What is the solution for y^{''}-2y^'-3y=4 ?

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

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y^{''}-2y^'-3y=4

y^{^{''}} - 2 y^{^{'}} - 3 y = 4

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{3}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 4

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{3}

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{3}

y^{^{''}} + 2 y^{^{'}} - 8 y = 3 e^{- 2 x} - e^{- x}

y^{^{''}} - a^{2} y = sec (a x)

y^{^{''}} - 4 y^{^{'}} - 5 y = x^{2}

y^{^{''}} + 2 y^{^{'}} + 2 y = t^{2}, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

y^{^{'''}} + 265 y^{^{'}} = tan (16 x)

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