how do I get F(y/x)?

Hi

Please have a look on the attachment. It has my question there. Please help me with it. Thank you.

As you can see, the book gives two formulas to solve homogeneous equation: Formulas.

Regards
PG

Hi,

One way you could do this is as follows.

Starting with the original equation:
dy/dx=-(x^2+y^2)/(x^2-x*y)

Expand the right side using partial fractions in x:
dy/dx=-2*y/(x-y)+y/x-1

Divide both the top and bottom of the first term on the right by x:
dy/dx=-2*(y/x)/((x/x)-(y/x))+y/x-1

Simplify and group occurrences of y/x:
dy/dx=-2*(y/x)/(1-(y/x))+(y/x)-1

If desired, change into this form:
dy/dx=((y/x)^2+1)/((y/x)-1)

This is now in the form:
dy/dx=F(y/x)

You could also start with the original and divide top and bottom of right hand side by x^2.

Either way once you subtract y/x you end up with:
(y/x+1)/(y/x-1)

MrAl: Many, many thanks. It was very kind of you.

Actually I have also solved it myself. You can have a look on the attachment.

L8: [latex]\ln x-2\ln \left\vert x\right\vert +2\ln \left\vert x+y\right\vert -\frac{1}{x}y+c=0[/latex]

Best wishes
PG

Hi again,

Oh yes, very good PG. You used the identity v=y/x more directly for a substitution. Good idea.

Thank you, MrAl.

As you can see here, there are two formulae given for homogeneous equation, 18.5 and 18.6. The form of the homogeneous differential equation I have encountered often so far is the 18.5 one. I see for 18.6 form, we have to take v=xy instead of v=y/x as was the case for the 18.5.

While solving ODE the main problem is realize which form of the equation we are working with, i.e. if the equation is separable, exact or what else. Other than a lot of practice, , is there any other way to judge the form of equation quickly?

With warmest regards
PG

Hi,

As i was saying once before, it's a matter of trying to get the equation into a recognizable form. That becomes the whole ball game

Sometimes what happens is you begin to see the same forms coming over and over again while doing certain tasks, so you learn to do them more easily.