What is the solution for y^{''}+y^'-6y=12t+10e^{-3t} ?

The solution for y^{''}+y^'-6y=12t+10e^{-3t} is y=c_{1}e^{2t}+c_{2}e^{-3t}-2t-1/3-2te^{-3t}

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y^{''}+y^'-6y=12t+10e^{-3t}

y^{^{''}} + y^{^{'}} - 6 y = 12 t + 10 e^{- 3 t}

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

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = 12 t + 10 e^{- 3 t}

Solve linear ODE:

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

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

\frac{d}{d θ} (θ cos (θ) - sin (θ))

\frac{d}{d x} (e^{x + 2 y})

\frac{d}{d x} (ln (\frac{( 2 x - 1 )}{x - 1}))

\int e^{s i n (x)} \cdot cos (x) d x

\int \frac{x - 11}{x ^{2} + 3 x - 4} d x

What is the solution for y^{''}+y^'-6y=12t+10e^{-3t} ?
The solution for y^{''}+y^'-6y=12t+10e^{-3t} is y=c_{1}e^{2t}+c_{2}e^{-3t}-2t-1/3-2te^{-3t}