0
$\begingroup$

I'm trying to copy the method from this video https://www.youtube.com/watch?v=FTHLz2vgj2I

to solve the following differential equation: $\frac{dr}{d\theta}+r\tan \theta=\frac{1}{\cos \theta}$

Basically what the video uploader did:

He had a differential equation, he brought it to the form $y'+P(x)y=Q(x)$ and the then said:

$$u=e^{\int P(x)dx}, (uy)'=uQ$$

And then he solved for $y$.

I tried to recreate that solution with my question.

$u=e^{\int P(x)dx}=e^{\int \tan \theta d\theta}=e^{-\ln( \cos \theta)}=\frac{1}{\cos \theta}$

$$\frac {d(ur)}{d\theta}=\frac{d(\frac{r}{\cos \theta})}{d\theta}=\frac{r'\cos \theta+r\sin \theta}{\cos^2 \theta}=\frac{1}{\cos ^2 \theta}$$

this implies that $r' \cos \theta+r\sin \theta=1$, which is the exact same ODE that I was asked to solve...So the method in the video did not work here, why?

$\endgroup$
1
$\begingroup$

Hint:

The method works, you just have to integrate it up. If $$\frac{d(r/\cos\theta)}{d\theta} = \frac{1}{\cos^2\theta}$$

then

$$\frac{r}{\cos\theta} = \int\frac{d\theta}{\cos^2\theta}$$

Solve the integral (remember to include the integration constant) and you get the solution. This integral is well known, but if you don't know it try to compute $\frac{d\tan\theta}{d\theta}$ and compare. In the end if you do everything correctly you should get an answer on the form $r = A\cos\theta + B\sin\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.