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

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

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2y^'+4y=2te^{-2t}

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

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

\int \frac{3 - x}{3 + x} d x

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\int \frac{3}{x 9 - ( ln ( x ) ) ^{2}} d x

f (x) = x^{2} - 10 x + 2, at x = 0

x \to 0 lim (csc (x) - cot (x))

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