What is the solution for y^{''}-8y^'+20y=te^t ?

The solution for y^{''}-8y^'+20y=te^t is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))+(e^tt)/(13)+6/169 e^t

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y^{''}-8y^'+20y=te^t

y^{^{''}} - 8 y^{^{'}} + 20 y = t e^{t}

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t}{13} + \frac{6}{169} e^{t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 20 y = t e^{t}

Solve linear ODE:

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t}{13} + \frac{6}{169} e^{t}

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{e ^{t} t}{13} + \frac{6}{169} e^{t}

y^{^{''}} + y = x sin (x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = - 8 e x + 6 e - x

y^{^{''}} + 16 y = csc (4 x)

y^{^{''}} - 4 y = 2 x + e^{- 2 x}

y^{^{''}} - y^{^{'}} = - 9

What is the solution for y^{''}-8y^'+20y=te^t ?
The solution for y^{''}-8y^'+20y=te^t is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))+(e^tt)/(13)+6/169 e^t