What is the solution for y^{''}-12y^'+35y=-40e^{-3t} ?

The solution for y^{''}-12y^'+35y=-40e^{-3t} is y=c_{1}e^{7t}+c_{2}e^{5t}-1/2 e^{-3t}

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y^{''}-12y^'+35y=-40e^{-3t}

y^{^{''}} - 12 y^{^{'}} + 35 y = - 40 e^{- 3 t}

y = c_{1} e^{7 t} + c_{2} e^{5 t} - \frac{1}{2} e^{- 3 t}

Solution steps

y^{''} (t) - 12 y^{'} (t) + 35 y = - 40 e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{5 t} - \frac{1}{2} e^{- 3 t}

y = c_{1} e^{7 t} + c_{2} e^{5 t} - \frac{1}{2} e^{- 3 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (1 + x^{2})

y^{^{''}} + y^{^{'}} - 6 y = e^{- 3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 9

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (x)

y^{^{''}} + 16 y = cos (4 t)

What is the solution for y^{''}-12y^'+35y=-40e^{-3t} ?
The solution for y^{''}-12y^'+35y=-40e^{-3t} is y=c_{1}e^{7t}+c_{2}e^{5t}-1/2 e^{-3t}