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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{( v ^{2} + v ) ( v - 1 ) ^{2}} d v

\int 9 z^{3} e^{z} d z

f (x) = \frac{x ^{2} - 1}{( x ^{2} + x + 1 )}

\int ln (5)

y^{^{''}} + 16 y = 3 sin (5 t), y (0) = 0, y^{^{'}} (0) = 0

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