Chapter 4.7, Problem 33P

By choosing the lower limit of integration in Eq. (27) in the text as the initial point $t_0$, show that $Y(t)$ becomes

$Y\left(t\right)={\displaystyle \underset{t_0}{\overset{t}{\int }}\frac{y_1\left(\tau \right)y_2\left(t\right)-y_1\left(t\right)y_2\left(\tau \right)}{y_1\left(\tau \right)y_2\text{'}\left(\tau \right)-y_1\text{'}\left(\tau \right)y_2\left(\tau \right)}}g\left(\tau \right)d\tau$

Show that $Y(t)$ is a solution of the initial value problem

$L[y]=g\left(t\right)$,      $y\left(t_0\right)=0$,      $y\text{'}\left(t_0\right)=0$

Thus $Y$ can be identified with $v$ in problem 32.

$Y\left(t\right)=-y_1\left(t\right){\displaystyle \int \frac{y_2\left(t\right)g\left(t\right)}{W[y_1,y_2]\left(t\right)}}dt+y_2\left(t\right){\displaystyle \int \frac{y_1\left(t\right)g\left(t\right)}{W[y_1,y_2]\left(t\right)}}dt$     (27)

32. Show that the solution of the initial value problem

$L[y]=y\text{'}\text{'}+p\left(t\right)y\text{'}+q\left(t\right)y=g\left(t\right)$,

$y\left(t_0\right)=y_0$,      $y\text{'}\left(t_0\right)=y_1$

can be written as $y=u\left(t\right)+v\left(t\right)$, where $u$ and $v$ are solutions of the two initial value problems

$L[u]=0$, $u\left(t_0\right)=y_0$, $u\text{'}\left(t_0\right)=y_1$,

$L[v]=g\left(t\right)$, $v\left(t_0\right)=0$, $v\text{'}\left(t_0\right)=0$,

respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately. Observe that $u$ is easy to find if a fundamental set of solutions of $L[u]=0$ is known.