What is the solution for y^{''}-3y^'-40y=(t+1)e^t ?

The solution for y^{''}-3y^'-40y=(t+1)e^t is y=c_{1}e^{8t}+c_{2}e^{-5t}-(e^tt)/(42)-41/1764 e^t

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y^{''}-3y^'-40y=(t+1)e^t

y^{^{''}} - 3 y^{^{'}} - 40 y = (t + 1) e^{t}

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - \frac{e ^{t} t}{42} - \frac{41}{1764} e^{t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 40 y = (t + 1) e^{t}

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - \frac{e ^{t} t}{42} - \frac{41}{1764} e^{t}

y = c_{1} e^{8 t} + c_{2} e^{- 5 t} - \frac{e ^{t} t}{42} - \frac{41}{1764} e^{t}

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}

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

y^{^{''}} + 0.7 y + 4 = cos (3 t)

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

y^{^{''}} + y^{^{'}} = - 16 cos (3 x)

What is the solution for y^{''}-3y^'-40y=(t+1)e^t ?
The solution for y^{''}-3y^'-40y=(t+1)e^t is y=c_{1}e^{8t}+c_{2}e^{-5t}-(e^tt)/(42)-41/1764 e^t