What is the solution for 4(3x+y-2)dx-(3x+y)dy=0 ?

The solution for 4(3x+y-2)dx-(3x+y)dy=0 is y=(8\W(1/(8e^{(-49x-c_{1)-6)/8}})-6)/(7)-3x+2

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4(3x+y-2)dx-(3x+y)dy=0

4 (3 x + y - 2) d x - (3 x + y) d y = 0

y = \frac{8 W ( \frac{1}{8 e ^{\frac{- 49 x - c _{1} - 6}{8}}} ) - 6}{7} - 3 x + 2

Solution steps

4 (3 x + y - 2) d x - (3 x + y) d y = 0

Solve using substitution:

y = \frac{8 W ( \frac{1}{8 e ^{\frac{- 49 x - c _{1} - 6}{8}}} ) - 6}{7} - 3 x + 2

y = \frac{8 W ( \frac{1}{8 e ^{\frac{- 49 x - c _{1} - 6}{8}}} ) - 6}{7} - 3 x + 2

\frac{d}{d x} (arcsin (x) - 1 - x^{2})

y^{^{'}} = \frac{- y}{40 - t}

\int 2 (1 + tan^{2} (a)) d a

x \to 0 lim (sec^{2} (x))

\frac{d}{d x} (380000 x - \frac{1}{32} x^{2})

What is the solution for 4(3x+y-2)dx-(3x+y)dy=0 ?
The solution for 4(3x+y-2)dx-(3x+y)dy=0 is y=(8\W(1/(8e^{(-49x-c_{1)-6)/8}})-6)/(7)-3x+2