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

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

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

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

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

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

y^{^{''''}} - 5 y^{^{''}} + 4 y = e^{x} (- 122 - 48 x + 72 x^{2})

y^{^{''}} + y^{^{'}} - 12 y = e^{t} + e^{2 t} - 1

y^{^{'''}} + 100 y^{^{'}} = tan (10 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = 5 x^{2} sin (4 x)

1 x^{^{''}} - 3 x^{^{'}} + 2 x = 20 sin (2 t)

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