What is the solution for y^{''}+4y^'+3y=1,y^'(0)=1 ?

The solution for y^{''}+4y^'+3y=1,y^'(0)=1 is y=(-1-3c_{2})e^{-t}+c_{2}e^{-3t}+1/3

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y^{''}+4y^'+3y=1,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 3 y = 1, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 3 D + 2) y = e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{''}} + y^{^{'}} + 2 y - x - 3 = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 4 y = e^{t}

x^{^{''}} + 4 x^{^{'}} + 4 x = 4 t^{2} + 6 e^{t}

What is the solution for y^{''}+4y^'+3y=1,y^'(0)=1 ?
The solution for y^{''}+4y^'+3y=1,y^'(0)=1 is y=(-1-3c_{2})e^{-t}+c_{2}e^{-3t}+1/3