Equation of the type mdx+ndy=0

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SUMMARY

The discussion revolves around solving the differential equation of the form (3x^4sin(y) - y^3)dx + (x^5cos(y) + 3xy^2)dy = 0. Participants identify that the equation is not exact and explore the use of integrating factors to convert it into an exact equation. The integrating factor can be derived using the formulas \(\frac{My - Nx}{N}\) or \(\frac{Nx - My}{M}\), depending on whether the resulting function is solely a function of x or y. The conversation emphasizes the importance of understanding exact equations and integrating factors in solving differential equations.

PREREQUISITES
  • Understanding of differential equations (DEs)
  • Familiarity with exact equations
  • Knowledge of integrating factors
  • Basic calculus skills
NEXT STEPS
  • Study the method of finding integrating factors for non-exact differential equations
  • Learn how to apply the exactness test (dM/dy = dN/dx)
  • Explore examples of exact equations and their solutions
  • Review the concept of linear differential equations (LDEs) and their integrating factors
USEFUL FOR

Students preparing for exams in differential equations, educators teaching calculus, and anyone seeking to deepen their understanding of exact equations and integrating factors.

chaoseverlasting
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Homework Statement



Solve:

(3x^4siny-y^3)dx+(x^5cosy+3xy^2)dy=0

Homework Equations





The Attempt at a Solution



At first I thought it was a simple equation of the type mdx+ndy=0, but when I integrated m wrt x and ignored all terms containing x in n (all of them in this case), and added, I didnt get the solution.
 
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The problem looks suspiciously like it was intended to be an exact equation but something messed up in transcribing it. Because it's not.
 
Yes...it does look like it would be an exact equation, but it's not quite there.

Is there anything we can do to make it so?

how about finding an integrating factor to multiply through in order to convert it to exact?

if...

\frac{My-Nx}{N}

is a function of x only, then the solution to:

\frac{d\mu}{dx} = \frac{My-Nx}{N} \mu

gives \mu as the appropriate integrating factor

or if

\frac{Nx-My}{M}

is a function of y only, then the solution to:

\frac{d\mu}{dx} = \frac{Nx-My}{M} \mu

gives you the integrating factor, \mu
 
Last edited:
Sure enough. There IS an integrating factor. Can you find it, chaoseverlasting?
 
No man. I have no idea what you guys are doing. The only integrating factor I know of is in the Linear DE. I don't think I can reduce this to a LDE
 
Then I strongly suggest you go back to your textbook and review "exact equations" and "integrating factors"!
 
The thing is, I am not even in college yet. I have a basic understanding of what DE's are but this one came in an exam. We haven't done this sort of thing.
 
chaoseverlasting said:
The thing is, I am not even in college yet. I have a basic understanding of what DE's are but this one came in an exam. We haven't done this sort of thing.

We believe you. :rolleyes:
 
Well, then just try and multiply the equation by x^n. Then apply the exactness test (dM/dy=dN/dx) and determine an n that works. Then go back and try your integration again. That's an example of an integrating factor.
 

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