What is the solution for yy^'+3y=8 ?

The solution for yy^'+3y=8 is y=(8\W(-1/(8e^{(9t+c_{1)+8)/8}}))/(3)+8/3

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yy^'+3y=8

y y^{^{'}} + 3 y = 8

y = \frac{8 W ( - \frac{1}{8 e ^{\frac{9 t + c _{1} + 8}{8}}} )}{3} + \frac{8}{3}

Solution steps

y y^{'} (t) + 3 y = 8

Solve separable ODE:

y = \frac{8 W ( - \frac{1}{8 e ^{\frac{9 t + c _{1} + 8}{8}}} )}{3} + \frac{8}{3}

y = \frac{8 W ( - \frac{1}{8 e ^{\frac{9 t + c _{1} + 8}{8}}} )}{3} + \frac{8}{3}

\frac{d x}{d t} = \frac{a}{x} + b, x (0) = 0

\frac{d y}{d x} y = e^{x^{2}}

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{4} e^{- x} d y = 0

x^{^{'}} + c x = 0

\frac{d y}{d x} = 15 x y

What is the solution for yy^'+3y=8 ?
The solution for yy^'+3y=8 is y=(8\W(-1/(8e^{(9t+c_{1)+8)/8}}))/(3)+8/3