What is the solution for y^{''}+y^'=(3t^2+2t)e^t ?

The solution for y^{''}+y^'=(3t^2+2t)e^t is y=c_{1}+c_{2}e^{-t}+(3e^tt^2)/2-(7e^tt)/2+15/4 e^t

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y^{''}+y^'=(3t^2+2t)e^t

y^{^{''}} + y^{^{'}} = (3 t^{2} + 2 t) e^{t}

y = c_{1} + c_{2} e^{- t} + \frac{3 e ^{t} t ^{2}}{2} - \frac{7 e ^{t} t}{2} + \frac{15}{4} e^{t}

Solution steps

y^{''} (t) + y^{'} (t) = (3 t^{2} + 2 t) e^{t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{3 e ^{t} t ^{2}}{2} - \frac{7 e ^{t} t}{2} + \frac{15}{4} e^{t}

y = c_{1} + c_{2} e^{- t} + \frac{3 e ^{t} t ^{2}}{2} - \frac{7 e ^{t} t}{2} + \frac{15}{4} e^{t}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 3 sin (2 x)

y^{^{''}} + 4 y = 5 e^{x} - 4 x^{2}

y^{^{''}} + x + y = 0

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

y^{^{''}} + 12 y^{^{'}} + 27 y = 64 e^{- 1 t}

What is the solution for y^{''}+y^'=(3t^2+2t)e^t ?
The solution for y^{''}+y^'=(3t^2+2t)e^t is y=c_{1}+c_{2}e^{-t}+(3e^tt^2)/2-(7e^tt)/2+15/4 e^t