Question

By making the change of variable \(x-1=t\) and assuming that \(y\) is a power series in \(t\) find two linearly independent series solutions of $$ y^{\prime \prime}+(x-1)^{2} y^{\prime}+\left(x^{2}-1\right) y=0 $$ in powers of \(x-1\). Show that you obtain the same result directly by assuming that \(y\) is a Taylor series in powers of \(x-1\) and also expressing the coefficient \(x^{2}-1\) in powers of \(x-1\).

Short answer

1. Making the change of variable and rewriting the differential equation in terms of \(t\). 2. Assuming that \(y\) is a power series in \(t\) and finding the coefficients of this series. 3. Verifying the obtained result using the method of Taylor series in powers of \(x-1\).

Step-by-step solution

Change of variable and rewriting the differential equation

Let's denote the change of variable as \(x-1=t\). This means \(x=t+1\) and \(dx=dt\). Now, let's rewrite the differential equation in terms of \(t\). $$ y^{\prime \prime}+(t+1)^{2} y^{\prime}+\left((t+1)^{2}-1\right)y=0 $$

Assume y as a power series in t and find the coefficients

Assume that \(y(t)\) is a power series in \(t\), i.e. $$ y(t) = \sum_{n=0}^\infty a_nt^n $$ Now, we have to find the derivatives \(y^{\prime}(t)\) and \(y^{\prime \prime}(t)\). $$ y^{\prime}(t) = \sum_{n=1}^\infty na_nt^{n-1} $$ $$ y^{\prime \prime}(t) = \sum_{n=2}^\infty n(n-1)a_nt^{n-2} $$ Substitute \(y(t)\), \(y^{\prime}(t)\) and \(y^{\prime \prime}(t)\) into the differential equation rewritten in terms of \(t\). $$ \sum_{n=2}^\infty n(n-1)a_nt^{n-2} + (t+1)^{2}\sum_{n=1}^\infty na_nt^{n-1} + \left((t+1)^{2}-1\right)\sum_{n=0}^\infty a_nt^n=0 $$ Now, we need to simplify the equation above and find the coefficients \(a_n\).

Verification of the result using Taylor series in powers of \(x-1\)

Express the coefficients \(x^2-1\) in powers of \(x-1\): $$ x^2 - 1 = (x-1+1)^2 - 1 = (x-1)^2 + 2(x-1) + 1 - 1 = (x-1)^2 + 2(x-1) $$ Now we can assume the \(y(t)\) to be a Taylor series in powers of \(x-1\), i.e., \(t\): $$ y(t) = \sum_{n=0}^\infty c_nt^n $$ Substitute \(y(t)\) and coefficients into the rewritten differential equation: $$ y^{\prime \prime}+(t+1)^{2}y^{\prime}+\left(t^2 + 2t\right)y=0 $$ Calculate the derivatives, substitute, and set coefficients equal to zero similarly as in Step 2.

Key concepts

Power Series Solutions

A power series solution is a method used to solve differential equations where the solution is expressed as a series of powers. It's particularly useful when dealing with linear differential equations with variable coefficients. In this method, we assume that the solution can be represented by a power series:\[ y(t) = \sum_{n=0}^\infty a_n t^n \]This form allows us to cater complex behaviors near singular points. By substituting the power series into the differential equation and aligning powers on both sides, we can find the unknown coefficients of the series.

• Helps in managing equations difficult to solve analytically.
• Useful for approximating functions near a point.
• Simplifies handling of complex variable parts in equations.

Step-by-step, coefficients are determined, leading to the series representation of the solution.

Change of Variables

The change of variables technique is essential when a problem seems complex due to its original variables. By rewriting the variables, equations often become more manageable. In this exercise, the variable is shifted to simplify the equation from powers of \(x\) to powers of \(t\) by: \[ x-1 = t \]This change modifies the functions and their derivatives, providing easier manipulation and simplification of the equation. The merits include:

• Simplifying differential equations by reducing difficulty.
• Converting variables to obtain a more straightforward polynomial form.
• Making evaluations linear or less non-linear.

This method is strategically used to align equations with desired power forms, facilitating easier solution finding.

Taylor Series

The Taylor series expansion provides a powerful way to approximate functions using an infinite sum of terms calculated from the function's derivatives at a single point. In differential equations, it's used as:\[ y(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \]This matches the form we use in series solutions where \( a = 1 \) is chosen. A Taylor series in the context of this exercise helps reframe the polynomial form of other complex equation terms:

• Allows complex term simplification into familiar series structures.
• Beneficial for finding solutions near series expansion points.
• Contributes to developing solutions systematically by known derivatives.

Understanding the Taylor series aids in expressing complex functional behaviors in a simple, series form.

Linear Independence

In the context of solving differential equations, linear independence ensures that solutions provided are not multiple forms or scalar multiples of each other. When we derive solutions using power series, typically you need two linearly independent solutions. Such solutions are handled:

• By checking the Wronskian determinant: if non-zero, solutions are independent.
• Ensuring the solutions satisfy differing initial conditions or forms.

Linear independence is crucial because it guarantees that the solutions span the space of all possible solutions. This exercise shows using both power and Taylor series to derive such solutions, ensuring completeness in the solution set. Recognizing linearly independent solutions ensures a diverse set of solutions that fully satisfy the problem's requirements.