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}}

\int_{- 3}^{4} ((y + 12) - (y^{2})) d y

y^{^{''}} + 7 y^{^{'}} + 6 y = 0

f (x) = 2 (x)^{3} ln (x)

x \to - 2 lim (\frac{x ^{2} + 4}{x + 2})

y = - x^{2} + 10, y = - 1.5 x + 7

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)}