The function $y(x)$ satisfies the linear equation

$$y^{\prime \prime}(x)+p(x) y^{\prime}(x)+q(x) y(x)=0$$

The Wronskian, $W(x)$, of two independent solutions denoted $y_{1}(x)$ and $y_{2}(x)$ is defined to be

$$W(x)=\left|\begin{array}{cc} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{array}\right| .$$

Let $y_{1}(x)$ be given. In this case, show that the expression for $W(x)$ can be interpreted as a first-order inhomogeneous differential equation for $y_{2}(x)$. Hence, by explicit derivation, show that $y_{2}(x)$ may be expressed as

$$y_{2}(x)=y_{1}(x) \int_{x_{0}}^{x} \frac{W(t)}{y_{1}(t)^{2}} d t,$$

where the rôle of $x_{0}$ should be briefly elucidated.

Show that $W(x)$ satisfies

$$\frac{d W(x)}{d x}+p(x) W(x)=0 .$$

Verify that $y_{1}(x)=1-x$ is a solution of

$$x y^{\prime \prime}(x)-\left(1-x^{2}\right) y^{\prime}(x)-(1+x) y(x)=0 .$$

Hence, using $(*)$ with $x_{0}=0$ and expanding the integrand in powers of $t$ to order $t^{3}$, find the first three non-zero terms in the power series expansion for a solution, $y_{2}(x)$, of ( $\dagger$ ) that is independent of $y_{1}(x)$ and satisfies $y_{2}(0)=0, y_{2}{ }^{\prime \prime}(0)=1$.