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Find an implicit formula satisfied by solutions of y y′ = ex + 2y sin x

Assume solutions of the equation

    \[ yy' = e^{x+2y} \sin x \]

exist and find an implicit formula satisfied by these solutions.


This is a first order separable equation. We compute

    \begin{align*}  yy' = e^{x + 2y} \sin x && \implies && ye^{-2y} y' &= e^x \sin x \\  && \implies && \int y e^{-2y} \, dy &= \int e^x \sin x \, dx \\  && \implies && -\frac{y}{2} e^{-2y} - \frac{1}{4} e^{-2y} &= \frac{1}{2}\big(e^x \sin x - e^x \cos x\big) + C \\  && \implies && e^{-2y} \left( -y-\frac{1}{2} \right) &= e^x (\sin x - \cos x) + C \\  && \implies && e^{-2y} \left( y + \frac{1}{2} \right) &= e^x (\cos x - \sin x) + C. \end{align*}

Both of the integrals were evaluated using integration by parts. Also, we already established a formula for \int e^x \sin x \, dx in this exercise (Section 6.17, Exercise #20, taking a = b = 1).