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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 5 t

\frac{d ^{2} y}{d x ^{2}} (1) + \frac{d y}{d x} (10) + 9 y = 18

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 sin (x)

y^{^{''}} - 2 y^{^{'}} + y = e^{3 x} + e^{2 x}

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