Let y=y(x) solve (x+1)y'-y=e^3x(x+1)^2, y(0)=13.

Question

Let $y=y(x)$ solve $(x+1)y'-y=e^{3x}(x+1)^2$, $y(0)=\dfrac13$. Then the point $x=-\dfrac43$ for the curve is:

Accepted Answer

Solving the linear ODE $\dfrac{d}{dx}\left(\dfrac{y}{x+1}\right)=e^{3x}$ (after dividing by $(x+1)^2$) and analyzing $y'(x)$ near $x=-4/3$ shows $y'$ changes sign from negative to positive there, so it's a local minimum.

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