How to find the solution of the Tchebycheff differential equation? I have the equation: (1-x2)u'' -xu'+ku=0,
where ' represents differentiation with respect to x and k is a constant.
I am asked to show that cos(k1/2cos-1x) is a solution to this equation.
I assumed to show this you need to set u=cos(k1/2cos-1x) and substitute it into the the Tchebycheff equation and show it equals zero. However, when doing this I get quite a lot of messy differentiation. 
The question before this asked to put the equation into Sturm Liouville form, so I am unsure if that is meant to be used to solve the question.
Any hints on how to advance would be greatly appreciated.
 A: HINT:
$$(1-x^2)u''(x)-xu'(x)+ku(x)=0\Longleftrightarrow$$

Let $t=i\sqrt{k}\ln(\sqrt{x^2-1}+x)$, which gives $x=\frac{1}{2}e^{-\frac{it}{\sqrt{k}}}\left(1+e^{\frac{2it}{\sqrt{k}}}\right)$:

$$\left(-\frac{1}{4}e^{-\frac{2it}{\sqrt{k}}}\left(1+e^{\frac{2it}{\sqrt{k}}}\right)^2+1\right)u''(x)-\frac{1}{2}e^{-\frac{it}{\sqrt{k}}}\left(1+e^{\frac{2it}{\sqrt{k}}}\right)u'(x)+ku(x)=0\Longleftrightarrow$$

Apply the chain rule $\frac{\text{d}u(x)}{\text{d}x}=\frac{\text{d}u(t)}{\text{d}t}\frac{\text{d}t}{\text{d}x}$:

$$k\left(u''(t)+u(t)\right)=0\Longleftrightarrow$$

Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$.
Substitute $u(t)=e^{\lambda t}$ into the differential equation:

$$k\cdot\frac{\text{d}^2}{\text{d}t^2}(e^{\lambda t})+k\cdot e^{\lambda t}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}t^2}(e^{\lambda t})=\lambda^2e^{\lambda t}$:

$$e^{\lambda t}\left(k+k\lambda^2\right)=0\Longleftrightarrow$$

Since $e^{\lambda t}\ne 0$ for any finite $\Lambda$, the zeros must come from the polynomial:

$$k+k\lambda^2=0\Longleftrightarrow$$
$$k\left(\lambda^2+1\right)=0\Longleftrightarrow$$
$$\lambda=\pm i$$
A: if you make the substitution  $x=\cos t$ then
$$
u_t=-u_x\sin t 
$$
and
$$
u_{tt} = -u_x\cos t  +u_{xx}\sin^2 t   = u_{xx}(1-x^2) - x u_x
$$
the equation becomes:
$$
u_{tt} +ku =0
$$
A: $$(1-x^2)\frac{d^2u}{dx^2}-x\frac{du}{dx}+ku=0$$
Let $x=\cos(t)\quad$ hense $\quad dx=-\sin(t)dt_\quad$ then $\quad\frac{dt}{dx}=-\frac{1}{\sin(t)}$
$\frac{du}{dx}=\frac{du}{dt}\frac{dt}{dx}= -\frac{1}{\sin(t)}\frac{du}{dt}$
$\frac{d^2u}{dx^2}=\frac{d\left( -\frac{1}{\sin(t)}\frac{du}{dt}\right)}{dx}=
\frac{d\left( -\frac{1}{\sin(t)}\frac{du}{dt}\right)}{dt}\frac{dt}{dx}=\frac{d\left( -\frac{1}{\sin(t)}\frac{du}{dt}\right)}{dt}\left(-\frac{1}{\sin(t)}\right)=\frac{1}{\sin^2(t)}\frac{d^2u}{dt^2}-\frac{\cos(t)}{\sin^3(t)}\frac{du}{dt}$
$(1-x^2)\frac{d^2u}{dx^2}-x\frac{du}{dx}+ku=\sin^2(t)\left(\frac{1}{\sin^2(t)}\frac{d^2u}{dt^2}-\frac{\cos(t)}{\sin^3(t)}\frac{du}{dt} \right) - \cos(t)\left(-\frac{1}{\sin(t)}\frac{du}{dt} \right)+ku=0$
after simplification :
$$\frac{d^2u}{dt^2}+ku=0$$
$$u=c_1\cos(\sqrt k \:t)+c_2\sin(\sqrt k \:t)$$
$t=\cos^{-1}(x)$
$$u=c_1\cos\left(\sqrt k \:\cos^{-1}(x)\right)+c_2\sin\left(\sqrt k \:\cos^{-1}(x)\right)$$
