What is the solution for y^{''}-y^'=9te^{2t} ?

The solution for y^{''}-y^'=9te^{2t} is y=c_{1}e^t+c_{2}+(9e^{2t}t)/2-27/4 e^{2t}

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y^{''}-y^'=9te^{2t}

y^{^{''}} - y^{^{'}} = 9 t e^{2 t}

y = c_{1} e^{t} + c_{2} + \frac{9 e ^{2 t} t}{2} - \frac{27}{4} e^{2 t}

Solution steps

y^{''} (t) - y^{'} (t) = 9 t e^{2 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} + \frac{9 e ^{2 t} t}{2} - \frac{27}{4} e^{2 t}

y = c_{1} e^{t} + c_{2} + \frac{9 e ^{2 t} t}{2} - \frac{27}{4} e^{2 t}

q^{^{''}} + 20 q^{^{'}} + 100 q = 24, q (0) = 0, q^{^{'}} (0) = 0

(D^{2} + 1) y = e^{i x}

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

y^{^{''}} - 4 y = 4 cos (2 x)

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} + 4 y = (\frac{x}{4} - \frac{1}{8}) e^{4 x}

What is the solution for y^{''}-y^'=9te^{2t} ?
The solution for y^{''}-y^'=9te^{2t} is y=c_{1}e^t+c_{2}+(9e^{2t}t)/2-27/4 e^{2t}