Solve the initial value problem

$dy/dx=8({x}^{3})\ast {e}^{-2y}$

y(1)=0

$dy/dx=8({x}^{3})\ast {e}^{-2y}$

y(1)=0

Elsa Brewer
2022-07-28
Answered

Solve the initial value problem

$dy/dx=8({x}^{3})\ast {e}^{-2y}$

y(1)=0

$dy/dx=8({x}^{3})\ast {e}^{-2y}$

y(1)=0

You can still ask an expert for help

Kaiden Weeks

Answered 2022-07-29
Author has **14** answers

$\frac{dy}{dx}=8{x}^{3}{e}^{-2y}$

$\Rightarrow {e}^{2y}dy=8{x}^{3}dx$

$\int {e}^{2y}dy=\int 8{x}^{3}dx$

$\frac{1}{2}{e}^{2y}=2{x}^{4}+C$

y(1)=0

$\Rightarrow \frac{1}{2}{e}^{2\ast 0}=2\ast (1{)}^{4}+C$

$\Rightarrow \frac{1}{2}=2+C,C=-\frac{3}{2}$

$\frac{1}{2}{e}^{2y}=2{x}^{4}-\frac{3}{2}$

${e}^{2y}=4{x}^{4}-3$

$2y=\mathrm{ln}(4{x}^{4}-3)$

$y=\frac{1}{2}\mathrm{ln}(4{x}^{4}-3)$

$\Rightarrow {e}^{2y}dy=8{x}^{3}dx$

$\int {e}^{2y}dy=\int 8{x}^{3}dx$

$\frac{1}{2}{e}^{2y}=2{x}^{4}+C$

y(1)=0

$\Rightarrow \frac{1}{2}{e}^{2\ast 0}=2\ast (1{)}^{4}+C$

$\Rightarrow \frac{1}{2}=2+C,C=-\frac{3}{2}$

$\frac{1}{2}{e}^{2y}=2{x}^{4}-\frac{3}{2}$

${e}^{2y}=4{x}^{4}-3$

$2y=\mathrm{ln}(4{x}^{4}-3)$

$y=\frac{1}{2}\mathrm{ln}(4{x}^{4}-3)$

Ethen Blackwell

Answered 2022-07-30
Author has **3** answers

$\int {e}^{2y}dy=8\int {x}^{3}dx$

$\frac{1}{2}{e}^{2y}=2{x}^{4}+c$

$y=\frac{1}{2}\mathrm{ln}(4{x}^{4}+A)$

for the initial value y(1)=0 ie. y = 0, x = 1,

$0=0.5\mathrm{ln}(4+A)$

$\mathrm{ln}(4+A)=0$

${e}^{0}=1=4+A$

$\therefore A=-3$

thus the solution is:

$y=\frac{1}{2}\mathrm{ln}(4{x}^{3}-4)$

$\frac{1}{2}{e}^{2y}=2{x}^{4}+c$

$y=\frac{1}{2}\mathrm{ln}(4{x}^{4}+A)$

for the initial value y(1)=0 ie. y = 0, x = 1,

$0=0.5\mathrm{ln}(4+A)$

$\mathrm{ln}(4+A)=0$

${e}^{0}=1=4+A$

$\therefore A=-3$

thus the solution is:

$y=\frac{1}{2}\mathrm{ln}(4{x}^{3}-4)$

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