What is the solution for y^'+2y=xe^{-2x} ?

The solution for y^'+2y=xe^{-2x} is y=(x^2)/(2e^{2x)}+(c_{1})/(e^{2x)}

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y^'+2y=xe^{-2x}

y^{^{'}} + 2 y = x e^{- 2 x}

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

Solution steps

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

Solve linear ODE:

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

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

(4 x - 2)^{3} (x^{2} + 1)^{2}

f (x) = \frac{45}{0.01 x ^{2} + 1}

cos^{2} (5 t)

x^{^{''''}} + 8 x^{^{''}} + 16 x = 0

y = x^{2} - 33, y = 7 - 3 x

What is the solution for y^'+2y=xe^{-2x} ?
The solution for y^'+2y=xe^{-2x} is y=(x^2)/(2e^{2x)}+(c_{1})/(e^{2x)}