Question

Solve
xy*dx=(x^2+3y^2)*dy

Answer 1

Alright! Lets first rewrite the DE as:
\[\frac{dy}{dx}=\frac{xy}{x^2+3y^2}\]
Use a substitution \(y=xu\), where u is a function of x, which implies \(\frac{dy}{dx}=xu'+u\):
\[xu'+u=\frac{x(xu)}{x^2+3(x^2u^2)} \implies xu'=\frac{u}{1+3u^2}-u\]
\[\implies x\frac{du}{dx}=\frac{u-3u^3}{1+3u^2} \implies \frac{1+3u^2}{u-3u^3}du=\frac{dx}{x}.\]
Now, integrate both sides and then re-substitute for y.

Answer 2

I made a mistake on the last line, it should be:
\[x\frac{du}{dx}=\frac{-3u^3}{1+3u^2} \implies \frac{1+3u^2}{u^3}=-3\frac{dx}{x}.\]
Looks much better! :D

Answer 3

thank you very much!

Answer 4

You're welcome!

Answer 5

one more question:

i came to \[3* lnu ^{3} -2/u ^{2}= -3\ln x + \ln c\] so this is
\[3\ln y ^{3}/x ^{3}-2x ^{2}/y ^{2}= -3\ln x+\ln c\]
Now how do i get
\[x ^{2}=6y ^{2}\ln \left| Cy \right|\] which is the result

Answer 6

I gotta go now. I'll be back!

Answer 7

ok, i will wait for you:)

Answer 8

I can't really get to the result you gave, assuming it's a correct form.