What is the solution for (d^2y)/(dt^2)+3(dy)/(dt)-40y=1-4t ?

The solution for (d^2y)/(dt^2)+3(dy)/(dt)-40y=1-4t is y=c_{1}e^{5t}+c_{2}e^{-8t}+t/(10)-7/400

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(d^2y)/(dt^2)+3(dy)/(dt)-40y=1-4t

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 40 y = 1 - 4 t

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} + \frac{t}{10} - \frac{7}{400}

Solution steps

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 40 y = 1 - 4 t

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} + \frac{t}{10} - \frac{7}{400}

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} + \frac{t}{10} - \frac{7}{400}

\int_{2}^{\infty} x e^{- 8 x} d x

e^{x}, x e^{x}, 0, 1

y = \frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

\int - \frac{1}{2} cos (2 x) d x

\frac{( 2 s + 1 )}{( s ^{2} - 2 s + 2 )}

What is the solution for (d^2y)/(dt^2)+3(dy)/(dt)-40y=1-4t ?
The solution for (d^2y)/(dt^2)+3(dy)/(dt)-40y=1-4t is y=c_{1}e^{5t}+c_{2}e^{-8t}+t/(10)-7/400