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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = x^{3} - x, (- 1, 0)

f (x) = \frac{2}{x}, at x = - 1

\frac{1}{x} d y - \frac{y}{x ^{2}} d x = 0

n = 1 \sum \infty (\frac{2}{7})^{n}

y^{^{'}} = \frac{- 2 x}{y}

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