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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 2})

\int_{- 2}^{- 1} \frac{1}{x} d x

\int_{- 1}^{1} x^{2} + 1 d x

f (x) = x^{2} - 2, at x = a

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

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