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#### rsoy

##### New member

- Mar 3, 2013

- 14

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

\(\displaystyle (x+y)dy=(y-x)dx\)

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

\(\displaystyle (x-y)dx+(x+y)dy=0\)

Next, I compute the partials:

\(\displaystyle \frac{\delta M}{\delta y}=-1\ne1=\frac{\delta N}{\delta x}\)

Thus, I have found the equation is not exact.

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

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

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

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

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

\(\displaystyle \frac{dy}{dx}=\frac{\frac{y}{x}-1}{\frac{y}{x}+1}\)

So, I know I need to make the substitution:

\(\displaystyle v=\frac{y}{x}\,\therefore\,y=vx\,\therefore\,\frac{dy}{dx}=v+x\frac{dv}{dx}\)

and so the ODE becomes:

\(\displaystyle v+x\frac{dv}{dx}=\frac{v-1}{v+1}\)

I know now that I must have a separable ODE:

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

Hence:

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

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

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

and now integrate?

\(\displaystyle (x+y)dy=(y-x)dx\)

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

\(\displaystyle (x-y)dx+(x+y)dy=0\)

Next, I compute the partials:

\(\displaystyle \frac{\delta M}{\delta y}=-1\ne1=\frac{\delta N}{\delta x}\)

Thus, I have found the equation is not exact.

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

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

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

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

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

\(\displaystyle \frac{dy}{dx}=\frac{\frac{y}{x}-1}{\frac{y}{x}+1}\)

So, I know I need to make the substitution:

\(\displaystyle v=\frac{y}{x}\,\therefore\,y=vx\,\therefore\,\frac{dy}{dx}=v+x\frac{dv}{dx}\)

and so the ODE becomes:

\(\displaystyle v+x\frac{dv}{dx}=\frac{v-1}{v+1}\)

I know now that I must have a separable ODE:

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

Hence:

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

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

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

and now integrate?

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