# Thread: [SOLVED] First order non-linear diffeq from G.F.Simmons

1. ## [SOLVED] First order non-linear diffeq from G.F.Simmons

Solve this differential equation:
$\displaystyle x dy = (y + x^2 + 9 y^2) dx$

I suspect that the trick is to extract the factor:
$\displaystyle \frac {3 y dx - x dy}{x^2 + 9 y^2}$

so as to equate the LHS to $\displaystyle d \left({\tan^{-1} {\frac {3y} x}}\right)$

because the solution is given as:
$\displaystyle \tan^{-1} \frac {3y} x = 3x + c$

and I can't work out how to split it so there's something sensible on the RHS.

Sorry I've been away for so long, I've been busy on ProofWiki.

2. Originally Posted by Matt Westwood
Solve this differential equation:
$\displaystyle x dy = (y + x^2 + 9 y^2) dx$

I suspect that the trick is to extract the factor:
$\displaystyle \frac {3 y dx - x dy}{x^2 + 9 y^2}$

so as to equate the LHS to $\displaystyle d \left({\tan^{-1} {\frac {3y} x}}\right)$

because the solution is given as:
$\displaystyle \tan^{-1} \frac {3y} x = 3x + c$

and I can't work out how to split it so there's something sensible on the RHS.

Sorry I've been away for so long, I've been busy on ProofWiki.
The "trick" here is to apply a substitution:

Divide both sides of the equation by $\displaystyle x^2\,dx$ to get $\displaystyle \frac{1}{x}\frac{\,dy}{\,dx}=\frac{1}{x}\left(\fra c{y}{x}\right)+1+9\left(\frac{y}{x}\right)^2$

Now apply the substitution $\displaystyle y=ux$.

This implies then that $\displaystyle \frac{\,dy}{\,dx}=u+x\frac{\,du}{\,dx}$

Now substitute everything into the DE to get $\displaystyle \frac{1}{x}\left(u+x\frac{\,du}{\,dx}\right)=\frac {1}{x}u+1+9u^2\implies \frac{\,du}{\,dx}=1+9u^2$

Now it becomes separable and we end up with $\displaystyle \frac{\,du}{9u^2+1}=\,dx\implies \tfrac{1}{3}\tan^{-1}\left(3u\right)=x+C\implies\tan^{-1}\left(3u\right)=3x+K$

Now back-substitute to get $\displaystyle \tan^{-1}\left(\frac{3y}{x}\right)=3x+K$

Does this make sense?

Good to see you back!

3. Yep, it's simple, gets the job done. Nice one.

Interestingly it's in the section of the book (George F. Simmons: "Differential Equations with Applications and Historical Notes, 1972) in which the emphasis is on first order, non-linear, non-exact, non-homogeneous diffeqs where the object of the exercise is to find an integrating factor. This last section of problems seems to concern diffeqs where the i.f. is not a factor of x, y, x+y or xy and require the user to recognise differential formulas in the forms d(x/y) = (y dx - x dy)/y^2, d(xy) = x dy + y dx, d(x^2 + y^2) = 2(x dx + y dy) and so on. One of the cited options is for d(tan^-1 (x/y)) = (y dx - x dy) / (x^2 + y^2) and I can see that one is probably being guided towards finding an i.f. in that particular form.

Not easy, but then this is a challenging book and I'm way out of practice on diffeqs and so on - spent far too long in the gauzy reaches of abstract alg.

Good to be back.

4. Right, I've got it.

If you let $\displaystyle z = \tan^{-1} \left({\frac {3y} x}\right)$ and calculate $\displaystyle \frac {\partial z} {\partial x}$ and $\displaystyle \frac {\partial z} {\partial y}$ and sum them, you get:

$\displaystyle d z = \frac {3 x dy - 3 y dx} {x^2 + 9 y^2}$
$\displaystyle d \left({\tan^{-1} \left({\frac {3y} x}\right)}\right) = 3 d x$