Solve the initial value problem dy/dx=8(x^3)*e^(-2y) y(1)=0

Solve the initial value problem
$dy/dx=8\left({x}^{3}\right)\ast {e}^{-2y}$
y(1)=0
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Kaiden Weeks
$\frac{dy}{dx}=8{x}^{3}{e}^{-2y}$
$⇒{e}^{2y}dy=8{x}^{3}dx$
$\int {e}^{2y}dy=\int 8{x}^{3}dx$
$\frac{1}{2}{e}^{2y}=2{x}^{4}+C$
y(1)=0
$⇒\frac{1}{2}{e}^{2\ast 0}=2\ast \left(1{\right)}^{4}+C$
$⇒\frac{1}{2}=2+C,C=-\frac{3}{2}$
$\frac{1}{2}{e}^{2y}=2{x}^{4}-\frac{3}{2}$
${e}^{2y}=4{x}^{4}-3$
$2y=\mathrm{ln}\left(4{x}^{4}-3\right)$
$y=\frac{1}{2}\mathrm{ln}\left(4{x}^{4}-3\right)$
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Ethen Blackwell
$\int {e}^{2y}dy=8\int {x}^{3}dx$
$\frac{1}{2}{e}^{2y}=2{x}^{4}+c$
$y=\frac{1}{2}\mathrm{ln}\left(4{x}^{4}+A\right)$
for the initial value y(1)=0 ie. y = 0, x = 1,
$0=0.5\mathrm{ln}\left(4+A\right)$
$\mathrm{ln}\left(4+A\right)=0$
${e}^{0}=1=4+A$
$\therefore A=-3$
thus the solution is:
$y=\frac{1}{2}\mathrm{ln}\left(4{x}^{3}-4\right)$