Let y=y(x) solve dy=e^ x+y,dx, N.

Question

Let $y=y(x)$ solve $dy=e^{\alpha x+y}\,dx$, $\alpha\in\mathbb{N}$. If $y(\log_e2)=\log_e2$ and $y(0)=\log_e\dfrac12$, then $\alpha$ is \_\_\_\_\_\_.

Answer: 2

Accepted Answer

$\int e^{-y}\,dy=\int e^{\alpha x}\,dx \implies -e^{-y}=\dfrac{e^{\alpha x}}{\alpha}+c$.

Using both points and solving the simultaneous equations gives $\alpha=2$ (as $\alpha\in\mathbb{N}$).

Solutions · 1