What is the solution for x^{''}+x=-e^t ?

The solution for x^{''}+x=-e^t is x=c_{1}cos(t)+c_{2}sin(t)-1/2 e^t

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x^{''}+x=-e^t

x^{^{''}} + x = - e^{t}

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

Solution steps

x^{''} (t) + x = - e^{t}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 9 y = 5 sin (4 x)

y^{^{''}} + y = cos (t), y (0) = 1, y^{^{'}} (0) = - 3

2 y^{^{''}} + 27 y = e^{t} - e^{- t}, y^{t} (0) = 2, y (0) = 0

y^{^{''}} - \frac{2}{5} y = - x^{2}

3 y^{^{'}} + 2 y^{^{''}} = - cos (3.14 \cdot \frac{t}{0.2})

What is the solution for x^{''}+x=-e^t ?
The solution for x^{''}+x=-e^t is x=c_{1}cos(t)+c_{2}sin(t)-1/2 e^t