What is the solution for y^{''}-y=e^t(2t+5) ?

The solution for y^{''}-y=e^t(2t+5) is y=c_{1}e^t+c_{2}e^{-t}+(e^tt^2)/2+2te^t

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y^{''}-y=e^t(2t+5)

y^{^{''}} - y = e^{t} (2 t + 5)

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

Solution steps

y^{''} (t) - y = e^{t} (2 t + 5)

Solve linear ODE:

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

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

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = - 6 e^{x} + 10 cos (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 6 x e^{- 3 x} + 9 + 50 sin (x)

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

y^{^{''}} + 4 y = 3 sin (2 t), y (0) = 2, y^{^{'}} (0) = - 1

x^{^{''}} + 9 x = 5 sin (3 t)

What is the solution for y^{''}-y=e^t(2t+5) ?
The solution for y^{''}-y=e^t(2t+5) is y=c_{1}e^t+c_{2}e^{-t}+(e^tt^2)/2+2te^t