What is the solution for y^{''}+y=2xe^x ?

The solution for y^{''}+y=2xe^x is y=c_{1}cos(x)+c_{2}sin(x)+e^xx-e^x

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y^{''}+y=2xe^x

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

y = c_{1} cos (x) + c_{2} sin (x) + e^{x} x - e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + e^{x} x - e^{x}

y = c_{1} cos (x) + c_{2} sin (x) + e^{x} x - e^{x}

\frac{d}{d x} (ln (6 - x))

\frac{d}{d x} (\frac{( x ^{2} - 8 x + 15 )}{x ^{2} + 3 x + 2})

\frac{\partial}{\partial x} (4 x^{2} y^{3})

\int \frac{4 x}{4 x ^{4} + 9} d x

y = 3 x, y = 2 x - 3, y = 0

What is the solution for y^{''}+y=2xe^x ?
The solution for y^{''}+y=2xe^x is y=c_{1}cos(x)+c_{2}sin(x)+e^xx-e^x