# General rule regarding solving each differnetial

1. Jan 30, 2010

### hangainlover

1. The problem statement, all variables and given/known data

(3y^2 + 2y)y' = xcosx
xyy'=ln(x) ; y(1)=2

2. Relevant equations

3. The attempt at a solution
for the first one i get y^3 + y^2= (x^2) sin(x) +cos(x) + C (constant)
and the second one is y^2 + ln(x)^2 +4
the problem is there is no way i can define those two as function y of x or function x of y
because of the exponents.

So my question is, if a question is asking for a solution to a differential equation, can we just leave it as that above?

2. Jan 30, 2010

### Staff: Mentor

The left side is correct, but you made a mistake when you integrated xcos(x). You can check what you have by differentiating x2sin x + cos x. If your integration is correct, you should get d/dx(x2sin x + cos x) = xcosx.

Once you fix your error, you're done. It's not always possible to get the solution so that you have y as a function of x. In this case you will have y as an implicit function of x.
You might have made a typo here. The solution should be y2 = (ln x)2 + 4. It's probably OK to leave it this way, but if you need to, you can solve for y, as y(x) = +/-sqrt( (ln x)2 + 4)

3. Jan 30, 2010

### hangainlover

yeah i made a typo in both solutions. Im sorry. I was really rushed to type it in

the first one should be y^3+y^2 = xsin(x) +cos(x) + C and the second one is y^2 = (ln x)^2 + 4 not y^2=ln(x)^2 +4

Last edited: Jan 30, 2010
4. Jan 30, 2010

### Staff: Mentor

Both look fine now.