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

The solution for y^{''}-y=t^2e^t is y=c_{1}e^t+c_{2}e^{-t}+(e^tt^3)/6-(e^tt^2)/4+(e^tt)/4

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

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

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{e ^{t} t ^{3}}{6} - \frac{e ^{t} t ^{2}}{4} + \frac{e ^{t} t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{e ^{t} t ^{3}}{6} - \frac{e ^{t} t ^{2}}{4} + \frac{e ^{t} t}{4}

y = c_{1} e^{t} + c_{2} e^{- t} + \frac{e ^{t} t ^{3}}{6} - \frac{e ^{t} t ^{2}}{4} + \frac{e ^{t} t}{4}

\int \frac{x ^{2}}{x ^{3}} d x

\frac{d y}{( 6 + y ) ^{2}} = \frac{d x}{( 10 + 9 x ) ^{4}}

\frac{df}{d θ} = sin (θ) + csc (θ)

\int (2 x^{3} - 5 x + 7) d x

\frac{d}{d v} (u \cdot v)

What is the solution for y^{''}-y=t^2e^t ?
The solution for y^{''}-y=t^2e^t is y=c_{1}e^t+c_{2}e^{-t}+(e^tt^3)/6-(e^tt^2)/4+(e^tt)/4