# Thread: Differential Equation Substitution help.

1. ## Differential Equation Substitution help.

Problem goes as follows :

(y^2 + x^2)dx + (2xy)dy = 0

Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.

Once I subbed it all in, I was left with

u + du/dx = - (1 + u^2) / (2u)

Can someone help me proceed from here?

2. Originally Posted by JonathanEyoon
Problem goes as follows :

(y^2 + x^2)dx + (2xy)dy = 0

Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.

Once I subbed it all in, I was left with

u + du/dx = - (1 + u^2) / (2u)

Can someone help me proceed from here?
Sure, move the u from the left to the right hand side and simplfiy. Now you can separate.

3. Originally Posted by danny arrigo
Sure, move the u from the left to the right hand side and simplfiy. Now you can separate.

But that would net me two terms on the right side which isn't separable. ?

4. LOL nevermind. Can't believe I forgot to simplify . Thanks dood

5. Originally Posted by JonathanEyoon
But that would net me two terms on the right side which isn't separable. ?
But all the terms are in the variable v which you can simplify.

Edit - You got it!

6. Hate to ask of this but can anyone give me a final answer for me? This problem doesn't have an answer in the back of my textbook . I got it down to

(-1/3)(ln(3(y^2/x^2) + 1) = x + C

7. Originally Posted by JonathanEyoon
Problem goes as follows :

(y^2 + x^2)dx + (2xy)dy = 0

Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.

Once I subbed it all in, I was left with

u + du/dx = - (1 + u^2) / (2u)

Can someone help me proceed from here?
Another way (a little quicker) is to re-write

$
(y^2 + x^2)dx + (2xy)dy = 0
$

as

$d(x \,y^2) + x^2 dx = 0$

which integrates giving

$x \,y^2 + \frac{1}{3} x^3 = c$

8. Originally Posted by danny arrigo
Another way (a little quicker) is to re-write

$
(y^2 + x^2)dx + (2xy)dy = 0
$

as

$d(x \,y^2) + x^2 dx = 0$

which integrates giving

$x \,y^2 + \frac{1}{3} x^3 = c$

Could you show me how you simplified that? Also the way I did it, I was left with what I gave on the post above yours. I'm not sure where i'm going wrong here. Any help is appreciated.

9. Originally Posted by JonathanEyoon
Problem goes as follows :

(y^2 + x^2)dx + (2xy)dy = 0

Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.

Once I subbed it all in, I was left with

u + du/dx = - (1 + u^2) / (2u)

Can someone help me proceed from here?
Let's pick it up from here. When we isolate u' we get

$\frac{du}{dx} =- \frac{1+3u^2}{2u}$.

Separating gives

$\frac{2u}{1+3u^2}\,du = - \frac{dx}{x}$

integrating

$\frac{1}{3} \ln |\, 1 + 3u^2\, | = - \ln |\,x\,| + \frac{1}{3} \ln c$

multiply by 3, moving the $\ln |x|$ to the RHS and exponentiating

$(1+3u^2)x^3 = c$

which after substituting $u = \frac{y}{x}$ and simplifying gives the answer

$x^3 + 3xy^2 = c$.

Hope it helps.

10. That helped more than you can imagine . I really appreciate it!

11. Ok sorry but I have no idea where you got the

+ 1/3 lnC

Shouldn't it be just lnX + C?

12. Originally Posted by JonathanEyoon
Ok sorry but I have no idea where you got the

+ 1/3 lnC

Shouldn't it be just lnX + C?
Sure. So integrating

$\frac{1}{3} \ln |\, 1 + 3u^2\, | = - \ln |\,x\,| + c$

multiply by 3, moving the $\ln |x|$ to the RHS and exponentiating

$(1+3u^2)x^3 = e^{3c}$

which after substituting $u = \frac{y}{x}$ and simplifying gives the answer

$x^3 + 3xy^2 = e^{3c}$.

Since $c$ is a constant then $e^{3c}$ which we'll call $C$

$(1+3u^2)x^3 = C$ same answer. So why did I use $\frac{1}{3} \ln c$. I saw that I was going to multiply by 3 and exponentiate.

13. I hate to ask this but could you show me in more detailed steps or least tell me in more detail how you went from after integrating and then getting

(1 + 3u^2)x^3.

I get that e^3c is just another constant so that'll equal C later but i'm at an absolute loss as to how you got what I typed above. If you multiply both sides by 3, you should end up with

ln(1+3u^2) = -3lnx + 3c

If you take e to the both sides, you should get

(1 + 3u^2) = - 1/x^3 + e^3c

now I can get this far... and I just can't see how you're able to get

(1 + 3u^2)x^3 on the left hand side and e^3c on the right hand side.

I mean supposing you DID multiply both sides by x^3, THEN I can see you getting that for the left side but on the right side you'll end up with -1 + (x^3)(e^3c)

.......If you don't mind could you show it to me in clearer steps? I've been racking my brains for over a day on this problem and I can't seem to get it to how you're able to get it to. THanks.

14. Originally Posted by JonathanEyoon
I hate to ask this but could you show me in more detailed steps or least tell me in more detail how you went from after integrating and then getting

(1 + 3u^2)x^3.

I get that e^3c is just another constant so that'll equal C later but i'm at an absolute loss as to how you got what I typed above. If you multiply both sides by 3, you should end up with

ln(1+3u^2) = -3lnx + 3c

If you take e to the both sides, you should get

(1 + 3u^2) = - 1/x^3 + e^3c Mr F says: No. See below.

now I can get this far... and I just can't see how you're able to get

(1 + 3u^2)x^3 on the left hand side and e^3c on the right hand side.

I mean supposing you DID multiply both sides by x^3, THEN I can see you getting that for the left side but on the right side you'll end up with -1 + (x^3)(e^3c)

.......If you don't mind could you show it to me in clearer steps? I've been racking my brains for over a day on this problem and I can't seem to get it to how you're able to get it to. THanks.
$e^{\ln |x|^{-3} + 3C} = e^{\ln |x|^{-3}} \cdot e^{3C} = \frac{1}{x^3} \cdot e^{3C}$ NOT $\frac{1}{x^3} {\color{red}+} e^{3C}$.