What is the solution for y^{''}+4y^'+3y=1 ?

The solution for y^{''}+4y^'+3y=1 is y=c_{1}e^{-t}+c_{2}e^{-3t}+1/3

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y^{''}+4y^'+3y=1

y^{^{''}} + 4 y^{^{'}} + 3 y = 1

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 1

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

y^{^{''}} + 4 y = 9 sin (t), y (0) = 1, y^{^{'}} (0) = - 1

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

9 y^{^{''}} - 16 y = sin (x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 cos^{2} (2 x)

y^{^{''}} + 25 y = 3, y (0) = 1, y^{^{'}} (0) = 5

What is the solution for y^{''}+4y^'+3y=1 ?
The solution for y^{''}+4y^'+3y=1 is y=c_{1}e^{-t}+c_{2}e^{-3t}+1/3