JEE Main 2022 — Differential Equations Question with Solution

Given

$\frac{dy}{dx}+e^x\left(x^2-2\right)y=\left(x^2-2x\right)\left(x^2-2\right)e^{2x}$

It is linear differential equation so,

$IF=e^{\int \left(x^2-2\right)e^{x}dx}$

$=e^{x^2{e}^x-2\int x{e}^{x}dx-2e^x}$  $=e^{x^2{e}^x-2\left[x{e}^x-e^x\right]-2e^x}$

$IF=e^{\left(x^2-2x\right)e^x}$

Now solution is given by

$y\times e^{\left(x^2-2x\right)e^x}=\int e^{\left(x^2-2x\right)e^x}\times \left(x^2-2x\right)\left(x^2-2\right)e^{2x}$

$=\int e^{\left(x^2-2x\right)e^x}\times \left(x^2-2x\right)e^x\left(x^2-2\right)e^x$

Now let  $\left(x^2-2x\right)e^x=t$

We get $e^x\left(x^2-2x\right)+e^x\left(2x-2\right)dx=dt$

$e^x\left(x^2-2x+2x-2\right)dx=dt$ 

$e^x\left(x^2-2\right)dx=dt$

$y\times e^{\left(x^2-2x\right)e^x}=\int e^t\times t\times dt$

$y\times e^{\left(x^2-2x\right)e^x}=e^t\left(t-1\right)+c$

$y\times e^{\left(x^2-2x\right)e^x}=e^{\left(x^2-2x\right)e^x}\left(\left(x^2-2x\right)e^x-1\right)+c$

Now at $x=0 y=0$

$0\times e^0=e^0\left(0x{e}^x-1\right)+c$

$0=-1+c$  $c=1$

Putting the value of $c=1$ we get the equation as

$\Rightarrow y\times e^{\left(x^2-2x\right)e^x}=e^{\left(x^2-2x\right)e^x}\left(\left(x^2-2x\right)e^x-1\right)+1$

Now putting $x=2$ in equation we get

$y\times e^{\left(4-4\right)e^2}=e^{\left(4-4\right)e^2}\left(\left(4-4\right)e^2-1\right)+1$

$\Rightarrow y\times e^0=e^0\left(\left(0\right)e^2-1\right)+1$

$\Rightarrow y=-1+1=0$