# Thread: Solve the following ODE .. #6

1. ## Solve the following ODE .. #6

Hello .

Question:

Solve the following ODE :
$y^2dx + (x^2+3y+4y^2)dy = 0$

Solution:

It is not Linear or Bernoulli in x or y.
It is not Exact.
It is not Separable.
It can not be converted it to an exact equation by the integrating factor.

I tried to substitute $t=y^2 \implies dy=\dfrac{dt}{2\sqrt{t}}$ to get:

$2t^{3/2}dx+(x^2+3\sqrt{t}+4t)dt=0$

Which is ,again, I do not know what is its type!

Maybe it can be solved by inspection, but I can not figure it out.
Any help?

2. After some algebraic manipulations the DE becomes...

$\displaystyle \frac{dx}{dy} = -\frac{x^{2}}{y^{2}} - \frac{3}{y} - 4$ (1)

... where y is the independent variable. The (1) is a Riccati's equation, the solution of which is in most cases a little hard...

Kind regards

$\chi$ $\sigma$

3. Mathematica obtains the following solution of the Ricatti equation that chisigma mentioned:

Let

$f_{1}(y)=\sqrt{y}J_{i\sqrt{15}}\left(\dfrac{2\sqrt {3}}{\sqrt{y}}\right)+\sqrt{3}J_{-1+i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\rig ht),$

$f_{2}(y)=-\sqrt{3}J_{1+i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\s qrt{y}}\right)+\sqrt{y}J_{-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right )C_{1},$

$f_{3}(y)=\sqrt{3}J_{-1-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right )C_{1}-\sqrt{3}J_{1-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right )C_{1}.$

Then

$x(y)=-\dfrac{\sqrt{y}\left(f_{1}(y)+f_{2}(y)+f_{3}(y)\ri ght)}{2\left(J_{i\sqrt{15}}\left(\dfrac{2\sqrt{3}} {\sqrt{y}}\right)+J_{-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right )C_{1}\right)}.$

WolframAlpha gives this.

4. Just to add a bit to this thread. If you let

$x = f(y) \dfrac{u'}{u}$

in Chisigma equation (1) and pick $f(y)$ accordingly, you can linearize your ODE.

5. Just to add (part 2). Under the change of variables

$x = \dfrac{12}{X^2},\;\;\; y = \dfrac{XY}{\sqrt{6}}$

this linear ODE becomes

$X^2 Y'' + X Y' +\left(X^2 + 15\right)Y = 0$

which explains the odd Bessel function solutions that Adrian picked up.

6. This is too advanced for me.
Thanks.