What is the solution for y^'+4xy=x^3e^{x^2},y(0)=-1 ?

The solution for y^'+4xy=x^3e^{x^2},y(0)=-1 is y=(e^{x^2}(3x^2-1))/(18)-(17)/(18e^{2x^2)}

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y^'+4xy=x^3e^{x^2},y(0)=-1

y^{^{'}} + 4 x y = x^{3} e^{x^{2}}, y (0) = - 1

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} - \frac{17}{18 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{17}{18 e ^{2 x^{2}}}

y = \frac{e ^{x^{2}} ( 3 x ^{2} - 1 )}{18} - \frac{17}{18 e ^{2 x^{2}}}

\int x cosh (2 x^{2} - 3) d x

\frac{e ^{- s}}{( s + 1 ) ^{2} + 1}

\int_{0}^{π} 2 sin^{4} (x) d x

\int \frac{x ^{3}}{x ^{2} + x + 1} d x

x^{2} y^{^{''}} - 7 x y^{^{'}} - 20 y = 0

What is the solution for y^'+4xy=x^3e^{x^2},y(0)=-1 ?
The solution for y^'+4xy=x^3e^{x^2},y(0)=-1 is y=(e^{x^2}(3x^2-1))/(18)-(17)/(18e^{2x^2)}