Question

Question: Solve (12.12)andto get (12.15). Hint: Use Cramer's rule (Chapter 3, Section 3); note that the denominator determinant is the Wronskian [Chapter 3 , equation (8.5)) of the functionssin xand cos x.

Short answer

The value of$A\left(x^{\text{'}}\right)$ and$B\left(x^{\text{'}}\right)$ from equation$A\left(x^{\text{'}}\right)\sin x^{\text{'}}-B\left(x^{\text{'}}\right)\cos x^{\text{'}}=0$ and$-A\left(x^{\text{'}}\right)\cos x^{\text{'}}-B\left(x^{\text{'}}\right)\sin x^{\text{'}}=1$ is$A\left(x^{\text{'}}\right)=-\cos x^{\text{'}}$ is $B\left(x^{\text{'}}\right)=-\sin x^{\text{'}}$.

Step-by-step solution

Given information

The given expressions are $A\left(x^{\text{'}}\right)\sin x^{\text{'}}-B\left(x^{\text{'}}\right)\cos x^{\text{'}}=0-A\left(x^{\text{'}}\right)\cos x^{\text{'}}-B\left(x^{\text{'}}\right)\sin x^{\text{'}}=1$.

Definition of Green Function

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

Solve the given function

Let's first rewrite the equations in a way:

$\begin{array}{l}A\left(x^{\text{'}}\right)\sin x^{\text{'}}-B\left(x^{\text{'}}\right)\cos x^{\text{'}}=0\\ A\left(x^{\text{'}}\right)\cos x^{\text{'}}+B\left(x^{\text{'}}\right)\sin x^{\text{'}}=-1\end{array}$

Now we can use the Cramer's rule:

$\begin{array}{c}A\left(x^{\text{'}}\right)=\frac{\left|\begin{array}{cc}0& -\cos x^{\text{'}}\\ -1& \sin x^{\text{'}}\end{array}\right|}{\left|\begin{array}{cc}\sin x^{\text{'}}& -\cos x^{\text{'}}\\ \cos x^{\text{'}}& \sin x^{\text{'}}\end{array}\right|}\\ =\frac{-\cos x^{\text{'}}}{{\sin }^2{x}^{\text{'}}+{\cos }^2{x}^{\text{'}}}\\ =-\cos x^{\text{'}}\end{array}$

$\begin{array}{c}B\left(x^{\text{'}}\right)=\frac{\left|\begin{array}{cc}\sin x^{\text{'}}& 0\\ \cos x^{\text{'}}& -1\end{array}\right|}{\left|\begin{array}{cc}\sin x^{\text{'}}& -\cos x^{\text{'}}\\ \cos x^{\text{'}}& \sin x^{\text{'}}\end{array}\right|}\\ =-\sin x^{\text{'}}\end{array}$

Notice that the denominator is the Wronskian of $\sin x^{\text{'}}$and $\cos x^{\text{'}}$because $\text{d}\left(\sin x^{\text{'}}\right)/\text{d}{x}^{\text{'}}=\cos x^{\text{'}}$ and $\text{d}\left(-\cos x^{\text{'}}\right)/\text{d}{x}^{\text{'}}=\sin x^{\text{'}}$.