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$$xy'=y \cos \bigg(\ln{y\over x}\bigg)$$

In the first I tried a substitution to no avail.

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    $\begingroup$ It's $\cos \bigg (\ln \frac{y}{x}\bigg)$, right? $\endgroup$ Mar 6, 2015 at 19:11
  • $\begingroup$ yes ,your correction is correct.. $\endgroup$ Mar 6, 2015 at 19:13
  • $\begingroup$ is there a source to this equation? where did you get it from? $\endgroup$
    – tzoorp
    Mar 6, 2015 at 19:22
  • $\begingroup$ Its from a russian assignement book, the author filipov. Its in a series of homogenious equations and ones with bringing in a substitution ussually it being z=y/x.. $\endgroup$ Mar 6, 2015 at 19:25

4 Answers 4

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Hint: Let $z = \frac{y}{x}$ then $y = zx \implies y' = z'x + z$

$$y' = \frac{y}{x} \cos \bigg(\ln \frac{y}{x}\bigg) \implies z'x + z = z\cos (\ln z) $$

It's separable.

Edit: Spoiler:

$$\int \frac{1}{z\cos (\ln z) - z} dz = \cot \bigg(\frac{\ln z}{2}\bigg) + C'$$

Then

$$\cot \bigg(\frac{\ln z}{2}\bigg) = \ln x + C \implies z = e^{2 \ arccot ({\ln x+ C})}$$

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    $\begingroup$ you got to it first. i was going to solve that. $\endgroup$
    – abel
    Mar 6, 2015 at 19:39
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    $\begingroup$ @BozoVulicevic You've got any questions? Feel free to ask. $\endgroup$ Mar 6, 2015 at 19:46
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    $\begingroup$ @BozoVulicevic This is not in my answer. $\endgroup$ Mar 6, 2015 at 19:52
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    $\begingroup$ @BozoVulicevic Hey! You totally checked the spoilers! Yes, it was a typo. $\endgroup$ Mar 6, 2015 at 19:56
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    $\begingroup$ Take $u = \ln z$. And notice that $\cos t = 1 - 2\sin^2 \frac{t}{2}$. You need to show some work either, huh? $\endgroup$ Mar 6, 2015 at 20:02
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i am going to make a change of variable $$\frac{y}{x} = e^u, y = xe^u, \frac{dy}{dx} = xe^u \frac{du}{dx} + e^u.$$ the differential equation $xy' = y\cos \left(\ln (\frac yx)\right)$ turns into a simpler $$ xe^u \frac{du}{dx} + e^u = e^u\left(\cos u\right)$$ that is $$x\frac{du}{dx} + 1= \cos u$$ which is separable and $$\frac{du}{1-\cos u} =\frac 12 \csc^2 u/2 \, du= -\frac{dx}{x}$$ which intgerates to $$\cot(u/2) = \ln(x) + \ln C $$

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  • $\begingroup$ You're a close second if i could id give you the answer as well .. $\endgroup$ Mar 6, 2015 at 20:05
  • $\begingroup$ @BozoVulicevic, that is fine. i am glad i tried a little different subs. it is fun to come up with nice subs. $\endgroup$
    – abel
    Mar 6, 2015 at 20:11
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Rewrite the equation as follows:

$$\frac{dy}{y} = \cos{(\log{y}-\log{x})} \frac{dx}{x} $$

Let $u=\log{y}$ and $v=\log{x}$. Then

$$\frac{du}{dv} = \cos{(u-v)}$$

Let $u=q+v$; then

$$\frac{dq}{dv} = \cos{q}-1 = -2 \sin^2{\left ( \frac{q}{2} \right )}$$

Then

$$-\int dq \csc^2{\left ( \frac{q}{2} \right )} = 2 \cot{\left ( \frac{q}{2} \right )} = v + C$$

where $C$ is a constant of integration. Backsub and get

$$2 \cot{\left [ \frac12 \log{\left (\frac{y}{x}\right )} \right ]} = \log{x} + C$$

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  • $\begingroup$ Could you by anychance explain the second and fourth line here. I have no doubt that this is the best answer. $\endgroup$ Mar 6, 2015 at 19:36
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    $\begingroup$ @BozoVulicevic: $q=u+v \implies u' = q' + 1 = \cos{q}$. (Note the mistake I made pprior to the edit. Also, $$\int dx \, \csc^2{x} = \cot{x} + C$$ $\endgroup$
    – Ron Gordon
    Mar 6, 2015 at 19:38
  • $\begingroup$ Upvote Abels answer ! $\endgroup$ Mar 6, 2015 at 20:07
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    $\begingroup$ @BozoVulicevic: why? - it looks a lot like mine. $\endgroup$
    – Ron Gordon
    Mar 6, 2015 at 20:09
  • $\begingroup$ I want you all to get votes because only one can get the answer. I have no doubt that your answer is very adequete but we just did not use this technique where i study, and the answer i chose is much closer to our practice in class.. i very much appreciate your effort.. $\endgroup$ Mar 6, 2015 at 20:18
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$y=zx$, $y'=z'x+z$, $z'x^2+zx=zx cos(ln(z))$

Separation of variables: $\frac{dz}{z(cos(ln(z))-1)}= \frac{dx}{x}$.

Hint for the Integral (Integratin by parts with first factor $z^{-1}$ and second factor $(cos(ln(z))-1)^{-1}$): $\int \frac{dz}{z(cos(ln(z))-1)}= \frac{ln(z)}{cos(ln(z))-1}- \int \frac{ln(z)sin(ln(z))dz}{z(cos(ln(z))-1)^2}$

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