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I have a problem which I don't seem to be able to solve without help. It says:

Given the differential equation: $$x\cos\left(\dfrac yx\right)y^\prime=y\cos{\left(\dfrac yx\right)-x}$$ Find:

  1. An integrating factor of the form: $\mu_1=\mu_1(x)$
  2. An integrating factor of the form: $\mu_2=\mu_2(\frac yx)$

I am able to solve question number 1, but I don't have any means of answering question number 2, methodically. To solve question number 1, I first demonstrate that the differential equation is not exact: $$\underbrace{\left[y\cos{\left(\dfrac yx\right)-x}\right]}_{\equiv M(x,y)}dx+\underbrace{\left[-x\cos{\left(\dfrac yx\right)}\right]}_{\equiv N(x,y)}dy = 0$$

And, since $\partial M/\partial y \neq \partial N/\partial x$, it is not exact.

Therefore, I can try to find an integrating factor $\mu_1$ which:

  1. Is continuous, and
  2. Depends solely on x

using the formula: $$\mu_1(x)=\dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}=\ldots=-\dfrac{2}{x}$$

Which (albeit discontinuous on $x=0$) is an integrating factor of the oringinal differential equation.


My two questions are:

  1. How can I find an integrating factor which depends exclusively on $\frac yx$? (Note: I've tried performing the change of variable: $v=y/x$, but I reach a plateau).
  2. Is $(-2/x)$ an integrating factor, even if it has a discontinuity at $x=0$? (it was taught at school that it has to be continuous for all $x$).

Thank you and apologies, I am kind of new to this area of mathematics.


Edit 1

Following @B. Goddard's answer below, I reach the following equation: $$\left(\dfrac yx \cos{\dfrac yx}-1\right)\dfrac {\partial \mu}{\partial y}-\left(\dfrac yx\cos{\dfrac yx}\right)\dfrac{\partial \mu}{\partial x}+2\mu\cos{\dfrac yx}=0$$

The problem is that this exercise is from an elementary course on differential equations, and we didn't study PDE's yet. (Maybe the above equation simplifies in a way I don't see?)

Maybe I should have attempted the change of variable $(v\equiv y/x)$ before multipliying the whole equation by $\mu(y/x)$?

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Since the original DE is not defined at $x=0$, it doesn't matter that the integrating factor is discontinuous there. Everything is continuous on the domains of $M$ and $N$.

The answer to the first question is a bit tedious. So I'll just sketch the steps. First, you used a formula for the $\mu(x)$ integrating factor. Look at the derivation of that formula and follow the same steps.

Assume there is a $\mu(y/x)$ and multiply your DE by it. The new equation needs to be exact, so the (new) $\partial M/\partial y$ needs to equal the (new) $\partial N/\partial x.$

So that's two ugly product rules you have to compute. When you're taking the derivative of $\mu$ with respect to $y$, the chain rule makes you multiply by $1/x$. When you take the derivative with respect to $x$, the chain rule makes you multiply by $-y/x^2.$ Set $\partial M/\partial y = \partial N/\partial x.$

Put this last equation in the form $\mu' = \mu \times $ crud and simplify. The simplified crud should be a function of $y/x.$ If not, no integrating factor exists. (In the case of $\mu(x)$, your formula is the "crud".) If so, you can replace $y/x = z$ and solve the DE $\mu'(z) = \mu(z)\mbox{crud}(z)$ for $\mu$. Now you have the integrating factor.

Good luck.

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  • $\begingroup$ Thank you for the answer. Following your procedure, I obtained a Partial Differential Equation (contains both $\partial \mu/\partial y$ and $\partial \mu/\partial x$), and we didn't study PDE's yet. I will update the answer to give you the equation that $\mu=\mu(y/x)$ would need to satisfy. Again thank you for your time. $\endgroup$ – Jose Lopez Garcia Jun 14 '18 at 14:07
  • $\begingroup$ This won't lead to a PDE. You should get $\mu' = \mu x $junk. It's a separable ODE. $\endgroup$ – B. Goddard Jun 14 '18 at 15:33

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