solutions of Initial value problem of system of differential equations

Question

I have problems when reading this theorem from the book. I am confused that why here $$ u(\tau)=u^{\prime}(\tau)=\ldots=u^{(n-2)}(\tau)=0, \quad u^{(n-1)}(\tau)=1 $$ but $$y\left(t_0\right)=y^{\prime}\left(t_0\right)=\ldots=y^{(n-1)}\left(t_0\right)=0 $$ And why in the example below the theorem,$y(1)=y'(1)=0$ changed to $y(1)=2,y'(1)=5$ later? Could someone please explain to me! Thanks in advance. \\

THEOREM 6. Let $L(t, \lambda)=\lambda^n+a_{n-1}(t) \lambda^{n-1}+\ldots+a_1(t) \lambda+a_0(t) \quad$ and $\quad D=\frac{d}{d t}$. Denote by $E(t, \tau)$ the uniquely determined solution $u(t)$ of the initial value problem (16) $\quad L(t, D) u=0$ $$ u(\tau)=u^{\prime}(\tau)=\ldots=u^{(n-2)}(\tau)=0, \quad u^{(n-1)}(\tau)=1 $$ Then (17) $$ y(t)=\int_{t_0}^t E(t, \tau) g(\tau) d \tau $$ is the solution of the problem $$ \begin{aligned} &L(t, D) y=g(t) \\ &y\left(t_0\right)=y^{\prime}\left(t_0\right)=\ldots=y^{(n-1)}\left(t_0\right)=0 \end{aligned} $$ Example 6. Find the solution of the differential equation (18) $t^2 y^{\prime \prime}-2 t y^{\prime}+2 y=t^2 \sin t^4, \quad t>0$, which satisfies $y(1)=2, y^{\prime}(1)=5$. Solution: The homogeneous equation $t^2 y^{\prime \prime}-2 t y^{\prime}+2 y=0$ is an Euler equation. The indicial equation is $\lambda(\lambda-1)-2 \lambda+2=0$, with roots $\lambda=1, \lambda=2$. The general solution of the homogeneous equation is thus $$ y(t)=C t+D t^2 \quad(C, D \text { constants }) . $$ We now find the fundamental solution. By (16), this must satisfy the conditions $y(\tau)=0, y^{\prime}(\tau)=1$. We obtain the following system of equations for $C$ and $D$ : $$ \left\{\begin{array} { l } { C \tau + D \tau ^ { 2 } = 0 } \\ { C + 2 D \tau = 1 } \end{array} \quad \Longleftrightarrow \quad \left\{\begin{array}{l} C=-1 \\ D=1 / \tau \end{array}\right.\right. $$ It follows that $$ E(t, \tau)=-t+\frac{t^2}{\tau}, \quad t, \tau>0 $$ By theorem 6 $$ \bar{y}(t)=\int_1^t\left(-t+\frac{t^2}{\tau}\right) \sin \tau^4 d \tau, \quad t>0 $$ is that solution of equation (18) which satisfies $y(1)=y^{\prime}(1)=0$. (We have divided (18) by $t^2$ to make the coefficient of $y^{\prime \prime}$ equal to 1 .)

The general solution of $(18)$ is obtained by adding the general solution of the homogeneous equation, i.e., $$ y(t)=\bar{y}(t)+C t+D t^2 $$ The conditions $y(1)=2, y^{\prime}(1)=5$ lead to $C+D=2, C+2 D=5$, resulting in $C=-1, D=3$. Hence the required solution is $$ y(t)=\int_1^t\left(-t+\frac{t^2}{\tau}\right) \sin \tau^4 d \tau-t+3 t^2, \quad t>0 $$

Answer 1

In the theorem, the solution $u(t)$ of the homogeneous equation with initial conditions $u(\tau)=\dots=u^{(n-2)}(\tau)=u^{(n-1)}(\tau)-1=0$ is just a tool to construct the solution $\bar y$ of the non-homogeneous equation with initial conditions $\bar y(t_0)=\dots=\bar y^{(n-1)}(t_0)=0.$

In the example:

• we first solve $t^2u^{\prime \prime}-2 tu^{\prime}+2u=0\quad (t>0),\quad u(\tau)=u'(\tau)-1=0.$ The solution is $u(t)=-t+\frac{t^2}\tau$ (restricted to $\tau>0$ to keep $t,\tau$ in the same interval);
• then we apply the theorem with $t_0=1$ to deduce the solution $\bar y$ of $t^2\bar y^{\prime \prime}-2 t\bar y^{\prime}+2\bar y=t^2\sin t^4\quad (t>0),\quad \bar y(1)=\bar y'(1)=0.$
• finally, we add to $\bar y$ a solution $z$ of the homogeneous equation such that $z(1)=2,z'(1)=5.$ The resulting function $y=\bar y+z$ will be the solution of $t^2y^{\prime \prime}-2 ty^{\prime}+2y=t^2\sin t^4\quad (t>0),\quad y(1)=2,y'(1)=5.$