I have faced this task, when preparing to my university's "mathematical methods" exam, and I do not really understand how to solve it. The task content is as follows:

**Given**: $$(1-x^2)\cdot y''\_{xx} -x\cdot y'\_x +y = \frac{\sqrt{1-x^2}}{x}$$ Solve the equation by substituion of the independent variable: $x(t) = \cos(t), x\in(0,1).$

**My attempt**:

I have performed the suggested substitution:

* $(1-\cos^{2}(t))\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y}{\partial x(t)} \normalsize+ y = \large\frac{\sqrt{1-\cos^2(t)}}{cos(t)}.$
* $\sin^2(t)\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y / \partial t}{\partial x / \partial t} \normalsize+ y = \tan(t).$
* $\sin^2(t)\cdot\frac{\large\partial\left(\frac{\dot{y}}{-\sin(t)}\right) / \partial t}{\large\partial x / \partial t} + \large\frac{\cos(t)}{\sin(t)}\normalsize\dot{y} + y = \tan(t).$
* $-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$

That is now I have to solve: $-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$ And that is the point, where I am stuck, I have no idea how to proceed from here, and I did not manage to find the way to solve such a linear D.E. in my course book.

I would appreciate any help, hints or book references, thank you in advance!

**Post Scriptum**:

As correctly mentioned by @Aryadeva, I have made a mistake, when calculating $\large\frac{\partial^2 y}{\partial x^2}$, I simply lost "minus".

That is, after recalculating I had to solve: $$\ddot{y} + y = \tan(t), \text{ s.t. } x\in(0,1) \text{ and } x = \cos(t) \Leftrightarrow \forall x\in \mathbb{D}\_x: \text{ }t = \arccos{x}.\tag{1}$$

My solution has involved appying vatiation of parameters method:

$\underline{\text{Step I}}$: find characteristic polynomial over $y(t)$ to get a general solution $(\large y\_{\small G})$ for $(1)$:

$\lambda^2 + 1 = 0. \Rightarrow \lambda = \pm i.$

$\begin{bmatrix}
\lambda\_1\\
\lambda\_2\\
\end{bmatrix} = \begin{bmatrix}
i\\
-i\\
\end{bmatrix} \Rightarrow \large y\_{\small G} \normalsize = C\_1 e^{it}+C\_2 e^{-it} = \dot{\tilde {C}}\_1\cos(t) + \dot{\tilde{C}}\_2 \sin(t).$

$\underline{\text{Step II}}$: Finding particular solution $(\large y\_{\small P})$ for $(1)$, using Wronskian matrices:

I would omit my calculations here, would just note that for the given D.E.:

$\mathbb{W} = \begin{pmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}, \mathbb{W}\_{\small \Delta} = \begin{vmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{vmatrix} = 1.$

$\mathbb{W}\_{\small \Delta,\normalsize 1} = \begin{vmatrix}0 & \sin(t) \\ 1 & \cos(t) \end{vmatrix} = -\sin(t), \mathbb{W}\_{\small \Delta,\normalsize 2} = \begin{vmatrix}\cos(t) & 0 \\ -\sin(t) & 1 \end{vmatrix} = \cos(t).$

Thus:
$$\dot{\tilde{C}}\_1(t) = \int\frac{\tan(t)\cdot \mathbb{W}\_{\small \Delta,\normalsize 1}}{\mathbb{W}\_{\small \Delta}} \partial t = \sin(t) +\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}\_1,$$
$$\dot{\tilde{C}}\_2(t) = \int\frac{\tan(t)\cdot \mathbb{W}\_{\small \Delta,\normalsize 2}}{\mathbb{W}\_{\small \Delta}} \partial t = -\cos(t) + \bar{C}\_2.$$
Thereof:
$$\large y\_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}\_1\right)\cos(t) +\left(-\cos(t) +\bar{C}\_2\right)\sin(t).$$

$\underline{\text{Step III}}$: Now it just suffices to write down the whole solution and transfer from $t$ to $x$:

* $\large y \normalsize (t) = \large y\_{\small G} + y\_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\tilde{C}\_1\right)\cos(t) +\left(-\cos(t) +\tilde{C}\_2\right)\sin(t).$

Now, as $x \in (0, 1): \cos(\arccos{x}) = x, \text{ }\sin(\arccos{x}) = \sqrt{1-x^2}:$
$$\text{Answer: } \begin{array}{|c|}
\hline
y(x) = \left(\sqrt{1-x^2} + \frac{1}{2}\ln\left(\frac{\sqrt{1-x^2} - 1}{\sqrt{1-x^2} +1}\right)+\tilde{C}\_1 \right)x +\left(-x+\tilde{C}\_2\right)\sqrt{1-x^2}.\\
\hline
\end{array}$$