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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 7 sin (2 t) + 5 t cos (2 t)

\frac{24}{( 4 x - 17 ) ^{\frac{5}{2}}}

y = 8 9 x^{2} + 7

y = arccos (4 x)

x \to \infty lim (x + 1)

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