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#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

we have xdx + (2x+y)*dy=0
∂P/ ∂y is not equal to ∂Q/ ∂x so I have to find an integrating factor.
I write this as 0 +x*dμ/dy=2μ .............
x= 2μ/∂μ/∂y

2/x =∂μ/∂y/μ

lnμ=2lnx so μ=x^2...
I write the initial equation x^2*xdx+x^2(2x+y)dy=0 but still the partial derivatives arent equal where am I wrong :/
Here is a link to the question:

Differential equations math question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper

The equation has no integrating factor $\mu$ depending only on $x$, if so consider $\mu Pdx+\mu Qdy=0$. Then, $$(\mu P)_y=(\mu Q)_x\Leftrightarrow 0=\mu'(2x+y)+2\mu \Leftrightarrow \frac{\mu'}{\mu}=-\frac{2}{2x+y}\ne \mu(x)$$ and we get a contradiction. The given equation is homogeneous, so using $y=vx$ you'll get a separating variables equation.

#### Prove It

##### Well-known member
MHB Math Helper
Here is the question:

Here is a link to the question:

Differential equations math question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
\displaystyle \displaystyle \begin{align*} x\,dx + \left( 2x + y \right) \, dy &= 0 \\ x\,dx &= -\,\left( 2x + y \right)\, dy \\ \frac{dx}{dy} &= -\frac{2x + y}{x} \end{align*}

Now a substitution of the form \displaystyle \displaystyle \begin{align*} x = v\,y \implies \frac{dx}{dy} = v + y\,\frac{dv}{dy} \end{align*} is appropriate...

\displaystyle \displaystyle \begin{align*} \frac{dx}{dy} &= -\frac{2x+y}{x} \\ v+ y\,\frac{dv}{dy} &= -\frac{2v\,y + y}{v\,y} \\ v + y\,\frac{dv}{dy} &= -\frac{2v + 1}{v} \\ y\,\frac{dv}{dy} &= -\frac{2v+1}{v}-v \\ y\,\frac{dv}{dy} &= -\frac{v^2 + 2v + 1}{v} \\ \frac{v}{ \left( v + 1 \right) ^2} \, \frac{dv}{dy} &= -\frac{1}{y} \\ \int{ \frac{v}{ \left( v + 1 \right) ^2} \, \frac{dv}{dy} \, dy} &= \int{-\frac{1}{y}\,dy} \\ \int{ \frac{v}{ \left( v + 1 \right)^2} \, dv} &= -\ln{|y|} +C_1 \\ \int{ \frac{u - 1}{u^2}\,du} &= -\ln{|y|} +C_1 \textrm{ after making the substitution } u = v + 1 \implies du = dv \\ \int{u^{-1} - u^{-2}\,du} &= -\ln{|y|} + C \\ \ln{|u|} + u^{-1} + C_2 &= -\ln{|y|} + C_1 \\ \ln{|v+1|} + \frac{1}{v + 1} &= -\ln{|y|} + C \textrm{ where } C = C_1 - C_2 \\ \ln{ \left| \frac{x}{y} + 1 \right| } + \frac{1}{\frac{x}{y} + 1} &= -\ln{|y|} + C \\ \ln{ \left| \frac{x + y}{y} \right| }+ \frac{y}{x + y}&= -\ln{|y|} + C \\ \ln{ |x + y|} -\ln{|y|} + \frac{y}{x +y} &= -\ln{|y|} + C \\ \ln{|x+y|}+ \frac{y}{x+y} &= C \end{align*}

It is not possible to get either variable explicitly in terms of the other.