What is the solution for y^'+4xy=x^3e^{x^2} ?

The solution for y^'+4xy=x^3e^{x^2} is y=(e^{x^2}(3x^2-1))/(18)+(c_{1})/(e^{2x^2)}

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y^'+4xy=x^3e^{x^2}

y^{^{'}} + 4 x y = x^{3} e^{x^{2}}

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

Solution steps

y^{'} (x) + 4 x y = x^{3} e^{x^{2}}

Solve linear ODE:

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} + \frac{c _{1}}{e ^{2 x^{2}}}

d x = 4 (x^{2} + 1) d t

\frac{1}{( s + 5 )}

\frac{1}{5 s - 2}

\int \frac{x ^{2}}{2 x} d x

\int c^{x} d x

What is the solution for y^'+4xy=x^3e^{x^2} ?
The solution for y^'+4xy=x^3e^{x^2} is y=(e^{x^2}(3x^2-1))/(18)+(c_{1})/(e^{2x^2)}