{2}=2\text{y}^2\log\text{y}$ |
: |
$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$ |
AnswerThe given differential equation is
$\text{x}^2=2\text{y}^2\log\text{y}\ \ ...(1)$
$\therefore\ \ 2\text{x}=2\Big\{\text{y}^2\times\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big[2\text{y}\frac{\text{dy}}{\text{dx}}\Big]\Big\}$
$\text{or}\ \ 2\text{x}=2(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \ \text{x}=(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
Multiplying both sides by y, we get
$\text{xy}=(\text{y}^2+2\text{y}^2\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\text{or}\ \ \text{xy}=(\text{y}^2+\text{x}^2)\frac{\text{dy}}{\text{dx}}\ \ [\because \text{of}\ (1)]$
$\text{or}\ \ (\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$
Hence the result.
View full question & answer→Question 403 Marks
Find the solution of the differential equation $\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0.$
Answer$\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0$
$\Rightarrow \text{y}\sqrt{1+\text{x}^{2}}\text{dy}=-\text{x}\sqrt{1+\text{y}^{2}}\text{dx}=0$
$\Rightarrow \frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=\frac{-\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
$\Rightarrow \int\frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
Let $1+\text{y}^{2}=\text{t}^{2}=\text{t}^{2}, 1+\text{x}^{2}=\text{p}^{2}$
$\Rightarrow 2\text{y}\ \text{dy}=2\text{t}\ \text{dt}, 2\text{x}\ \text{dx}=2\text{p}\ \text{dp}$
$\Rightarrow \text{y}\ \text{dy}=\text{t}\ \text{dt}, \text{x}\ \text{dx}=\text{p}\ \text{dp}$
Substituting in above equation, we get
$\Rightarrow \int\text{dt}=-\int\text{dp}$
$\Rightarrow \text{t}=-\text{p}+\text{C}$
$\Rightarrow\sqrt{1+\text{x}^{2}}+\sqrt{1+\text{y}^{2}}=\text{C}$
View full question & answer→Question 413 Marks
Form the differential equation representing the family of curves given by $(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2,\ \text{where a}$ is an arbitrery constant.
AnswerEquation of the given family of curves is $(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2$
$\Rightarrow\ \ \text{x}^2+\text{a}^2-2\text{ax}+2\text{y}^2=\text{a}^2\ \ $ $\Rightarrow\ \ \text{x}^2-2\text{ax}+2\text{y}^2=0$
$\Rightarrow\ \ \text{x}^2+2\text{y}^2=2\text{ax}^\ \ ....\text{(i)}$
Here number of arbitrary constants is one only (a).
So we will differentiate both sides of equation only once w,r.t. x,
$\Rightarrow\ \ 2\text{x}+2.2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$ $\Rightarrow\ \ 2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}\ \ ...\text{(ii)}$
Dividing eq. (i) by eq. (ii), we have $\frac{\text{x}^2+2\text{y}^2}{2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}}=\frac{2\text{ax}}{2\text{a}}\ \ \Rightarrow\ \ \frac{\text{x}^2+2\text{y}^2}{2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}}=\text{x}$
$\Rightarrow\ \ \text{x}\Big(2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}^2+2\text{y}^2$ $\Rightarrow\ \ 2\text{x}^2+4\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+2\text{y}^2$
$\Rightarrow\ \ 4\text{xy}\frac{\text{dy}}{\text{dx}}=2\text{y}^2-\text{x}^2\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{y}^2-\text{x}^2}{4\text{xy}}$
View full question & answer→Question 423 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 433 Marks
Solve the following differential equations:$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy, y}\neq0$
Answer$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx = xe}^{\frac{\text{x}}{\text{y}}}\text{dy}+\text{y}^2\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}-\text{xe}^{\frac{\text{x}}{\text{y}}}\text{dy = y}^2\text{dy}$
$\Rightarrow(\text{ydx}-\text{xdy})\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}^2\text{dy}$
