What is the solution for (dy)/(dx)=(3x-9)/(2x+8) ?

The solution for (dy)/(dx)=(3x-9)/(2x+8) is y= 3/2 (x-7ln(2(x+4))+4)+c_{1}

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(dy)/(dx)=(3x-9)/(2x+8)

\frac{d y}{d x} = \frac{3 x - 9}{2 x + 8}

y = \frac{3}{2} (x - 7 ln (2 (x + 4)) + 4) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{3 x - 9}{2 x + 8}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{3 x - 9}{2 x + 8}

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

then

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

y = \int \frac{3 x - 9}{2 x + 8} d x

\int \frac{3 x - 9}{2 x + 8} d x = \frac{3}{2} (x - 7 ln (2 (x + 4)) + 4) + c_{1}

y = \frac{3}{2} (x - 7 ln (2 (x + 4)) + 4) + c_{1}

x \to 0 lim (\frac{1 - e ^{- 2 x}}{x})

\frac{d y}{d x} = 2 + e^{- 2 x + y + 1}

y = 2 x - 3 x^{2}

\frac{d}{d x} (cos^{3} (5 x))

\int_{0}^{1} \frac{2 t}{( 2 - t ) ^{2}} d t

What is the solution for (dy)/(dx)=(3x-9)/(2x+8) ?
The solution for (dy)/(dx)=(3x-9)/(2x+8) is y= 3/2 (x-7ln(2(x+4))+4)+c_{1}