Differential Equation: Cauchy-Euler Boundary Value Problem I'm a bit confused with how my text finds the constants:
$$
x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=24 x^{5}, \quad y(1)=0, \quad y(2)=0
$$
Auxiliary / Characteristic Equation: $m(m-1)-3 m+3=(m-1)(m-3)=0$. General solution of the associated homogeneous equation is $y=c_{1} x+c_{2} x^{3}$. And this is where I get a bit confused.
Applying $ y(1) = 0 $ to this solution 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 $y_{2}(x)=4 x-x^{3}$
It goes on to the Wronskian ($W\left(y_{1}(x), y_{2}(x)\right)$) and into the Green's function for the boundary-value problem and to the particular integral and solution.
My confusion is how values for $ c_1 $ and $ c_2$ are defined. It seems like arbitrary integers are are assigned corresponding to $y_1$ and $y_2$? How do I make sense of that?
EDIT: Rest of the problem:
$$
W\left(y_{1}(x), y_{2}(x)\right)=\left|\begin{array}{ll}x-x^{3} & 4 x-x^{3} \\ 1-3 x^{2} & 4-3 x^{2}\end{array}\right|=6 x^{3}
$$
Hence the Green’s function for the boundary-value problem is
$$
G(x, t)=\left\{\begin{array}{ll}\frac{\left(t-t^{3}\right)\left(4 x-x^{3}\right)}{6 t^{3}}, & 1 \leq t \leq x \\ \frac{\left(x-x^{3}\right)\left(4 t-t^{3}\right)}{6 t^{3}}, & x \leq t \leq 2\end{array}\right.
$$
In order to identify the correct forcing function f, we put into standard form:
$$
y^{\prime \prime}-\frac{3}{x} y^{\prime}+\frac{3}{x^{2}} y=24 x^{3}
$$
From this equation we see that $f(t)=24 t^{3}$ and so $y_{p}(x)$ is
$$
\begin{aligned} y_{p}(x) &=24 \int_{1}^{2} G(x, t) t^{3} d t \\ &=4\left(4 x-x^{3}\right) \int_{1}^{x}\left(t-t^{3}\right) d t+4\left(x-x^{3}\right) \int_{x}^{2}\left(4 t-t^{3}\right) d t \end{aligned}
$$
Simplification leads to
$$
y_{p}(x)=3 x^{5}-15 x^{3}+12 x
$$
 A: Obviously, the linear system
$$
c_1+c_2=0\\2c_1+8c_2=0
$$
only has the trivial solution $c_1=c_2=0$. You should have left a parameter after the first step, $y(x)=c(x-x^3)$ so that then $0=y(2)=-6c$ directly gives $c=0$.
But in your task you first have to find a particular solution before determining the constants of the complementary or homogeneous part. The method of undetermined coefficients gives $y_p(x)=ax^5$ with some constant $a$. Then find the coefficients so that the boundary conditions are satisfied for $$y(x)=ax^5+c_1x+c_2x^3.$$

In the solution with variation-of-constants, to construct the Green function you need basis solutions $y_1$ satisfying the left boundary condition and $y_2$ satisfying the right boundary condition. Then the product $y_1(\min(t,x))y_2(\max(t,x))$ satisfies the homogeneous DE for $t\ne x$ and considering the jump in the derivative, has a right side $W(x)\delta(t-x)$ in the normalized equation. As it is a product, any factor in the basis solution cancels out by dividing with the Wronski determinant, so indeed the choice of the second initial condition is arbitrary.
A: General solution of
$$x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=24 x^{5}$$
is $$y=c_1x+c_2x^3+3x^5.$$
From boundary conditions we get
$$c_1+c_2=3,\\ 2c_1+8c_2+96=0.$$
Then $c_1=12$, $c_2=-15$.
Final answer is
$$y=3 {{x}^{5}}-15 {{x}^{3}}+12 x$$
