Thread: ODE help

Hi..i got this equation (2xy^2 - x)dx + (x^2y - y)dy=0 ;I know its not exact as i ve partially differentiated it to get M=4xy(wrt y) and N= 2xy(wrt x) After multipying it by h(y)..i got h= i/sqrt(2y^2-1)..after that i cant prove that its exact..can someone please give me a lead on this..

the part is h=1/sqrt(2y^2-1) sorry..

Originally Posted by ashes

Hi..i got this equation (2xy^2 - x)dx + (x^2y - y)dy=0 ;I know its not exact as i ve partially differentiated it to get M=4xy(wrt y) and N= 2xy(wrt x) After multipying it by h(y)..i got h= i/sqrt(2y^2-1)..after that i cant prove that its exact..can someone please give me a lead on this..

Factorise it as $\displaystyle 0 = (2xy^2 - x)dx + (x^2y - y)dy = x(2y^2 - 1)dx + y(x^2 - 1)dy$, and it separates!

$\displaystyle \int\frac{x}{x^2-1}dx + \int\frac{y}{2y^2-1}dy = 0$.

Originally Posted by Opalg

Factorise it as $\displaystyle 0 = (2xy^2 - x)dx + (x^2y - y)dy = x(2y^2 - 1)dx + y(x^2 - 1)dy$, and it separates!

$\displaystyle \int\frac{x}{x^2-1}dx + \int\frac{y}{2y^2-1}dy = 0$.

And it separated here too.

Originally Posted by ashes

Hi..i got this equation (2xy^2 - x)dx + (x^2y - y)dy=0 ;I know its not exact as i ve partially differentiated it to get M=4xy(wrt y) and N= 2xy(wrt x) After multipying it by h(y)..i got h= i/sqrt(2y^2-1)..after that i cant prove that its exact..can someone please give me a lead on this..

I try to solve this differential equation as:
$\displaystyle
P(x,y)=2 x y^{2} \;\;\;; \; Q(x,y)= x^{2 y} -y
$
You have to find $\displaystyle \mu=\mu(z)$ for z = z(x,y) and solve this DE:
$\displaystyle
\frac{\mu'(z)}{\mu(z)} = \frac{\frac{\partial{P}}{\partial{y}}-\frac{\partial{Q}}{\partial{x}}}{\frac{\partial{z} }{\partial{x}}\; Q - \frac{\partial{z}}{\partial{y}}\; P}
$
This DE can be solved only by numerical approach; it is very difficult solve because when you differentiate Q you get something ugly to calculate. Note:
if $\displaystyle \frac{\partial{P}}{\partial{y}} = \frac{\partial{Q}}{\partial{x}}$ would be exact and it would be very simple to solve. How to manipulate this equation they would be exact- insert:
$\displaystyle
\frac{\partial{(\mu P)}}{\partial{y}}=\frac{\partial{(\mu Q)}}{x}
$

Originally Posted by snoopy

I try to solve this differential equation as:
$\displaystyle
P(x,y)=2 x y^{2} \;\;\;; \; Q(x,y)= x^{2 y} -y
$
You have to find $\displaystyle \mu=\mu(z)$ for z = z(x,y) and solve this DE:
$\displaystyle
\frac{\mu'(z)}{\mu(z)} = \frac{\frac{\partial{P}}{\partial{y}}-\frac{\partial{Q}}{\partial{x}}}{\frac{\partial{z} }{\partial{x}}\; Q - \frac{\partial{z}}{\partial{y}}\; P}
$
This DE can be solved only by numerical approach; it is very difficult solve because when you differentiate Q you get something ugly to calculate. Note:
if $\displaystyle \frac{\partial{P}}{\partial{y}} = \frac{\partial{Q}}{\partial{x}}$ would be exact and it would be very simple to solve. How to manipulate this equation they would be exact- insert:
$\displaystyle
\frac{\partial{(\mu P)}}{\partial{y}}=\frac{\partial{(\mu Q)}}{x}
$

Why would you do this when it's so clearly seperable?

Originally Posted by mr fantastic

And it separated here too.

Double posters should be banned from the forum!

Originally Posted by mr fantastic

Why would you do this when it's so clearly seperable?

Dear mr fantastic.

Only problem is that this is differential equation and not integral equation.
When you have problem like
$\displaystyle \frac{dy(x)}{y(x)} = -{\omega}^2\; x$
you integrate in both sides. When you can do this is only when differential equation is linear and homogen that is when is satisfy $\displaystyle y(kx,ky)=ky(x,y)$
Most important is when you calculate something you have to have very good theoretical background !

Think about that.

Originally Posted by Opalg

Double posters should be banned from the forum!

It is mistake. This things cannot be solved this way !! If it is I wanna see proof.
You have to manipulate DE to become homogene and it is not LINEAR.

Originally Posted by snoopy

Dear mr fantastic.

Only problem is that this is differential equation and not integral equation.
When you have problem like
$\displaystyle \frac{dy(x)}{y(x)} = -{\omega}^2\; x$
you integrate in both sides. When you can do this is only when differential equation is linear and homogen that is when is satisfy $\displaystyle y(kx,ky)=ky(x,y)$
Most important is when you calculate something you have to have very good theoretical background !

Think about that.

Dear snoopy.

The DE is seperable and the solution can be found in a straightforward way. No need to muddy the water with white noise.

I suggest you review the basic theory.

Originally Posted by mr fantastic

Dear snoopy.

The DE is seperable and the solution can be found in a straightforward way. No need to muddy the water with white noise.

I suggest you review the basic theory.

Dear mr fantastic
I would be very PLEASED if you calculate that DE and show it to me. However when you do any step of calculations I would be very happy to tell me from where it comes from and also any lemma, definition, statement and theorem you used to calculate that DE. That I very important in mathematics as you already know.

Good luck.

Originally Posted by mr fantastic

Dear snoopy.

The DE is seperable and the solution can be found in a straightforward way. No need to muddy the water with white noise.

I suggest you review the basic theory.

However I would be very PLEASED if you exact solution of this DE. Be careful when you want so something proof: conclamation on definitions, lemmas and theorems. As you already know that the proof is the most important and it count everything.


Good luck

well..so which is which?the separable one sounds easy and i manage to do it but i did learn that if the dfq is not exact, we have to multiply it by the Integrating factor to make it exact which I tried but its too difficult..someone pls give me the confirmed way to approach it...im starting to get confused..

Originally Posted by ashes

well..so which is which?the separable one sounds easy and i manage to do it but i did learn that if the dfq is not exact, we have to multiply it by the Integrating factor to make it exact which I tried but its too difficult..someone pls give me the confirmed way to approach it...im starting to get confused..

Opalg and myself have both given you the same advice - the DE is seperable. You used that approach and got the answer. The End.

alright