# Exact ODE

I missed a course and now I have to solve an ODE, I don't freaking know how :)

$$(ax+by)dx+(kx+ly)dy=0$$

So, I have to know on what condition this equation is exact (1), and I have to find the exact equation (2). Well if a, b, k and l = 0 then it's certainly over but i guess it's not really the point.

Let's say M(x,y) = ax+by and N(x,y) = kx+ly, IF i'm right, the ODE is exact when;

$$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$

Then;

$$\frac{\partial (ax+by)}{\partial y}=\frac{\partial (kx+ly)}{\partial x}$$

$$b=k$$

So my answer to the first question is; is it exact when b = k. And to show how sure I am (not...) I'll say c = b = k for the next part of this question. However, intuitively, i'd say the winning condition is when a and l = 0, and b = -k, (b=c,k=-c) then by integrating you'll got bxy+kxy = 0, so cxy-cxy=0, 0=0.

And then, I don't know what method to use to find the exact ODE.

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HallsofIvy
Homework Helper
Yes, it is true that the differential M(x,y)dx+ N(x,y)dy is "exact" when there exist some function f(x,y) such that
$$\frac{\partial f}{\partial x}= M(x,y)$$
and
$$\fract{\partial f}{\partial y}= N(x,y)$$
and that requires, in order that
$$\frac{\partial^2f}{\partial x\partial y}= \frac{\partial^2f}{\partial y\partial x}$$
that
$$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$

will be an exact equation if and only if b= k. But I don't know what you mean by "find the exact equation"! Obviously, the exact equation is
(ax+ by)dx+ (bx+ ly)dy when b= k. Is it possible that you mean "find the general solution to the exact equation"??

An equation is exact if and only if the left hand side is an exact differential: For some f(x,y), df= M(x,y)dx+ N(x,y)dy which means that we must have $$\frac{\partial f}{\partial x}= M(x,y)$$ and that $$\frac{\partial f}{\partial y}= N(x,y)$$.

In particular, if df= (ax+ by)dx+ (bx+ ly)dy we must have
fx= ax+ by and fy= bx+ ly. Since the partial derivative wrt x is taken by treating y as a constant, we can "back out" by taking the anti-derivative treating y as a constant: f(x,y)= (1/2)ax2+ bxy+ g(y). "g(y)" is the constant of integration". Since we are treating y as a constant, it could be any function of y. Now differentiate that with respect to y: fy= bx+ g'(y)= by+ ly. Precisely because we set k= b, the x terms cancel and we have g'(y)= ly. Integrating that, g(y)= (1/2)ly+ C (since g is a function of y only, that C really is a constant). Putting that into the equation of f(x,y),
f(x,y)= (1/2)ax2+ bxy+ (1/2)ly. Since df= 0 according to the equation, f(x,y) must equal a constant:
(1/2)ax2+ bxy+ (1/2)ly= C is the general solution to the equation.

wow. Thanks, you really took time to answer and it's appreciated.

So...

$$(ax+by)dx + (kx+ly)dy = 0$$

To be exact, b must be equal to k

$$f(x,y) = \frac{ax^2}{2}+byx+\frac{ly^2}{2}$$

f(x,y) is equal to a constant because df = 0 (but it could still be written, f(x,y) = ... + C no ?)

$$C = \frac{ax^2}{2}+byx+\frac{ly^2}{2}$$

That's clear. However you wrote at the general answer "ly/2" and I got "ly^2/2". It doesn't make more sence ? If "ly/2", then how can you get "ly" when you derive f(x,y) ? And another thing, is there a way to test that answer numerically with Maple (don't know MatLab or Mathematica yet...) ?

Thanks a lot for your time. If you ever need a letter of recommendation from an unknown "non-math major" non-american undegraduate student with a lot of trouble writting in english, I'm your man.

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HallsofIvy
Homework Helper
"f(x,y) is equal to a constant because df = 0 (but it could still be written, f(x,y) = ... + C no ?"

No. f(x,y)= (1/2)ax2+ bxy+ (1/2)cy2+C is simply a function in two variables and does not define y, even implicitly, as a function of x. Either (1/2)ax2+ bxy+ (1/2)cy2+C= 0 or, same thing, (1/2)ax2+ bxy+ (1/2)cy2= C does.

"However you wrote at the general answer "ly/2" and I got "ly^2/2". It doesn't make more sence ? "

Yes! I messed up the superscripts in "(1/2)ax2+ bxy+ (1/2)ly= C"!

By the way, your English is far better than my (put practically any language here!).