Let y=y(x) solve _edydx=3x+4y, y(0)=0.

Question

Let $y=y(x)$ solve $\log_e\dfrac{dy}{dx}=3x+4y$, $y(0)=0$. If $y\left(-\dfrac23\log_e2\right)=\alpha\log_e2$, find $\alpha$:

Accepted Answer

$\dfrac{dy}{dx}=e^{3x}e^{4y}$. Separating: $\int e^{-4y}\,dy=\int e^{3x}\,dx \implies \dfrac{e^{-4y}}{-4}=\dfrac{e^{3x}}{3}+C$; using $(0,0)$: $C=-\dfrac{7}{12}$.

At $x=-\dfrac23\log_e2$: solving gives $4y=\ln\dfrac36=\ln\dfrac12 \implies y=-\dfrac14\ln2$, so $\alpha=-\dfrac14$.

Solutions · 1