MHB Show whether the given ODE is exact, then solve

[r-soy]

I am given the following ODE, and the instructions are to show whether it is exact or not, and then solve:

$$(x+y)dy=(y-x)dx$$

My first step, is to put the equation in the form $$M(x,y)\,dx+N(x,y)\,dy=0$$:

$$(x-y)dx+(x+y)dy=0$$

Next, I compute the partials:

$$\frac{\delta M}{\delta y}=-1\ne1=\frac{\delta N}{\delta x}$$

Thus, I have found the equation is not exact.

Next, I want to find an integrating factor. First I consider:

$$\frac{\frac{\delta M}{\delta y}-\frac{\delta N}{\delta x}}{N(x,y)}=-\frac{2}{x+y}$$

Since this is not a function of just $x$, I next consider:

$$\frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M(x,y)}=\frac{2}{x-y}$$

Since this is not a function of just $y$, I next observe that the given ODE is homogeneous:

$$\frac{dy}{dx}=\frac{\frac{y}{x}-1}{\frac{y}{x}+1}$$

So, I know I need to make the substitution:

$$v=\frac{y}{x}\,\therefore\,y=vx\,\therefore\,\frac{dy}{dx}=v+x\frac{dv}{dx}$$

and so the ODE becomes:

$$v+x\frac{dv}{dx}=\frac{v-1}{v+1}$$

I know now that I must have a separable ODE:

$$x\frac{dv}{dx}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}=-\frac{v^2+1}{v+1}$$

Hence:

$$-\frac{v+1}{v^2+1}\,dv=\frac{1}{x}\,dx$$

At this point, I am unsure how to proceed. Should I write:

$$\left(\frac{1}{2}\cdot\frac{2v}{v^2+1}+\frac{1}{v^2+1} \right)dv=-\frac{1}{x}\,dx$$

and now integrate?

 

[Prove It]

http://www14.0zz0.com/2013/03/14/07/459161749.jpeg

[math]\displaystyle \begin{align*} \left( \frac{x-y}{x^2 + y^2} \right) dx + \left( \frac{x + y}{x^2 + y^2} \right)dy &= 0 \\ \left( \frac{y - x}{x^2 + y^2}\right) dx &= \left( \frac{x + y}{x^2 + y^2} \right) dy \\ \frac{y - x}{x + y} &= \frac{dy}{dx} \end{align*}[/math]

Now make the substitution [math]\displaystyle \begin{align*} v = \frac{y}{x} \implies y = x\,v \implies \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*}[/math] and the DE becomes

[math]\displaystyle \begin{align*} \frac{x\,v - x}{x + x\,v} &= v + x\,\frac{dv}{dx} \\ \frac{x\,v - x - v\left( x + x\,v \right)}{x + x\,v} &= x\,\frac{dv}{dx} \\ \frac{x\,v - x - x\,v - x\,v^2 }{x + x\,v} &= x\,\frac{dv}{dx} \\ \frac{-x - x\,v^2}{x + x\,v} &= x\,\frac{dv}{dx} \\ \frac{-1 - v^2}{1 + v} &= x\,\frac{dv}{dx} \\ \frac{1}{x} &= -\left( \frac{1 + v}{1 + v^2} \right) \frac{dv}{dx} \end{align*}[/math]

Now you can integrate both sides.

 

[MarkFL]

I want to apologize here for removing the image originally posted and causing confusion and wasted time on the part of two of our helpers. I consider the time of our helpers to be very valuable, and hate to have squandered it.(Blush)

I had advised the OP that a link to an image of hand-written work probably would not get help, and boy was I was ever wrong. I should know by now our helpers will go the extra mile! I thought I was helping, but instead hindered. I am very sorry for this.

 

[alyafey22]

I want to apologize here for removing the image originally posted and causing confusion and wasted time on the part of two of our helpers. I consider the time of our helpers to be very valuable, and hate to have squandered it.(Blush)

I had advised the OP that a link to an image of hand-written work probably would not get help, and boy was I was ever wrong. I should know by now our helpers will go the extra mile! I thought I was helping, but instead hindered. I am very sorry for this.

I think you posted your solution in the question itself !

 

[MarkFL]

I posted what the OP can do on his own, and wanted to leave the last steps so he would actually have something to ask. I did not handle this well at all. (Giggle)

 

[r-soy]

In fact , to solve only one line

I don't now why you used y=vx

relly I am confued now ..

my answer was correct but last step for solve

 

[Prove It]

No harm done. I wonder if the OP can go any further now...

- - - Updated - - -

In fact , to solve only one line

I don't now why you used y=vx

relly I am confued now ..

my answer was correct but last step for solve

It's a standard method. If your derivative is equal to some function which is a fractional combination of x and y, a substitution of the form v = y/x, or y = xv, is appropriate. You can see how it turned the DE into a separable one. You can now get a function v in terms of x through integrating, which you can then use to get y in terms of x.

 

[I like Serena]

I want to apologize here for removing the image originally posted and causing confusion and wasted time on the part of two of our helpers. I consider the time of our helpers to be very valuable, and hate to have squandered it.(Blush)

I had advised the OP that a link to an image of hand-written work probably would not get help, and boy was I was ever wrong. I should know by now our helpers will go the extra mile! I thought I was helping, but instead hindered. I am very sorry for this.

I was pretty impressed on multiple counts.

I had started my reply and since there was only a link to an external image, I kept it short.
Then, when I clicked the submit button, the quoted OP in my post suddenly changed, metamorphosing in a long well formatted series of formulas.
When I looked at my answer, it suddenly looked kind of crappy. ;)
I quickly deleted it before anyone could see.

It was a fun experience!