What is the solution for y^'=t^2e^{-4t}-4y ?

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

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

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

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

Solution steps

y^{'} (t) = t^{2} e^{- 4 t} - 4 y

Solve linear ODE:

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

f (x) = 4 x (x^{2} - 4 x + 5)^{7}, at x = 2

\int (x^{2} - 3)^{4} (2 x) d x

f (x) = e^{4 x}

\frac{s}{s ^{2} + 6 s + 25}

\frac{d}{d x} (\frac{3 - 2 x - x ^{2}}{x ^{2} - 6})

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