What is the solution for (dy)/(dt)=t^2e^{-4t}-4y ?

The solution for (dy)/(dt)=t^2e^{-4t}-4y is y=(t^3)/(3e^{4t)}+(c_{1})/(e^{4t)}

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(dy)/(dt)=t^2e^{-4t}-4y

\frac{d y}{d t} = t^{2} e^{- 4 t} - 4 y

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

Solution steps

\frac{d y}{d 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}}

7 x y^{3} + 5 x y = 12, at x = 1

(2 x + 3) + (2 y - 2) y^{^{'}} = 0

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

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

x \to 0 lim (\frac{3 sin ( x )}{x})

What is the solution for (dy)/(dt)=t^2e^{-4t}-4y ?
The solution for (dy)/(dt)=t^2e^{-4t}-4y is y=(t^3)/(3e^{4t)}+(c_{1})/(e^{4t)}