Solving Exact Differential Equations: An Attempt at a Solution

[CINA]

Homework Statement

\frac{dy}{dx}=\frac{x+3y}{3x+y}

Homework Equations

M(x,y)dx+N(x,y)dy=0

Exact equation (I think)

The Attempt at a Solution

Ok, so I try to put it in the form of the of an exact equation, like above, but I end up with

-(x+3y)dx+(3x+y)dy=0

This results in the partial of M with respect to y being -3 and the Partial of N with respect to x being 3, which is not equal!

I tried investigating to see if there is an integrating factor but \frac{M_{y}-N_{x}}{N} and \frac{N_{x}-M_{y}}{M} are not functions of just one variable. What am I doing wrong? Different method? Algebra mistake?

Thanks

 

[rock.freak667]

Try using a substitution of y=vx.

 

[CINA]

ok, so

y=vx

and

dy=vdx+xdv



(-x-3vx)dx+(3x+vx)(vdx+xdv)=0 So far so good...



-xdx+xv^{2}dx+3x^{2}dv+vx^{2}dv=0 Collect like terms...



x(v^{2}-1)dx+x^{2}(3+v)dv=0 Factor out like terms...



\frac{1}{x}dx=\frac{-(3+v)}{(v^{2}-1)}dv Seperate variables...is this right? I tend to make a lot of simple errors, no matter how many times I check my work...

Thanks!

 

[rock.freak667]

That should be correct. Now just split the function in 'v' into partial fractions and integrate.

 

[CINA]

Ok so...

\frac{1}{x}=\frac{1}{v+1}-\frac{2}{v-1}

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

ln(x)=ln(v+1)-2ln(v-1)+ln(c) Ok I don't really know what to do next, here's a try...

ln(x)=ln(c\frac{v+1}{(v-1)^{2}})

Is this right? I'm not very good with natural logs, hopefully I didn't make a mistake...

Next step would be exponentiate both sides? (Assuming what I just did was right)

 

[rock.freak667]

Well you don't really have to exponentiate both sides since if lnx=lny then x=y.

So essentially, you can just remove the 'ln' and then replace 'v'.