$\Rightarrow\frac{(\text{ydx}-\text{xdy})}{\text{y}^2}\text{e}^{\frac{\text{x}}{\text{y}}}=\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\text{dy}$
$\Rightarrow\int\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\int\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y + C}$
View full question & answer→Question 443 Marks
Find the particular solution of the differential equation $(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}=0,\ \text{given that y}=1\ \text{when x}=0. $
AnswerThe given differential equation is
$(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^\text{x}\ \text{dx}=0$ $\text{or}\ \ (1+\text{e}^{2\text{x}})\text{dy}=-(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}$
$\therefore\ \ \frac{1}{1+\text{y}^2}\text{dy}=-\frac{\text{e}^{\text{x}}}{1+\text{e}^{2\text{x}}}\text{dx}$
$\Rightarrow\ \ \int\frac{1}{1+\text{y}^2}\text{dy}=-\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}\ \ ...(1)$
$\text{Let l}=\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}1+\text{dx}$
$\text{Put}\ \ \text{e}^\text{x}=\text{t},\ \therefore\ \text{e}^\text{x}\ \text{dx}=\text{dt}$
$\therefore\ \text{l}=\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}=\tan^{-1}(\text{e}^\text{x})$
$\therefore\ \text{form}\ (2),\ \tan^{-1}\text{y}=\tan(\text{e}^\text{x})+\text{c}\ \ ...(2)$
$\text{Now}, \ \text{x}=0,\ \text{y}=1$
$\therefore\ \tan^{-1}1=-\tan(\text{e}^0)+\text{c}$ $\Rightarrow\ \ \frac{\pi}{4}=-\tan^{-1}(1)+\text{c}$
$\Rightarrow\ \ \frac{\pi}{4}=-\frac{\pi}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{\pi}{2}$
$\therefore\ \ \text{form}\ (2), \tan^{-1}=-\tan\text{e}^{\text{x}}+\frac{\pi}{2},$ which is required solution.
View full question & answer→Question 453 Marks
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 463 Marks
Solve the differential equation $\bigg[\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\bigg]\frac{\text{dx}}{\text{dy}}=1(\text{x}\neq0).$
AnswerGiven: Differential equation: $\bigg(\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\bigg)\frac{\text{dx}}{\text{dy}}=1$
$\Rightarrow\ \Big(\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}-\frac{\text{y}}{\sqrt{\text{x}}}\Big)\frac{\text{dx}}{\text{dy}}=1$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\sqrt{\text{x}}}=\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}$
$\text{Comparing this equation with }\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{P}=\frac{1}{\sqrt{\text{x}}}\ \text{and}\ \text{Q}=\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}$
$\int\text{P}\ \text{dx}=\int\frac{1}{\sqrt{\text{x}}}\text{dx}=\int\text{x}^{\frac{-1}{2}}\text{dx}=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}=2\sqrt{\text{x}}$ $\text{I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{2\sqrt{\text{x}}}$
$\text{The general solution is}\ \ \text{y}(\text{I.F})=\int\text{Q}(\text{I.F})\text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}^{2\sqrt{\text{x}}}=\int\frac{\text{e}^{-2\sqrt{\text{x}}}}{\sqrt{\text{x}}}\text{e}^{2\sqrt{\text{x}}}\ \text{dx}+\text{c}=\int\frac{1}{\sqrt{\text{x}}}\text{dx}+\text{x}$
$\Rightarrow\ \ \text{y}^{2\sqrt{\text{x}}}=\int\text{x}^{\frac{-1}{2}}\ \text{dx}+\text{c}=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{c}=2\sqrt{\text{x}}+\text{c}$ $\Rightarrow\ \ \text{y}=\text{e}^{-2\sqrt{\text{x}}}\ (2\sqrt{\text{x}}+\text{c})$
View full question & answer→Question 473 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{x} \ \text{sin} \ \text{x}\ : \ \text{xy}' = \text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2} $ $(\text{x} \neq 0 \ \text{and} \ \text{x} > \text{y} \ \text{or} \ \text{x} < – \text{y})$
