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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{( x ^{2} - 1 )}{2 y}, y (3) = 1

\frac{\partial}{\partial y} (y^{2} e^{x yz})

\frac{\partial}{\partial x} (y^{5} - 4 x y)

y^{^{'}} = \frac{6 y ^{3}}{( 3 - 2 x ) ^{2}}, y (0) = - 1

\int_{0}^{1} ln (2 x) d x

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