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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - x = e^{t} sin (e^{- t}) cos (e^{- t})

y^{^{''}} + y^{^{'}} - 20 y = 4 e^{2} t

y^{^{''}} + 4 y^{^{'}} = 2 sin (2 x)

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

y^{^{''}} - 3 y^{^{'}} = 4 sin (x)

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