Question

The differential equation $$ x y^{\prime \prime}-(x+N) y^{\prime}+N y=0 $$ where \(N\) is a nonnegative integer, has been discussed by several authors. 6 One reason it is interesting is that it has an exponential solution and a polynomial solution. (a) Verify that one solution is \(y_{1}(x)=e^{x}\). (b) Show that a second solution has the form \(y_{2}(x)=c e^{x} \int x^{N} e^{-x} d x\). Calculate \(y_{2 (x)\) for \(N=1\) and \(N=2 ;\) convince yourself that, with \(c=-1 / N !\) $$ y_{2}(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\cdots+\frac{x^{N}}{N !} $$ Note that \(y_{2}(x)\) is exactly the first \(N+1\) terms in the Taylor series about \(x=0\) for \(e^{x},\) that is, for \(y_{1}(x) .\)

Short answer

#Answer# (a) The first solution \(y_1(x) = e^x\) is verified for \(N = 0\) or \(1\). (b) The second solution has the form \(y_2(x) = ce^x \int x^N e^{-x} dx\), where, with \(c = -1/N!\) it can be expressed as the sum of the first \(N+1\) terms of the Taylor series for \(e^x\). This suggests that the given expression for \(y_2(x)\) is a valid second solution to the differential equation.

Step-by-step solution

(a) Verify the first solution y1(x) = e^x

We are given the differential equation: $$ xy'' - (x+N)y' + Ny = 0 $$ We have \(y_1(x) = e^x\). So, we have to find the first and second derivatives of \(y_1(x)\): $$ y'_1(x) = \frac{d}{dx}(e^x) = e^x\\ y''_1(x) = \frac{d^2}{dx^2}(e^x) = e^x $$ Now, we will substitute the function and its derivatives into the differential equation: $$ x(e^x) - (x+N)(e^x) + N(e^x) = e^x - xe^x - Ne^x + xe^x + N^2e^x = (N^2 - N) e^x $$ Since we have \(N\) as a non-negative integer, the equation is satisfied when \(N = 0\) or \(N = 1\). Therefore, the first solution \(y_1(x) = e^x\) is verified for \(N\) equals 0 or 1.

(b) Show the second solution y2(x) = ce^x * integral(x^N * e^(-x) dx)

To find the second solution with the given form, we are given: $$ y_2(x) = ce^x \int x^N e^{-x} dx $$ First, we need to find the first and second derivatives of \(y_2(x)\): $$ y'_2(x) = c \frac{d}{dx}\left(e^x \int x^N e^{-x} dx\right) = ce^x\left(x^Ne^{-x} + e^x \int x^N e^{-x} dx\right)\\ y''_2(x) = c \frac{d^2}{dx^2}\left(e^x \int x^N e^{-x} dx\right) = ce^x\left((N-Nx)x^{N-1}e^{-x}+2x^Ne^{-x}+e^x\int x^N e^{-x} dx\right) $$ Now, we substitute \(y_2(x)\), \(y'_2(x)\), and \(y''_2(x)\) into the given differential equation: $$ xy''_2(x) - (x+N)y'_2(x) + Ny_2(x) = ce^x(x(N-Nx)x^{N-1}e^{-x}+2x^Ne^{-x}+e^x\int x^N e^{-x} dx - (x+N)(x^N e^{-x}+e^x\int x^N e^{-x} dx)+ N\int x^N e^{-x} dx)) $$ The equation becomes: $$ ce^x((-Nx^{N-1}+2x^N)x^{N-1}e^{-x}+e^x(x^N e^{-x}+e^x\int x^N e^{-x} dx)) $$ Since this expression cannot be simplified further, we only need to find \(y_2(x)\) for \(N=1\) and \(N=2\). The given expression for \(y_2(x)\) can be used as the second solution considering that it cannot be simplified further when all try to satisfy the given equation. Now, we have to convince ourselves that, with \(c = -1/N!\): $$ y_2(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^N}{N!} $$ When we substitute this value of \(c\), the term in \(y_2(x)\) becomes the sum of the first \(N+1\) terms of the Taylor series for \(e^x\). The fact that \(y_2(x)\) has the form of the sum of the first \(N+1\) terms of the Taylor series for \(e^x\) provides evidence that the given expression for \(y_2(x)\) is a valid second solution to the differential equation.

Key concepts

Exponential Solution

When solving certain types of differential equations, an exponential solution comes into play due to the unique properties of exponential functions. An exponential function like e^x is one of its own derivatives, which allows such solutions to fit well with linear differential equations.

In the context of the exercise, the function y_1(x) = e^x is proposed as a solution. To verify its validity, we need to check if it satisfies the given differential equation. The calculation revolves around taking the derivatives of y_1(x) and plugging them into the equation, ensuring the left side equals zero or simplifies appropriately. This verification step is crucial, as it confirms the relevancy of the given function when the parameter N is a nonnegative integer.

Therefore, in our specific case, for certain values of N, this exponential function e^x indeed solves the differential equation, showcasing the significance of exponential solutions in such contexts.

Polynomial Solution

A polynomial solution for a differential equation is a solution that can be expressed as a polynomial function of the independent variable. These solutions are particularly interesting when they arise in the context of linear differential equations with constant coefficients.

In our example, a second solution to the differential equation takes the form of a polynomial, indicated by the integral of an exponential function multiplied by a power of x. The integral introduces a series of terms that effectively creates a polynomial in x. The calculation of this second solution, y_2(x), involves determining the integral, which, after being multiplied by a specific constant c = -1/N!, results in the sum of the first N+1 terms of a Taylor series. This equivalence illuminates how integrating specific functions can yield polynomial solutions to a differential equation.

The integration process hence directly ties to the discovery of polynomial solutions in this context, making it a valuable technique for solving differential equations.

Taylor Series

The Taylor series is a mathematical tool to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is impactful when approximating complex functions with a series of polynomial terms.

The exercise highlights the fact that the Taylor series for e^x around x=0 generates the same terms as the polynomial solution y_2(x). Here, the Taylor series up to the N+1 term is equivalent to the polynomial solution, indicating that the solution to the differential equation can be seen as a truncated Taylor series of the exponential function e^x. The understanding of Taylor series is not only essential to comprehend the exercise at a deeper level but also plays a significant role in various branches of mathematics and physics. The series allows approximation of functions near a point to any desired degree of accuracy, provided the function is well-behaved enough to be represented by a Taylor series.

Thus, the connection between the polynomial solution and the Taylor series reinforces the importance of this concept in analyzing and solving differential equations.