What is the solution for 3y^'+7y=4e^{-x} ?

The solution for 3y^'+7y=4e^{-x} is y= 1/(e^x)+(c_{1}}{e^{\frac{7x)/3}}

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3y^'+7y=4e^{-x}

3 y^{^{'}} + 7 y = 4 e^{- x}

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{7 x}{3}}}

Solution steps

3 y^{'} (x) + 7 y = 4 e^{- x}

Solve linear ODE:

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{7 x}{3}}}

y = \frac{1}{e ^{x}} + \frac{c _{1}}{e ^{\frac{7 x}{3}}}

y = ln (6 (ln (x))^{8})

\frac{d y}{d x} = - 2 x tan (y)

x \to 8 - lim (x - 8)

f (x) = (e^{2 x} - 4)^{2}, (0, 9)

x \to 0 lim (\frac{2 cos ( x ) - 2}{x})

What is the solution for 3y^'+7y=4e^{-x} ?
The solution for 3y^'+7y=4e^{-x} is y= 1/(e^x)+(c_{1}}{e^{\frac{7x)/3}}