JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Differential Equations

The integrating factor is $I.F.=e^{\int e^x(x^2-2)dx}$.

Note that $\int e^x(x^2-2)dx=\int e^x(x^2-2x+2x-2)dx=e^x(x^2-2x)$ (using integration by parts).

So $I.F.=e^{e^x(x^2-2x)}$.

Multiplying the equation by I.F. and integrating:

$y\cdot e^{e^x(x^2-2x)}=\int e^{e^x(x^2-2x)}\cdot e^x(x^2-2x)(x^2-2)e^{x}dx$

Let $t=e^x(x^2-2x)$. Then the integral becomes $\int e^t\cdot tdt=t{e}^t-e^t+c$.

At $x=0$, $t=0$, so $0=0-1+c$, giving $c=1$.

At $x=2$, $t=e^2(4-4)=0$, so $y(2)\cdot 1=0\cdot 1-1+1=0$.

Therefore, $y(2)=0$.