What is the solution for -4x+(2xdy)/(dx)=(2dy)/(dx)+3x^2+2 ?

The solution for -4x+(2xdy)/(dx)=(2dy)/(dx)+3x^2+2 is y= 1/4 (3x^2+14x+18ln(x-1)-17)+c_{1}

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-4x+(2xdy)/(dx)=(2dy)/(dx)+3x^2+2

- 4 x + \frac{2 x d y}{d x} = \frac{2 d y}{d x} + 3 x^{2} + 2

y = \frac{1}{4} (3 x^{2} + 14 x + 18 ln (x - 1) - 17) + c_{1}

Solution steps

- 4 x + \frac{2 x d y}{d x} = \frac{2 d y}{d x} + 3 x^{2} + 2

Let

y

be the dependent variable. Divide by

d x

- 4 x + \frac{2 x \frac{d y}{d x}}{1} = \frac{2 \cdot \frac{d y}{d x}}{1} + 3 x^{2} + 2

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

y^{'} (x) : y^{'} (x) = \frac{3 x ^{2} + 2 + 4 x}{2 ( - 1 + x )}

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

then

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

y = \int \frac{3 x ^{2} + 2 + 4 x}{2 ( - 1 + x )} d x

\int \frac{3 x ^{2} + 2 + 4 x}{2 ( - 1 + x )} d x = \frac{1}{4} (3 x^{2} + 14 x + 18 ln (x - 1) - 17) + c_{1}

y = \frac{1}{4} (3 x^{2} + 14 x + 18 ln (x - 1) - 17) + c_{1}

t an g e n t y = \frac{1}{( 1 + x ^{2} ) , at ( 1 , \frac{1}{2} )}

\int_{0}^{1} 1 + 7 x d x

\frac{\partial}{\partial x} (x - y^{2})

d er i v a t i v e 4 x^{- 2}

\frac{\partial}{\partial x} (\frac{y}{y + 4})

What is the solution for -4x+(2xdy)/(dx)=(2dy)/(dx)+3x^2+2 ?
The solution for -4x+(2xdy)/(dx)=(2dy)/(dx)+3x^2+2 is y= 1/4 (3x^2+14x+18ln(x-1)-17)+c_{1}