What is the solution for -y^{''}+2y^'+3y=1 ?

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

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-y^{''}+2y^'+3y=1

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1

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

Solution steps

- y^{''} (t) + 2 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^{^{''}} - y^{^{'}} = e^{x} (3 sin (x) + cos (x))

4 \frac{d ^{2} y}{d x ^{2}} + 9 y = 15

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 x + 5

3 x^{^{''}} + 30 x^{^{'}} + 125 x = 350

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