Solve the following initial value problems ( dy dx )=2x +x^2-y, 0 given that y = 0 when x= 2

We have, ( dy dx )=2x +x^2-y dy dx +1 y= 2x +x^2 dy dx +( )y=2x+x^2 This is a linear differential equation of the form dy dx +Py=Q Integrating factor, I.F.=e^ =e^ dx =e^ = The solution of the given differential equation is given by y (I.F.)= (I.F.) dx+C = (2x+x^2 ) dx+C = 2x dx+ ^2 dx+C = 2x dx+ [x^2 dx- ( d dx x^2 dx )dx ] = 2x dx +x^2 - 2x dx+C =x^2 +C =x^2+coses x ....(1) It is given that y = 0 when x= 2 0= ( 2 )^2+coses 2 =- ^2 4 Puttuing C=- ^2 4 in (1) we get y=x^2- ^2 4 coses x Hence, y=x^2- ^2 4 coses x is the required solution.

Solve the following initial value problems ( dy dx )=2x +x^2-y, 0…

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Question

Solve the following initial value problems $\tan\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=2\text{x}\tan\text{x}+\text{x}^2-\text{y},\tan\text{x}\neq0$ given that y = 0 when $\text{x}=\frac{\pi}{2}$

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Answer

We have,
$\tan\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=2\text{x}\tan\text{x}+\text{x}^2-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{1}{\tan\text{x}}\text{y}=\frac{2\text{x}\tan\text{x}+\text{x}^2}{\tan\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+(\cot\text{x})\text{y}=2\text{x}+\text{x}^2\cot\text{x}$
This is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Integrating factor,
$\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{x dx}}$
$=\text{e}^{\log\sin\text{x}}$
$=\sin\text{x}$
The solution of the given differential equation is given by
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{ dx}+\text{C}$
$\Rightarrow\text{y}\times\sin\text{x}=\int(2\text{x}+\text{x}^2\cot\text{x})\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx}+\int\text{x}^2\cos\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx}+\Big[\text{x}^2\int\cos\text{x dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}^2\times\int\cos\text{x dx} \Big)\text{dx}\Big]$
$\Rightarrow\text{y}\sin\text{x}=\int2\text{x}\sin\text{x dx} +\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=\text{x}^2\sin\text{x}+\text{C}$
$\Rightarrow\text{y}=\text{x}^2+\text{coses x}\times\text{C}\ ....(1)$
It is given that y = 0 when $\text{x}=\frac{\pi}{2}$
$\therefore\ 0=\big(\frac{\pi}{2}\Big)^2+\text{coses}\frac{\pi}{2}\times\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{4}$
Puttuing $\text{C}=-\frac{\pi^2}{4}$ in (1) we get
$\text{y}=\text{x}^2-\frac{\pi^2}{4}\text{coses x}$
Hence, $\text{y}=\text{x}^2-\frac{\pi^2}{4}\text{coses x}$ is the required solution.

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