I found some difficulties with one exercise.
I have differential equation $$(1-x^2)^2y''=y$$ and I have got changes: $x=tg(t)$ and $y=\frac{u}{cos(t)}$ where $u=u(t)$.
I am not sure if I'm doing my steps right. I want to find $\frac{dy}{dt}$. And here are my steps:
- $dx=\frac{1}{cos^2(t)}dt => \frac{dt}{dx}=cos^2(t)$
- $dy=d(\frac{u}{cos(t)})=\frac{ducos(t)+usin(t)dt}{cos^2(t)dt}$
And now I'm confused what to do now. I don't even know if my derivative no. 2 is OK.
$ x= \tan(x)$ so $ \frac{dx}{dt}= \sec^2(t)$. $y=u \sec(t)$ so $ \frac{dy}{dt} = \frac{du}{dt} \sec(t)+ u \sec(t) \tan(t)$. Now by the chain rule \begin{eqnarray*} \frac{dy}{dx} =\frac{dt}{dx} \frac{dy}{dt} = \frac{\frac{du}{dt} \sec(t)+ u \sec(t) \tan(t)}{\sec(t)} \end{eqnarray*} Now differentiate again (chain rule, quotient rule, etc ... \begin{eqnarray*} \frac{d^2 y}{dx^2} =\frac{dt}{dx} \frac{d}{dt} \left( \frac{\frac{du}{dt} \sec(t)+ u \sec(t) \tan(t)}{\sec(t)} \right) \\ = \frac{ \sec(t) ( \frac{ d^2 u}{dt^2}+\frac{du}{dt} \tan(t)+u \sec^2(t) ) -\sec(t) \tan(t) ( \frac{du}{dt} +u \tan(t))}{\sec^4(t)} \\ =\frac{ \sec(t) ( \frac{d^2 u}{dt^2} +u)}{ \sec^4(t)} \end{eqnarray*} At this stage, one gets the feeling that your first equation should have been $(1 \color{red}{+} x^2)^2y''=y$ and you were expected to conclude that $ \color{blue}{u''=0}$.