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

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

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y^{''}+y^'+y=k

y^{^{''}} + y^{^{'}} + y = k

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + k

Solution steps

y^{''} (t) + y^{'} (t) + y = k

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + k

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + k

4 y^{^{''}} - y = e^{\frac{1}{2} x} + 7

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = x^{2} e^{- 3 x}

y^{^{''}} - y^{^{'}} + y = 2

y^{^{''}} + 3 y^{^{'}} - 2 = 0

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3}), y (0) = 0, y^{^{'}} (0) = 0

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