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

The solution for y^'+4xy=xe^{x^2},y(0)=1 is y=(e^{x^2})/6+5/(6e^{2x^2)}

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

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

y = \frac{e ^{x^{2}}}{6} + \frac{5}{6 e ^{2 x^{2}}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x^{2}}}{6} + \frac{5}{6 e ^{2 x^{2}}}

y = \frac{e ^{x^{2}}}{6} + \frac{5}{6 e ^{2 x^{2}}}

\frac{\partial}{\partial x} (ln (5 x^{2} + 8 y^{2} + 4))

(\frac{x ^{2} - 1}{x ^{2} + 1})^{^{'}}

\int \frac{2}{x ^{3} - 2 x ^{2} + 9 x - 18} d x

3 e^{- x} + e^{4 x}

\frac{d}{d x} (\frac{4 x + 5}{4 x - 5})

What is the solution for y^'+4xy=xe^{x^2},y(0)=1 ?
The solution for y^'+4xy=xe^{x^2},y(0)=1 is y=(e^{x^2})/6+5/(6e^{2x^2)}