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Differential Equation HW

  • Thread starter Beez
  • Start date
32
0
Hello, I have tried to solve the following problem but did not succeed to do so.

[y(y^3 - x)]dx + [x (y^3 + x)]dy = 0
I sense that the key factor here is (y^3 - x ) and (y^3 +x), but could not figure out how to lead the equation to

dy/dx + P(x)y = Q(x) form.

The general answer for the problem is 2xy^3 - x^2 = Cy^2.

Once I can change the equation to dy/dx + P(x)y = Q(x) form, I can do the rest (probably anybody can...)

Thanks for your help in advance.
 
651
1
This is of the form f(xy)ydx +F(xy)xdy=0
here the Integrating factor is ::
[tex] \frac{1}{xy[f(xy)-F(xy)]}[/tex]
and general Integral equation is ::
[tex]\int \frac{f(xy)+F(xy)}{f(xy)-F(xy)} \frac{d(xy)}{xy} + log\frac{x}{y} = c[/tex]
 
458
0
Beez said:
[y(y3 - x)]dx + [x (y3 + x)]dy = 0
that looks awfully a lot like exact DE: you'll have to get partial derivatives and the whole nine yards. Try that.
 
651
1
Well , there is an alternative if you use differentials
now
d(xy)= ydx +xdy
and
[tex]d(\frac{y}{x}) = \frac{xdy-ydx}{x^2}[/tex]

Here u can rearrange ur diff eqn to

:: y3 d(xy) + x3d(y/x)=0
divide by y3 u will get the required answer after integrating
 
32
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Exact D.E.

I believe it is an exact DE since I have just learned that part. But when I did
\partial M (x, y) / \partial y = 4y^3 - x and
\partial N (x, y) / \partial x = y^3 + 2x so they are not the same.

But I could not find integrating facutor to make their answers equal. What should I do now?
 
32
0
I will try that

Thank you "himanshu121". I will try that for now to see if I can understand that formula.


Well, I couldn't get it.

I have just started this differential equation class (independent). I thought I understood them well. However, when it comes to solve problems, I am experiencing a hard time. For example, this equation, I could not see why it Have some suggestion to improve my understanding?
 
Last edited:

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