Let y=y(x) solve e^ydydx-2e^y x+ x^2x=0, y(2)=0.

Question

Let $y=y(x)$ solve $e^y\dfrac{dy}{dx}-2e^y\sin x+\sin x\cos^2x=0$, $y\left(\dfrac{\pi}{2}\right)=0$. If $y(0)=\log_e(\alpha+\beta e^{-2})$, then $4(\alpha+\beta)$ is \_\_\_\_\_\_.

Answer: 4

Accepted Answer

Let $t=e^y$: $\dfrac{dt}{dx}-2t\sin x = -\sin x\cos^2x$. I.F.$=e^{2\cos x}$.

$$te^{2\cos x} = \int e^{2z}z^2\,dz,\quad z=\cos x.$$

Using $y(\pi/2)=0$ fixes the constant; at $x=0$, $y(0) = \log\left(\dfrac14+\dfrac34e^{-2}\right)$, giving $\alpha=\dfrac14,\beta=\dfrac34$, so $4(\alpha+\beta)=4$.

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