What is the solution for y^{''}-2y^'-3y=4t ?

The solution for y^{''}-2y^'-3y=4t is y=c_{1}e^{3t}+c_{2}e^{-t}-(4t)/3+8/9

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

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

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4 t}{3} + \frac{8}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4 t}{3} + \frac{8}{9}

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4 t}{3} + \frac{8}{9}

What is the solution for y^{''}-2y^'-3y=4t ?
The solution for y^{''}-2y^'-3y=4t is y=c_{1}e^{3t}+c_{2}e^{-t}-(4t)/3+8/9