Equation of the type mdx+ndy=0

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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
 
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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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