# How to solve this homogeneous partial differential equation

I'm trying to solve this second order differential equation, but it seems my solution isn’t accurate, since i could not find the correct $$\sigma$$ in $$\ddot \sigma - p e^\sigma - q e^{2\sigma} =0\qquad\qquad(1)$$ or $$\frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0$$ where $$p$$ and $$q$$ are constants. So any help is appreciated.

Here's what I have tried:

Let $$\sigma = \log ~ r$$, then: $$\dot \sigma= \frac{\dot r}{ r}$$, and $$\ddot\sigma= \frac{\ddot r}{ r} - \frac{\dot r^2}{r^2}$$. Sub in (1)

$$\frac{\ddot r}{ r} - \frac{\dot r^2}{r^2} - r^2 q - r p =0\qquad\qquad(2)$$

Now to solve (2), will I use something like $$r = e^{\lambda t}$$ again?

Then (2) becomes:

$$\lambda^2 - \lambda^2 - e^{\lambda t} q - p =0 \qquad\qquad(3)$$

therefore $$\lambda = \frac{1}{t}~ \log~ \frac{p}{q}$$, or $$r = \frac{p}{q}$$ and $$\sigma = \log \frac{p}{q}$$.
This solution can not be, cause it means $$\dot \sigma = \ddot \sigma =0!!!$$

Have I missed something??

Thanks.

• This is not linear, forget about homogeneous. You cannot use the ansatz $r=\exp(\lambda t)$ – Shubham Johri May 15 at 15:30

You can try to reduce the order of the DE: $$\ddot \sigma - p e^\sigma - q e^{2\sigma} =0$$ $$2\dot\sigma\ddot \sigma - 2\dot\sigma p e^\sigma -2\dot \sigma q e^{2\sigma} =0$$ Integrate: $$\dot \sigma^2- 2p e^\sigma -q e^{2\sigma} =C_1$$ $$\dot \sigma=\pm \sqrt { 2p e^\sigma +q e^{2\sigma} +C_1}$$ $$\int \dfrac {d\sigma}{ \sqrt { 2p e^\sigma +q e^{2\sigma} +C_1}}=\pm\int dt$$
Then substitute $$e^{\sigma}=u$$.
• Then I will have again, $\frac{\dot u^2}{u^2} -2 p u + 2 q u =0$ it's nonlinear DE as “shubham mentioned before, so I think I will not use the exponential again to proceed in solving. – Dr. phy May 16 at 17:35