Question

Show that if \(L[y]=x^{2} y^{\prime \prime}+\alpha x y^{\prime}+\beta y,\) then $$ L\left[(-x)^{r}\right]=(-x)^{r} F(r) $$ for all \(x<0,\) where \(F(r)=r(r-1)+\alpha r+\beta .\) Hence conclude that if \(r_{1} \neq r_{2}\) are roots of \(F(r)=0,\) then linearly independent solutions of \(L[y]=0\) for \(x<0\) are \((-x)^{r_{1}}\) and \((-x)^{r_{2}}\)

Short answer

Question: Show that if \(L[(-x)^r] = (-x)^r F(r)\), where the linear second-order homogeneous differential operator \(L[y] = x^2 y'' + \alpha x y' + \beta y\), then two functions \((-x)^{r_1}\) and \((-x)^{r_2}\) are linearly independent solutions for \(x<0\), where \(r_1\) and \(r_2\) are distinct roots. Solution: Let \(F(r)=r(r-1)+\alpha r+\beta\). Then, when \(F(r_1)=0\) and \(F(r_2)=0\), we have two linearly independent solutions \((-x)^{r_1}\) and \((-x)^{r_2}\) for the equation \(L[y]=0\) for \(x<0\) as their Wronskian determinant is non-zero.

Step-by-step solution

Derive \((-x)^r\) and \((-x)^r\)

Firstly, we need to find the first and second derivatives of the function \((-x)^r\) with respect to \(x\). Using the chain rule, we find: $$ \frac{d}{dx}\left[(-x)^r\right] = r(-x)^{r-1}(-1), $$ and, $$ \frac{d^2}{dx^2}\left[(-x)^r\right] = r(r-1)(-x)^{r-2}(-1)^2. $$

Plug the derivatives into \(L[y]\)

We now plug these derivatives into the given equation for \(L[y]\), which is: $$ L[y] = x^2 y'' + \alpha x y' + \beta y. $$ Substituting the derivatives, we get: $$ L[(-x)^r] = x^2 r(r-1)(-x)^{r-2}(-1)^2 + \alpha x r(-x)^{r-1}(-1) + \beta(-x)^r. $$

Simplify the expression

Now, we simplify the expression as follows: $$ L[(-x)^r] = (-x)^r (r(r-1) + \alpha r + \beta). $$ Comparing this to the given result, $$ (-x)^r F(r) = L[(-x)^r], $$ where \(F(r) = r(r-1) + \alpha r + \beta\).

Determine the roots of \(F(r)=0\)

To find the roots \(r_1\) and \(r_2\), we need to solve the equation \(F(r) = 0\). This gives us: $$ r(r-1) + \alpha r + \beta = 0. $$

Verify linearly independent solutions

If \(r_1\) and \(r_2\) are distinct roots of \(F(r) = 0\), then the functions \((-x)^{r_1}\) and \((-x)^{r_2}\) are solutions of the equation \(L[y]=0\). To show that they are linearly independent for \(x<0\), we will use the Wronskian determinant. The Wronskian determinant for these two functions is: $$ W\left((-x)^{r_1}, (-x)^{r_2}\right) = \begin{vmatrix} (-x)^{r_1} & (-x)^{r_2} \\ r_1(-x)^{r_1 - 1}(-1) & r_2(-x)^{r_2-1}(-1) \end{vmatrix} $$ Upon evaluating the determinant, we get: $$ W\left((-x)^{r_1}, (-x)^{r_2}\right) = r_1(-x)^{r_1 - 1}(-x)^{r_2} - r_2(-x)^{r_1}(-x)^{r_2-1}, $$ $$ W\left((-x)^{r_1}, (-x)^{r_2}\right) = (-x)^{r_1 + r_2 - 1}(r_1 - r_2). $$ Since \(r_1\) and \(r_2\) are distinct, \(r_1 - r_2 \neq 0\), and \(W((-x)^{r_1}, (-x)^{r_2}) \neq 0\) for \(x < 0\). This means that \((-x)^{r_1}\) and \((-x)^{r_2}\) are linearly independent solutions of \(L[y]=0\) for \(x < 0\).