## RolyPoly 2 years ago Solve: $(2xy)dx + (y^2-3x^2)dy=0$

1. sirm3d

$\frac{ \partial }{ \partial y }(2xy)=2x,\frac{ \partial }{ \partial x }(y^2-3x^2)=-6x$$\frac{ -(\frac{ \partial M }{ \partial y } - \frac{ \partial N }{ \partial x })}{ 2xy }=-\frac{ 4 }{ y }$the integrating factor that will make the DE exact is $\large e^{\int\limits_{}^{}\frac{ -4 }{ y }dy}=\frac{ 1 }{ y^4 }$

2. sirm3d

if you have not reached this part of the course, an alternate solution is $\large x=vy$ and $\large dx = vdy + ydv$ where the original DE is now separable in v and y.

3. sirm3d

have you solved this problem @RolyPoly ?

4. RolyPoly

I'm on the way solving it.. Sorry, please give me some more time...

5. RolyPoly

I'm not quite sure if I have learnt this way before. I have no memory of seeing this, but it appears in my book..

6. sirm3d

homogeneous function, doesn't it ring a bell?

7. RolyPoly

Nope :(

8. sirm3d

my comment on your attached file: $\huge F=\frac{ x^2 }{ y^3 }+g_1(y)$ NOT $\huge F_x$ and $\huge F=-\frac{1}{y}+\frac{x^2}{y^3}+g_2(x)$ instead of $\huge F_y$

9. RolyPoly

:O That's how my teacher writes... $$F_x$$ means integrating w.r.t. x, I think :S

10. sirm3d

about that alternate solution $\large \deg(2xy)=2;\deg(y^2)=2;\deg(-3x^2)=2$ the degrees of the terms are identical. a substitution $\large y = ux \text{ or }x=vy$ will transform the DE that is solvable by separation

11. sirm3d

taking $x=vy \text{ and } dx=vdy+ydv$$\large 2(vy)y(vdy+ydv)+(y^2-v^2y^2)dy=0$

12. RolyPoly

I suppose it is separable, but somehow, I can't separate it. I guess I have to check my calculations again first :/

13. sirm3d

collecting terms of dy and dv, $\large (2v^2y^2+y^2-v^2y^2)dy+2vy^3dv=0$or$\large y^2(v^2+1)dy+2vy^3dv=0$divide by $\large y^3(v^2+1)$ and the DE becomes $\large \frac{dy}{y}+\frac{2vdv}{v^2+1}=0$

14. RolyPoly

How come... you can get it but I can't... :'(

15. sirm3d

expand after replacing x by vy and dx by vdy + ydv collect terms of dy and dv, simplify and factor.

16. RolyPoly

I tried w = y/x but it doesn't seem to be working for me :(

17. RolyPoly

Sorry, I got to go now.. I'd keep trying until I got the answer. Thanks for your help!!

18. RolyPoly

@sirm3d You're right about the F and Fx thing! Sorry for my earlier mistakes!!! When I tried w=y/x, I got something quite ugly :S