What is the solution for x^{''}+x^'+4x=-3 ?

The solution for x^{''}+x^'+4x=-3 is x=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))-3/4

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x^{''}+x^'+4x=-3

x^{^{''}} + x^{^{'}} + 4 x = - 3

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) - \frac{3}{4}

Solution steps

x^{''} (t) + x^{'} (t) + 4 x = - 3

Solve linear ODE:

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) - \frac{3}{4}

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) - \frac{3}{4}

10 y^{^{''}} + 1000 y = 160 cos (8 t), y (0) = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 2 e^{x} - 3

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x + 3

y^{^{''}} + y = 9 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 5 y = - 9 e^{2 x}

What is the solution for x^{''}+x^'+4x=-3 ?
The solution for x^{''}+x^'+4x=-3 is x=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))-3/4