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 \(x < 0\).

Key concepts

Wronskian Determinant

The Wronskian determinant is a valuable tool in the study of differential equations, particularly for determining whether two functions are linearly independent. When you have two functions, say \( f(x) \) and \( g(x) \), their Wronskian is given by the determinant of the following matrix:

• The first row contains the functions: \( f(x) \) and \( g(x) \),
• The second row contains their derivatives: \( f'(x) \) and \( g'(x) \).

The Wronskian is then expressed as: \[ W(f,g) = \begin{vmatrix} f(x) & g(x) \ f'(x) & g'(x) \end{vmatrix} = f(x)g'(x) - g(x)f'(x). \] If the Wronskian of two functions does not equal zero in an interval, it suggests that these functions are linearly independent on that interval. This concept is crucial when analyzing solutions to differential equations since linear independence means that the functions form a complete set of solutions or a basis. To apply this to our specific cases \((-x)^{r_1}\) and \((-x)^{r_2}\), we calculate their Wronskian and find it to be non-zero provided \(r_1 eq r_2\). Thus, this affirms their linear independence for \(x < 0\).

Linearly Independent Solutions

Linearly independent solutions are essential when dealing with differential equations because they form the building blocks for general solutions. In simple terms, two functions are linearly independent if neither function is a constant multiple of the other. In the context of differential equations, particularly those of second order like the one in this exercise, having two linearly independent solutions means that we can express the general solution as a linear combination of these two functions. This principle plays a key role when analyzing second-order linear differential equations. If \( r_1 eq r_2 \) are roots of the characteristic equation \( F(r) = 0 \), the solutions \((-x)^{r_1}\) and \((-x)^{r_2}\) are linearly independent. This is because their Wronskian, a measure of their linear dependence, is non-zero. As a result, the general solution to the associated differential equation can be expressed as: \[ y(x) = C_1 (-x)^{r_1} + C_2 (-x)^{r_2}, \] where \(C_1\) and \(C_2\) are constants determined by boundary or initial conditions.

Roots of Polynomial Equations

Understanding the roots of polynomial equations is crucial when solving differential equations. The polynomial equation in this exercise is \( F(r) = r(r-1) + \alpha r + \beta = 0 \). Finding the roots of this polynomial, namely \( r_1 \) and \( r_2 \), helps in identifying solutions to the differential equation \( L[y]=0 \). Solving polynomial equations often involves factoring or using the quadratic formula. For quadratic polynomials, the solution \( r \) can be found using: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a, b, \) and \( c \) are coefficients from the polynomial equation. In our context, the polynomial \( F(r) \) is of quadratic form, thus having potentially two distinct roots \( r_1 \) and \( r_2 \). Identifying these roots is key because they determine the form of the linearly independent solutions to the differential equation. If the roots are distinct, then the solutions \((-x)^{r_1}\) and \((-x)^{r_2}\) provide a fundamental set of solutions, making it easier to express the most general solution of the differential equation. This demonstrates the interrelationship between the algebraic property of roots and their implications in solving differential equations.