What is the solution for y^{''}-3y^'+2y=4t+12e^{-t} ?

The solution for y^{''}-3y^'+2y=4t+12e^{-t} is y=c_{1}e^{2t}+c_{2}e^t+2t+3+2e^{-t}

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y^{''}-3y^'+2y=4t+12e^{-t}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + 12 e^{- t}

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

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 4 t + 12 e^{- t}

Solve linear ODE:

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

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

y^{^{''}} - 64 y = cosh (8 x)

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 6 y = x e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{5 e ^{t}}{1 + t ^{2}}

y^{^{''}} + 4 y = 2 x sin (2 x)

y^{^{''}} + y^{^{'}} + y = 3 x + 8

What is the solution for y^{''}-3y^'+2y=4t+12e^{-t} ?
The solution for y^{''}-3y^'+2y=4t+12e^{-t} is y=c_{1}e^{2t}+c_{2}e^t+2t+3+2e^{-t}