Let a curve y=f(x) pass through (0,5) and (_e2,k).

Question

Let a curve $y=f(x)$ pass through $(0,5)$ and $(\log_e2,k)$. If the curve satisfies $2(3+y)e^{2x}\,dx-(7+e^{2x})\,dy=0$, then $k$ is:

Accepted Answer

$$\frac{dy}{dx}-\frac{2ye^{2x}}{7+e^{2x}} = \frac{6e^{2x}}{7+e^{2x}}.$$

I.F. $=\dfrac{1}{7+e^{2x}}$.
$$\frac{y}{7+e^{2x}} = \int\frac{6e^{2x}}{(7+e^{2x})^2}\,dx = \frac{-3}{7+e^{2x}}+C.$$

Using $(0,5)$: $\dfrac58 = -\dfrac38+C \Rightarrow C=1$. So $y=-3+7+e^{2x}=e^{2x}+4$. At $x=\ln2$: $k=8$.

Solutions · 1