Solve the given differential equation subject to the indicated initial condition.

$(1+{x}^{4})dy+x(1+4{y}^{2})dx=0,y(1)=0$

$(1+{x}^{4})dy+x(1+4{y}^{2})dx=0,y(1)=0$

Damien Horton
2022-07-25
Answered

Solve the given differential equation subject to the indicated initial condition.

$(1+{x}^{4})dy+x(1+4{y}^{2})dx=0,y(1)=0$

$(1+{x}^{4})dy+x(1+4{y}^{2})dx=0,y(1)=0$

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ab8s1k28q

Answered 2022-07-26
Author has **17** answers

$(1+{x}^{4})dy+x(1+4{y}^{2})dx=0,y(1)=0$ separable diff.eqn

$\frac{dy}{1+4{y}^{2}}=\frac{xdx}{1+{x}^{4}}$ integrate both sides

$\int \frac{1}{1+4{y}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$

$\int \frac{1}{1+(2y{)}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$

2y=v

2dy=dv

x=u

2xdx=du

$\frac{1}{2}\mathrm{arctan}(2y)=\frac{1}{2}\mathrm{arctan}({x}^{2})+c$

$\frac{dy}{1+4{y}^{2}}=\frac{xdx}{1+{x}^{4}}$ integrate both sides

$\int \frac{1}{1+4{y}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$

$\int \frac{1}{1+(2y{)}^{2}}dy=\int \frac{x}{1+{x}^{4}}dx$

2y=v

2dy=dv

x=u

2xdx=du

$\frac{1}{2}\mathrm{arctan}(2y)=\frac{1}{2}\mathrm{arctan}({x}^{2})+c$

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$$(E):\{\begin{array}{r}{y}^{\prime}=ysi{n}^{2}(y)\\ y(0)={x}_{0}\end{array}$$

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