# Solve differential equation dy/dx-12x^3y=x^3

Solve differential equation $dy/dx-12{x}^{3}y={x}^{3}$
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Benedict

$dy/dx={x}^{3}+12{x}^{3}y$
$dy/dx={x}^{3}\left(1+12y\right)$
$dy/\left(\left(1+12y\right)\right)={x}^{3}dx$
Taking Integration on both sides, we get
$\int dy/\left(1+12y\right)=\int {x}^{3}dx$
Using u− substitution rule
Let $u=1+12y$
$du=12dy$
$dy=1/12du$
Substituting the values above
$\int dy/\left(1+12y\right)=\int {x}^{3}dx$
$1/12\int \left(du\right)/u={x}^{4}/4+C$
Integrating
$1/12u={x}^{4}/4+C$
$1/12\mathrm{ln}\left(1+12y\right)={x}^{4}/4+C$
Isolating y
$\mathrm{ln}\left(1+12y\right)=3{x}^{4}+12C$
$\mathrm{ln}\left(1+12y\right)=3{x}^{4}+{C}_{1}$
$1+12y={e}^{3{x}^{4}+{C}_{1}}$
$y=\left({e}^{3{x}^{4}+{C}_{1}}-1\right)/12$
$y=\left({e}^{3{x}^{4}+{C}_{1}}\right)/12-1/12$