the first one is an exact ode
I'm having trouble figuring out which method I should be using on some of these problems. Any help in figuring out which methods would be best is much appreciated.
1. (x^3 + y^3)dx - (3xy^2)dy = 0
2. xdy - ydx = sqrt(x^2 - y^2)dx
Just these two for now. Thanks in advance.
Edit: I tried the first one with Homogeneous method but I get to a point after substitution I can't integrate.
The first one is equivalent to , which is a Bernouli equation. Make the substitution . Thus,
The DE then becomes , which is a linear equaton. Can you take it from here?
For the second one, it is equivalent to
Now apply the substitution . Thus,
Thus, the DE becomes . This is a separable equation. Can you take it from here?
Does this make sense?
Ok got some more. You don't need to solve it out for me since i'm quite familiar with the procedure's of the different methods but it's just rearranging the problems to make it look like any one of the 5 that i'm having trouble with . Saw the mistake~! Thanks for fixing it up.
1.(1+x^3)dy - ((x^2)y)dx = 0
2.(1+x)dy/dx - xy = x^3 + x^2
For the first one, That should be separable correct? If so, how about the 2nd?
Yes, these are both "homogeneous". In the first, if we replace x by ax, y by ay, it becomes (a^3x^3+ a^3y^3)dx- 3a^2xy^2dy and canceling the a^3 terms we get just what we started with. That means it can be rewritten as a differential equation for y/x.
Let u= y/x so y= xu. y'= xu'+ u so dy= xdu+ udx. The equation becomes
dividing through by
which is separable. it separates as
Is that what you got? The righthand side is tedious but doable. The denominator can be factored by solving . is a solution so .
Now write out to see that we must have and so that the denominator factors as
That quadratic term doesn't factor with real numbers because its zeros are imaginary but we can complete the square: .
Now the left hand side can be rewritten as
and that can be integrated using partial fractions.
For the second problem if we multiply both x and y by a we get and again we can cancel the "a"s so this is a homogeneous equation. Let u= y/x so that y= xu and dy= xdu+ udx and the equation becomes or . That is again homogeneous and certainly simpler than the previous problem! It separates as and both sides are easy to integrate.