anonymous
  • anonymous
How to find Solution to differential equation (x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0
Differential Equations
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
How to find Solution to not exact differential equation \[(x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0\]
anonymous
  • anonymous
Isn't \[\int\limits \frac{x-4}{x^3}dx=\int\limits \frac{y^2-3}{y^4}dy\] enough?
anonymous
  • anonymous
It's probably not, I'm still pants at DE

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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.
anonymous
  • anonymous
i got \[- \frac{ 1 }{ x } + \frac{ 2 }{ x^{2} }-\frac{ 1 }{ y } + \frac{ 1 }{ y^{3} } = C\] am i right?
anonymous
  • 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.
anonymous
  • 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
anonymous
  • anonymous
@ gerryliyana i m getting -1/x + 2/x^2 + 1/ y^2 - 1 /y^3 = C where C is constant
anonymous
  • anonymous
@gerryliyana right????
anonymous
  • anonymous
ok @niksva but i got -1/x + 2/x^2 + 1/ y - 1 /y^3 = C where C is constant
anonymous
  • anonymous
omg my mistake it is 1/y
anonymous
  • anonymous
no problem :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.