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

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

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 2

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{( x ^{2} + 1 )}

y^{^{''}} - 2 y^{^{'}} - 3 y = - 4 e^{x} + 3

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

y^{^{''}} + 2 y^{^{'}} + 2 y = x e^{- 2 x} + 3 x

y^{^{''}} - y^{^{'}} - 6 y = 12 x - e^{x}

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