What is the solution for (2x+3y-1)dx-4(x+1)dy=0 ?

The solution for (2x+3y-1)dx-4(x+1)dy=0 is y=(2(3x+4))/3+1/3+c_{1}(x+1)^{3/4}

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(2x+3y-1)dx-4(x+1)dy=0

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

Solution steps

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

Solve linear ODE:

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

\frac{e ^{- 2 s}}{s ( s - 2 )}

\frac{d}{d x} (- 2 e^{5 y} x - 2 sin (2 x))

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

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

6 x^{\frac{1}{2}}

What is the solution for (2x+3y-1)dx-4(x+1)dy=0 ?
The solution for (2x+3y-1)dx-4(x+1)dy=0 is y=(2(3x+4))/3+1/3+c_{1}(x+1)^{3/4}