What is the solution for y^{''}+y^'+y=t ?

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

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y^{''}+y^'+y=t

y^{^{''}} + y^{^{'}} + y = t

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (6 x^{2} - 4 x)

y^{^{''}} + 5 y = 0, y^{^{'}} (1) = 5

- \frac{1}{4 x ^{\frac{3}{2}}}

y^{^{''}} + 8 y^{^{'}} + 20 y = 0

\frac{d}{d x} (x^{\frac{1}{4}} cos (7 ln (x)))

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