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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(4, - 2), (- 8, - 1)

x \to 3 lim (25 - x^{2})

y = sin (tan (6 x))

\frac{\partial}{\partial y} (x y + ln (y) + 2 x^{2})

\frac{\partial}{\partial x} (cos^{3} (x) y - 1)

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