What is the solution for (2x-1)dx+(3x+7)dy=0 ?

The solution for (2x-1)dx+(3x+7)dy=0 is y=-2/3 x+17/9 ln(3x+7)+c_{1}

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(2x-1)dx+(3x+7)dy=0

(2 x - 1) d x + (3 x + 7) d y = 0

y = - \frac{2}{3} x + \frac{17}{9} ln (3 x + 7) + c_{1}

Solution steps

(2 x - 1) d x + (3 x + 7) d y = 0

Let

y

be the dependent variable. Divide by

d x

2 x - 1 + (3 x + 7) \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

2 x - 1 + (3 x + 7) y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = \frac{- 2 x + 1}{3 x + 7}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{- 2 x + 1}{3 x + 7} d x

\int \frac{- 2 x + 1}{3 x + 7} d x = - \frac{2}{3} x + \frac{17}{9} ln (3 x + 7) + c_{1}

y = - \frac{2}{3} x + \frac{17}{9} ln (3 x + 7) + c_{1}

What is the solution for (2x-1)dx+(3x+7)dy=0 ?
The solution for (2x-1)dx+(3x+7)dy=0 is y=-2/3 x+17/9 ln(3x+7)+c_{1}