# Finding an implicit solution to this differential equation

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1. Mar 16, 2019 at 10:28 PM

### echomochi

1. The problem statement, all variables and given/known data
Find an equation that defines IMPLICITLY the parameterized family of solutions y(x) of the differential equation:
5xy dy/dx = x2 + y2

2. Relevant equations
y=ux
dy/dx = u+xdu/dx
C as a constant of integration

3. The attempt at a solution
I saw a similar D.E. solved using the y=ux substitution, but using it for mine wasn't as clean, so halfway through I decided to use an integrating factor:

5xy dy/dx = x2 + y2
dy/dx = [ x2 + y2 ] / 5xy
u + x du/dx = [ x2 + u2x2 ] / 5x2u
u + x du/dx = [ 1 + u2 ] / 5u
Divide across by x (and switch the order of the LHS):

du/dx + u/x = [ 1 + u2 ] / 5ux​

Integrating factor: (the integral symbol won't show up in the superscript)

eINT[1/x]dx = eln(x) = x

du/dx ⋅ x + u/x ⋅ x = [ 1 + u2 ] ⋅ x / 5ux
d/dx [ux] = [ 1 + u2 ] / 5u dx
Integrate both sides gets us:

ux = x [ [ 1 + u2 ] / 5u ] + C
After this, it's just cleaning up so the u's are on one side. We get:

u - [ 1 + u2 ] / 5u = C/x
[4u2 - 1] / 5u = C/x
We can now resubstitute the u's for y/x:

[4[y/x]2 - 1] / 5[y/x] = C/x
4y/5x - x/5y = C/x​

After this, I'm pretty stuck. I don't know how to isolate y(x) on one side. If I messed up anywhere, or if there is a much easier way to do this problem that I am ignoring, please let me know! Looking forward to the suggestions.

Last edited: Mar 16, 2019 at 10:39 PM
2. Mar 16, 2019 at 10:59 PM

### LCKurtz

Stop right there. This is not a linear DE in u and x because you don't have just x's on the right side. You won't be able to integrate the right side with respect to x because u depends on x. The integrating factor method doesn't apply. But you should be able to separate variables and get an equation in the form $f(u)du = g(x)dx$ which you can hopefully integrate.
Lastly, when you finish, you are asked for an implicit solution so you aren't expected to solve for y.

3. Mar 16, 2019 at 11:38 PM

### echomochi

Okay, that makes sense.

Going back to before dividing by x:

u + x du/dx = [ 1 + u2 ] / 5u
Subtract the u:

x du/dx = [ 1 - 4u2 ] / 5u
Rearrange to get du and dx on each side:

[ 1/x ] dx = [ 5u / [ 1 - 4u2 ] ] du
Then integrating:

ln( x ) + C = [ -5/8 ] ln( 1-4u2 )​

Raise e to all terms to clear out the logarithms:

eln( x ) + eC = e[ -5/8 ] ln( 1-4u2 )
x + eC = 1 / [ 1-4u2 ]5/8

My professor would want me to rewrite eC as A, and we can rewrite this more:

1 = [ x + A ] [ 1-4u2 ]5/8

Is resubstituting u=y/x all that is left?

4. Mar 17, 2019 at 4:29 PM

### LCKurtz

Careful with your algebra there. $e^{\ln x + C}\ne e^{\ln x} + e^C$

5. Mar 18, 2019 at 10:59 AM

### echomochi

Ah, right. Thank you for your help!