# Solve the given differential equation subject to the indicated initial condition. (1 + x^4) dy + x(1 + 4y^2) dx = 0, y(1)= 0

Damien Horton 2022-07-25 Answered
Solve the given differential equation subject to the indicated initial condition.
$\left(1+{x}^{4}\right)dy+x\left(1+4{y}^{2}\right)dx=0,y\left(1\right)=0$
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## Answers (1)

ab8s1k28q
Answered 2022-07-26 Author has 17 answers
$\left(1+{x}^{4}\right)dy+x\left(1+4{y}^{2}\right)dx=0,y\left(1\right)=0$ separable diff.eqn
$\frac{dy}{1+4{y}^{2}}=\frac{xdx}{1+{x}^{4}}$ integrate both sides
$\int \frac{1}{1+4{y}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$
$\int \frac{1}{1+\left(2y{\right)}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$
2y=v
2dy=dv
x=u
2xdx=du
$\frac{1}{2}\mathrm{arctan}\left(2y\right)=\frac{1}{2}\mathrm{arctan}\left({x}^{2}\right)+c$
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