Finding an implicit solution to this differential equation

[echomochi]

Homework Statement

Find an equation that defines IMPLICITLY the parameterized family of solutions y(x) of the differential equation:
5xy dy/dx = x² + y²

Homework Equations

y=ux
dy/dx = u+xdu/dx
C as a constant of integration

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)

e^INT[1/x]dx = e^ln(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.

 

[LCKurtz]

Homework Statement

Find an equation that defines IMPLICITLY the parameterized family of solutions y(x) of the differential equation:
5xy dy/dx = x² + y²

Homework Equations

y=ux
dy/dx = u+xdu/dx
C as a constant of integration

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)

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.

 

[echomochi]

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.

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:

e^{ln( x )} + e^C = e^{[ -5/8 ] ln( 1-4u2 )}
x + e^C = 1 / [ 1-4u² ]^5/8​

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

1 = [ x + A ] [ 1-4u² ]^5/8​

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

 

[LCKurtz]

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:

e^{ln( x )} + e^C = e^{[ -5/8 ] ln( 1-4u2 )}
x + e^C = 1 / [ 1-4u² ]^5/8​

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

 

[echomochi]

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

Ah, right. Thank you for your help!

 

[epenguin]

Ah, right. Thank you for your help!

The best thanks would be to give your solution. Can you imagine there is much satisfaction to helpers if they give help to guys and don't hear what the effect is?
Also they naturally usually help with only the stage you are at. So by not replying you are missing what they might have said about the solution, e.g. connection with other things.Or in this case the fact that the job is only half finished and you have only have half benefited from the exercise when you have written the solution. ( I've noticed that solutions that satisfy students are often not as good as they imagine.)

You should always check the solution to a differential equation by differentiating your solution and checking that that works. Especially in a case as tricky and unexpected as this one. And you should check any salient features of solutions that the d.e. is already telling you against what your solution equation says. And at least outline here what you find. Preferably a picture of solution curves. Time spent this way will benefit you more in insights than doing an extra problem of the same kind.