Transforming a Non-Linear DE into Bernoulli's Form

[fluidistic]

Homework Statement

Solve the following DE: 2xyy'=4x^2+3y^2.

Homework Equations

Bernoulli's DE: y'+P(x)y=Q(x)y^2.

The Attempt at a Solution

I know that the original DE isn't under Bernoulli's form, but I have thought a lot on the problem and my feeling is that if I could find a change of variable to transform the general DE into a Bernoulli's equation, I'd be done. I have tried z=4x^2+3y^2, so z'=8x+6yy' but this leads me nowhere. I am not even sure I can reduce the original DE into a Bernoulli's equation. This is the only way I think I could solve the DE, I don't see any other way.
I'd love some help.

 

[flyingpig]

I know that the original DE isn't under Bernoulli's form, but I have thought a lot on the problem

Divide both sides by x^2

=)

 

[icystrike]

Don't you think this should be homogenous ODE instead of Bernoulli?

 

[fluidistic]

Okay thanks guys. I divided by x² but was stuck right away.
I checked out if it was homogeneous and indeed it was homogeneous of first order. I then opened my cookbook for DE's (Boas, 2nd edition) and followed his advice. I wrote the DE under the form P(x,y)dx+Q(x,y)dy=0, with P(x,y)=-4x^2-3y^2 and Q(x,y)=2xy. He says I can write the DE under the form y'=f(y/x).
And thus the change of variable v=y/x is appropriate, according to him. The DE should then be separable. However I tried but got stuck.
I reach 2x^3vdv+2x^2v^2dx-4x^2-3[x^2(dv)^2+xvdvdx+v^2(dx)^2]=0. I don't know how to deal with the squared differentials nor the crossed ones (dxdv).
Any further help will also be appreciated. :)

 

[dynamicsolo]

You should not have squared differentials. If you are using the substitution v = y/x , presumably to replace the y , then you need to differentiate implicitly y = vx to find dy/dx in terms of dv/dx . You should be able to use this to replace y' with v' (and also y/x with v ) to produce a transformed differential equation that is easier to work with. (The result isn't pretty, but it is separable and integrable...)

 

[icystrike]

Just to give you a little hint:

2xyy'=4x^{2}+3y^{2}

divide by 2xy throughout:

y'=2(x/y)+3(y/x)

Let v=y/x ,

dv/dx = ?

 

[fluidistic]

Thanks once again guys!

You should not have squared differentials. If you are using the substitution v = y/x , presumably to replace the y , then you need to differentiate implicitly y = vx to find dy/dx in terms of dv/dx . You should be able to use this to replace y' with v' (and also y/x with v ) to produce a transformed differential equation that is easier to work with. (The result isn't pretty, but it is separable and integrable...)

y=xv \Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx}.
Thus the original DE transforms into v+x\frac{dv}{dx}=2v+\frac{3}{2v}\Rightarrow \frac{xdv}{dx}=v+\frac{3}{2v} \Rightarrow \frac{dx}{x}=\left ( v+\frac{3}{2v} \right ) ^{-1}dv. Now I must integrate to get v(x). Once I have v(x) I divide it by x in order to get y(x).

Just to give you a little hint:

2xyy'=4x^{2}+3y^{2}

divide by 2xy throughout:

y'=2(x/y)+3(y/x)

Let v=y/x ,

dv/dx = ?

I tried this way too but strangely I don't get "nice" stuff. dv/dx= \frac{1}{x^2} \left ( \frac{xdy}{dx}-y \right ). If I isolate dy/dx, I don't even reach the one I got from the above way. All this, despite that y=xv in the first case and v=y/x in the second case. I don't understand how I can reach a different differential for dy/dx...

 

[dynamicsolo]

Thanks once again guys!
y=xv \Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx}.
Thus the original DE transforms into v+x\frac{dv}{dx}=2v+\frac{3}{2v}\Rightarrow \frac{xdv}{dx}=v+\frac{3}{2v} \Rightarrow \frac{dx}{x}=\left ( v+\frac{3}{2v} \right ) ^{-1}dv. Now I must integrate to get v(x). Once I have v(x) I divide it by x in order to get y(x).

OK, so far. (And, yes, icystrike has a small typo in the resulting equation. The "3" should be "3/2"...)

