## anonymous 3 years ago How to find Solution to differential equation (x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0

1. anonymous

How to find Solution to not exact differential equation $(x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0$

2. anonymous

Isn't $\int\limits \frac{x-4}{x^3}dx=\int\limits \frac{y^2-3}{y^4}dy$ enough?

3. anonymous

It's probably not, I'm still pants at DE

4. anonymous

Anyway: don't ask me HOW to integrate that (use Wolfram integrator if you just want the answer), it's probably some counter-intuitive trig substitution.

5. anonymous

i got $- \frac{ 1 }{ x } + \frac{ 2 }{ x^{2} }-\frac{ 1 }{ y } + \frac{ 1 }{ y^{3} } = C$ am i right?

6. anonymous

http://integrals.wolfram.com/index.jsp?expr=%28x-4%29%2Fx%5E3&random=false http://integrals.wolfram.com/index.jsp?expr=%28x%5E2-3%29%2Fx%5E4&random=false So your signs are wrong for either the x or y terms but it's fine other than that.

7. anonymous

@gerryliyana in this type of differntial eqution, we use separation of variable technique as we can see from this differential equation, variables can be separated easily

8. anonymous

@ gerryliyana i m getting -1/x + 2/x^2 + 1/ y^2 - 1 /y^3 = C where C is constant

9. anonymous

@gerryliyana right????

10. anonymous

ok @niksva but i got -1/x + 2/x^2 + 1/ y - 1 /y^3 = C where C is constant

11. anonymous

omg my mistake it is 1/y

12. anonymous

no problem :)