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I am having trouble solving the differential equation:

$\dfrac{dy}{d{\theta}} = \dfrac{\theta \sec\left(\dfrac{y}{\theta}\right) + y}{\theta}$

I realise I need to put it in the form $\dfrac{dy}{d\theta} + h(\theta)y = 0$ and then find the integrating factor, but I'm having trouble rearranging it. I don't think it will be too hard to solve after that!


My first attempt was to divide by $\theta$ giving $\dfrac{dy}{d\theta} = \sec\left(\dfrac{y}{\theta}\right) + \dfrac{y}{\theta}$

Then I tried taking $\sec^{-1}$ giving

$\sec^{-1}\left(\dfrac{dy}{d\theta}\right) = \dfrac{y}{\theta} + \sec^{-1}\left(\dfrac{y}{\theta}\right) $

To no avail.

Any help in rearranging this will be much appreciated!

  • 0
    Your last equation doesn't follow from the one before; it is not, in general, true that $\sec^{-1}(a+b)=\sec^{-1}a+\sec^{-1}b$.2011-08-09

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To solve this, write $y = v \times \theta$. Hence, we get $\frac{dy}{d \theta} = \theta \frac{dv}{d \theta} + v$ Now plug this into the differential equation to get $\theta \frac{dv}{d \theta} + v = \sec(v) + v$ Now rearrange to get $\cos(v) dv = \frac{d \theta}{\theta}$ Integrating, we get $\sin(v) = \log(\theta) + c$ $v = \sin^{-1} (\log(\theta) + c)$ Hence, $y = \theta \times \sin^{-1} (\log(\theta) + c)$