What is the solution for 3y^'+4y=e^{-x} ?

The solution for 3y^'+4y=e^{-x} is y= 1/(e^x)+(c_{1}}{e^{\frac{4x)/3}}

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3y^'+4y=e^{-x}

3 y^{^{'}} + 4 y = e^{- x}

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{4 x}{3}}}

Solution steps

3 y^{'} (x) + 4 y = e^{- x}

Solve linear ODE:

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{4 x}{3}}}

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{4 x}{3}}}

tan (5 - sin (2 t))

\frac{d y}{d x} + 3 x^{2} y = 2 x e^{- x^{3}}

y^{^{'}} + 2 x^{- 1} y = x^{- 1} sin (2 x)

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

\int \frac{1}{2 ^{x} + 3} d x

What is the solution for 3y^'+4y=e^{-x} ?
The solution for 3y^'+4y=e^{-x} is y= 1/(e^x)+(c_{1}}{e^{\frac{4x)/3}}