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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 24 e^{t} + 36 t^{2}

y^{^{'''}} + y^{^{'}} = - 2 + 2 sin (x)

y^{^{''}} - 2 y^{^{'}} + 2 = 0

y^{^{''}} - 2 y^{^{'}} + 82 y = e^{x} cos (9 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4 x}, y (0) = 1, y^{^{'}} (0) = 0

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