Finding general solution of first order

Question

Given $\frac{\mathrm{d}y}{\mathrm{d}x}=xe^{-2y}$, find a general solution that satisfies $y(0)=2$. My steps are: $$ \int\frac{\mathrm{d}y}{e^{-2y}}=\int x \mathrm{d}x \implies \ln e^{2y}=x^2 + C \implies y= \frac{\ln(x^2 + C)}{2}. $$ Then $y(0)=2$ gives $$ 4 = \ln(0 + C) \implies e^4 = C, $$ so that $$ y=\frac{\ln{(x^2+e^4)}}{2}. $$

Is this right?

Answer 1

Yes that's correct

$$\int\frac{dy}{e^{-2y}}=\int x dx \implies \int{e^{2y}}dy=\int x dx$$ $$\frac {e^{2y}}2=\frac {x^2}2+K \implies e^{2y}=x^2+C$$ $$ y(0)=2 \implies C=e^4$$ $$y(x)=\frac {\ln |x^2+e^4|}2$$

Answer 2

$\int \frac{1}{e^{-2y}} dy =\int x dx$

$\int e^{2y} dy=\int x dx$

$\frac{1}{2} e^{2y}=\frac{x^2}{2}+C$