AnswerGiven: y = x sin x .....(i)
To prove: y given by eq. (i) is a solution of differential equation $\text{xy}'=\text{y}+\text{x} \sqrt{\text{x}^2-\text{y}^2}\ ....(\text{ii})$
Proof: From eq. (i), $\frac{\text{dy}}{\text{dx}}(=\text{y}') = \text{x} \frac{\text{d}}{\text{dx}}\text{sin}\ \text{x}+ \text{sin}\ \text{x} \frac{\text{d}}{\text{dx}} \text{x} = \text{x}\ \text{cos}\ \text{x}\ + {\text{sin}} \ \text{x}$
L.H.S. of eq. $(ii) = xy' = x (x \cos x + \sin x) = x^2 \cos x + x \sin x$
R.H.S. of eq. (ii) $= \text{y} + \text{x}\sqrt{\text{x}^2-\text{y}^2} = \text{x}\ \text{sin} \ \text{x}+\text{x} \sqrt{\text{x}^2-\text{x}^2\ \text{sin}^2\text{x}}$ [From eq. (i)]
$= \text{x}\ \text{sin}\ \text{x}+ \text{x} \sqrt{\text{x}^2(1-\text{sin}^2\text{x})} = \text{x}\ \text{sin}\ \text{x}+\text{x}\sqrt{\text{x}^2\text{cos}^2\text{x}}$
$= x \sin x + x.x \cos x = x \sin x + x^2 \cos x$
$= x^2 \cos x + x \sin x$
$\therefore$ L.H.S. = R.H.S
Hence, y given by eq. (i) is a solution of $\text{xy}'=\text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2}.$
View full question & answer→Question 483 Marks
Solve the differential equation $\text{y e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}\ \text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy}(\text{y}\neq0).$
AnswerGiven: Differential equation $\text{y.e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}.\text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy},\ \text{y}\neq0$
$\Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}}{\text{y.e}^\frac{\text{x}}{\text{y}}}+\frac{\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}$ $\Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\frac{\text{x}}{\text{y}}+\text{y.e}^{\frac{-\text{x}}{\text{y}}}\ \ ...\text{(i)}$
It is not a homogeneous differential equation because of presence of only y as a factor, yet it can be solved by putting $\frac{\text{x}}{\text{y}}=\text{v},\ \text{i.e}\ \text{x}=\text{vy}.\ \ \Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}$
Putting these values in eq. (i), we get
$\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}=\text{v}+\text{y e}^{-\text{v}}\ \ \Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\text{y.e}^{-\text{x}}$ $\Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{y}}{\text{e}^{\text{v}}}$
$\Rightarrow\ \ \text{e}^\text{v}=\text{y}+\text{c}\ \ \Rightarrow\ \ \text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}+\text{c}$
View full question & answer→Question 493 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\sin2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\sin2\text{x dx}$
$\Rightarrow\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\text{C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=-\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}$
Substituting the values of C in (1), we get
$\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\frac{1}{2}$
$\Rightarrow\log|\text{y}|=\frac{1-\cos 2\text{x}}{2}$
$\Rightarrow\log|\text{y}|=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2\text{x}}$
Hence, $\text{y}=\text{e}^{\sin^2\text{x}}$ is the required solution
View full question & answer→Question 503 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}^3\frac{\text{d}{^2}\text{y}}{\text{dx}^2}=1$ |
$\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}$ |
AnswerWe have
$\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{a}-\frac{1}{2\text{x}^2}\ ...(2)$
Now, differentiating both sides of (2) with respect to x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(-\frac{1}{2}\Big)\times\Big(\frac{-2}{\text{x}^3}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$
Hence, the given function is the solution to the given differential equation.
View full question & answer→