What is the solution for y^{''}-2y^'+y=14.5e^t ?

The solution for y^{''}-2y^'+y=14.5e^t is y=c_{1}e^t+c_{2}te^t+(29e^tt^2)/4

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y^{''}-2y^'+y=14.5e^t

y^{^{''}} - 2 y^{^{'}} + y = 14.5 e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{29 e ^{t} t ^{2}}{4}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = 14.5 e^{t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{29 e ^{t} t ^{2}}{4}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{29 e ^{t} t ^{2}}{4}

\int (1 - y^{2})^{- \frac{1}{2}} d y

y^{^{'}} + \frac{y}{x} = 9 x^{3} y^{2}

\frac{d}{d x} (n^{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = 0

\frac{d y}{d t} - (\frac{3}{t}) y = 2 t^{3} e^{2 t}, y (1) = 0

What is the solution for y^{''}-2y^'+y=14.5e^t ?
The solution for y^{''}-2y^'+y=14.5e^t is y=c_{1}e^t+c_{2}te^t+(29e^tt^2)/4