What is the solution for y^{''}+y^'-6y=te^{2t} ?

The solution for y^{''}+y^'-6y=te^{2t} is y=c_{1}e^{2t}+c_{2}e^{-3t}+(e^{2t}t^2)/(10)-(e^{2t}t)/(25)

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y^{''}+y^'-6y=te^{2t}

y^{^{''}} + y^{^{'}} - 6 y = t e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} + \frac{e ^{2 t} t ^{2}}{10} - \frac{e ^{2 t} t}{25}

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = t e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} + \frac{e ^{2 t} t ^{2}}{10} - \frac{e ^{2 t} t}{25}

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} + \frac{e ^{2 t} t ^{2}}{10} - \frac{e ^{2 t} t}{25}

\frac{\partial}{\partial x} (1 + \frac{2}{y})

\int_{0}^{34} (2 x - x^{2}) d x

x^{2} \frac{d ^{2} y}{d x ^{2}} - 7 x \frac{d y}{d x} + 16 y = 0

y^{^{'}} + 0.5 y = 2

\int (2 t - t - \frac{9}{t ^{2}}) d t

What is the solution for y^{''}+y^'-6y=te^{2t} ?
The solution for y^{''}+y^'-6y=te^{2t} is y=c_{1}e^{2t}+c_{2}e^{-3t}+(e^{2t}t^2)/(10)-(e^{2t}t)/(25)