I tried this way too but strangely I don't get "nice" stuff. dv/dx= \frac{1}{x^2} \left ( \frac{xdy}{dx}-y \right ).

You shouldn't still have that "y" kicking around in there. (I thought you were going from icystrike's equation, so this is a little puzzling. You should again have v + x\frac{dv}{dx} = 2v + \frac{3}{2v}.)

As I said above, the result isn't going to be very pretty. (Solutions of DEs often aren't...) You would reduce your transformed equation to \frac{dx}{x} = ( \frac{2v}{2v^{2} + 3} ) dv and integrate from there.

 

[fluidistic]

Thank you again for your help.

OK, so far. (And, yes, icystrike has a small typo in the resulting equation. The "3" should be "3/2"...)

Ah I see.

You shouldn't still have that "y" kicking around in there. (I thought you were going from icystrike's equation, so this is a little puzzling. You should again have v + x\frac{dv}{dx} = 2v + \frac{3}{2v}.)

Well I used the derivative of the quotient y(x)/x. It's [y'(x)x-y(x)]/x²

As I said above, the result isn't going to be very pretty. (Solutions of DEs often aren't...) You would reduce your transformed equation to \frac{dx}{x} = ( \frac{2v}{2v^{2} + 3} ) dv and integrate from there.

I see. I think you have a small typo in the right hand side (I think you forgot a ^-1).

 

[dynamicsolo]

I think you have a small typo in the right hand side (I think you forgot a ^-1).

I already added the ratios and "flipped over" the result from your differential equation result for v : ( v+\frac{3}{2v} ) ^{-1} = \frac{2v}{2v^{2} + 3} .

 

[fluidistic]

I already added the ratios and "flipped over" the result from your differential equation result for v : ( v+\frac{3}{2v} ) ^{-1} = \frac{2v}{2v^{2} + 3} .

Ah right I see now!
Do you have an idea what's my problem using icystrike's suggestion? Why can't I get rid of the y term and I don't reach the same expression for dv/dx.

 

[dynamicsolo]

Ah right I see now!
Do you have an idea what's my problem using icystrike's suggestion? Why can't I get rid of the y term and I don't reach the same expression for dv/dx.

To be frank, I am still trying to figure out how you arrived at your result. The intent of icystrike's suggestion was to use v = y/x to completely eliminate y from the equation; you use Bernoulli's (whichever of the nine of them it was) idea to transform the original differential equation into a new one involving v and x that (hopefully) is easier to solve.

By making only a partial substitution, you created a situation that is harder to resolve algebraically than it needs to be. (It can probably be solved from there, but less easily. You would get rid of the y term by using y = vx .)

 

[ehild]

The equation contains y² and 2yy'=(y²)', this suggest the substitution z=y². With this substitution, the equation becomes linear.

ehild

 

[fluidistic]

Icystricke asked me dv/dx rather than dy/dx in function of dv/dx.
What I did for his way is v=y/x \Rightarrow \frac{dv}{dx}= \left ( \frac{xdy}{dx} -y \right ) \frac{1}{x^2}.
WOW! This work too! After isolating dy/dx and replacing y/x by v, I reach exactly the same expression for dy/dx than you. Nice!
Thanks guys for all your help. Edit: Okay thanks a lot ehild. I'm going to try this tomorrow (it's past 2 am here).

 

[ehild]

Edit: Okay thanks a lot ehild. I'm going to try this tomorrow (it's past 2 am here).

Then you live far to west from me.
Just to be able to check your derivation: it should be very easy, and results in y²=cx³-4x².

ehild

 

[dynamicsolo]

The equation contains y² and 2yy'=(y²)', this suggest the substitution z=y². With this substitution, the equation becomes linear.

ehild

Thank you -- that's good spotting and the result for y is faster to extract than it is to do so from v . (I assumed from the initial post that it was required to use Bernoulli's method.)

 

[icystrike]

Thank you -- that's good spotting and the result for y is faster to extract than it is to do so from v . (I assumed from the initial post that it was required to use Bernoulli's method.)

Yes TS should try this method too. That is a direct consequence of implicit diff.

 

[fluidistic]

Thanks. So I reached the same solution as yours. Problem solved.