If the solution of dydx+e^x(x^2-2)y=(x^2-2x)(x^2-2)e^2x satisfies y(0)=0, then y(2) is:

Question

If the solution of $\dfrac{dy}{dx}+e^x(x^2-2)y=(x^2-2x)(x^2-2)e^{2x}$ satisfies $y(0)=0$, then $y(2)$ is:

Accepted Answer

I.F. $=e^{e^x}\cdot(	ext{terms})$; using the specific structure of the RHS (which matches $\dfrac{d}{dx}[(x^2-2)e^x\cdot y]$-type form) and applying $y(0)=0$ gives $y(2)=0$.

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