What is the solution for y^{''}=-4y+e^x ?

The solution for y^{''}=-4y+e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+1/5 e^x

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{x}

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

x y - 6 y^{2} = 4, (14, 2)

\int x^{n - 1} a + b x^{n} d x

\frac{d}{d x} (x (x + 4)^{2})

\frac{\partial}{\partial x} (10 x^{2} y + 5 x y^{3})

What is the solution for y^{''}=-4y+e^x ?
The solution for y^{''}=-4y+e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+1/5 e^x