What is the solution for y^'+3y=x+e^{-2x} ?

The solution for y^'+3y=x+e^{-2x} is y=(3x-1)/9+1/(e^{2x)}+(c_{1})/(e^{3x)}

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y^'+3y=x+e^{-2x}

y^{^{'}} + 3 y = x + e^{- 2 x}

y = \frac{3 x - 1}{9} + \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

Solution steps

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

Solve linear ODE:

y = \frac{3 x - 1}{9} + \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

y = \frac{3 x - 1}{9} + \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

x \to \infty lim (x - 2^{x})

y^{^{''}} + 8 y^{^{'}} + 9 y = 0

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

x \to 1 lim (ln (1))

What is the solution for y^'+3y=x+e^{-2x} ?
The solution for y^'+3y=x+e^{-2x} is y=(3x-1)/9+1/(e^{2x)}+(c_{1})/(e^{3x)}