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

The solution for y^{''}+y=xe^x,y(0)=0,y^'(0)=1 is y= 1/2 cos(x)+sin(x)+(e^xx)/2-1/2 e^x

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + (- 4) y^{^{'}} + 3 y = - 12

7 y^{^{''}} - 20 y^{^{'}} + 12 y = 8 e^{2}

y^{^{''}} + y = 2 sin (t)

y^{^{''}} - 2 y^{^{'}} + 10 y = 18 e^{x} sec (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = x e^{- x}

What is the solution for y^{''}+y=xe^x,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+y=xe^x,y(0)=0,y^'(0)=1 is y= 1/2 cos(x)+sin(x)+(e^xx)/2-1/2 e^x