# Rearranging a First-Order Linear Homogeneous Differential Equation

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!

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Hint: Let $y=\theta u$. Rewrite the equation in terms of $u$ and $\theta$ alone. It will simplify a lot. – André Nicolas Aug 9 '11 at 2:13
Hint: find a differential equation satisfied by the function $z$ defined by $z(\theta)=y(\theta)/\theta$. – Did Aug 9 '11 at 2:13
Your last equation doesn't follow from the one before; it is not, in general, true that $\sec^{-1}(a+b)=\sec^{-1}a+\sec^{-1}b$. – Gerry Myerson Aug 9 '11 at 3:32
Thank you very much! – aligray Aug 9 '11 at 4:32

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, $$y = \theta \times \sin^{-1} (\log(\theta) + c)$$