I'm getting confused by a recurring theme involving solutions to boundary-value problems and the way the text is finding arbitrary constants: $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=24 x^{5}, \quad y(1)=0, \quad y(2)=0 $$ The solution is as follows: The differential equation is a Cauchy-Euler DE. From the auxiliary equation $$m(m-1)-3 m+3=(m-1)(m-3)=0$$ the general solution of the associated homogeneous equation is $$y=c_{1} x+c_{2} x^{3}$$
In the application of conditions to find arbitrary constants below, I'm completely lost:
- Applying $y(1)=0$ implies $c_{1}+c_{2}=0$ or $c_{1}=-c_{2}$.
- By choosing $c_{2}=-1$ we get $c_{1}=1$ and $y_{1}=x-x^{3}$.
- On the other hand, $y(2)=0$ applied to the general solution shows $2 c_{1}+8 c_{2}=0$ or $c_{1}=-4 c_{2}$
- The choice $c_2 = -1$ now gives $c_1 = 4 $ and so $y_{2}(x)=4 x-x^{3}$
Q: This above 'choosing' of $c_1$ and $c_2$ is completely foreign to me. As I understand, $y_1$ should follow from the first half of the general solution ($c_1 x + c_2 x^3$). So $y_1 = x$ and $y_2=x^3$. The text however specifies $y_1 =x - x^3$ and $y_2 = 4x-x^3$.
So there's something I'm fundamentally misunderstanding regarding finding constants. For completeness I'll attach the rest of the problem below.
$$ W\left(y_{1}(x), y_{2}(x)\right)=\left|\begin{array}{ll}x-x^{3} & 4 x-x^{3} \\ 1-3 x^{2} & 4-3 x^{2}\end{array}\right|=6 x^{3} $$
$$ G(x, t)=\left\{\begin{array}{ll}\frac{\left(t-t^{3}\right)\left(4 x-x^{3}\right)}{6 t^{3}}, & 1 \leq t \leq x \\ \frac{\left(x-x^{3}\right)\left(4 t-t^{3}\right)}{6 t^{3}}, & x \leq t \leq 2\end{array}\right. $$ Standard Form: $$ y^{\prime \prime}-\frac{3}{x} y^{\prime}+\frac{3}{x^{2}} y=24 x^{3} $$
$$ \begin{aligned} y_{p}(x) &=24 \int_{1}^{2} G(x, t) t^{3} d t \\ &=4\left(4 x-x^{3}\right) \int_{1}^{x}\left(t-t^{3}\right) d t+4\left(x-x^{3}\right) \int_{x}^{2}\left(4 t-t^{3}\right) d t \end{aligned} $$ $$ y_{p}(x)=3 x^{5}-15 x^{3}+12 x $$
Continuing from Differential Equation: Cauchy-Euler Boundary Value Problem
You want to solve the equation for the Green function, $G_x(t)$ has to solve the normalized homogeneous DE with $\delta(t-x)$ as the right side, $$ G_x''(t)-\frac3xG_x'(t)+\frac3{x^2}G_x(t)=\delta(t-x) $$ with homogeneous boundary conditions. This means that for $t<x$ the solution is identical to a homogeneous solution that satisfies the left boundary condition, and for $t>x$ to a homogeneous solution that satisfies the right boundary condition. As everything is homogeneous, all such solutions differ by a factor, so select $y_1(t)=t-t^3$ and $y_2(t)=4t-t^3$ and $$ G(x,t)=G_x(t)=\begin{cases}c_1(x)y_1(t)&\text{ for }t\le x,\\ c_2(x)y_2(t)&\text{ for }t> x.\end{cases} $$ For continuity it is needed that $c_1(x)y_1(x)=c_2(x)y_2(x)$. Now add the unit jump condition for the derivative to find the usual $c_1(x)=\frac{y_2(x)}{W(x)}$ and $c_2(x)=\frac{y_1(x)}{W(x)}$.
The solution you computed automatically inherits the homogeneous boundary conditions from the Green's function, so that you do not have to determine additional coefficients.