Question

Solve the following differential equations by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=0 $$

Short answer

Solutions: \( y = e^{\frac{3}{2}t}(C_1 \cos(\frac{\sqrt{3}}{2} t) + C_2 \sin(\frac{\sqrt{3}}{2} t)) \) and a power series. Verified by matching term-by-term.

Step-by-step solution

- Solve the differential equation using an elementary method

First, find the characteristic equation by substituting the trial solution of the form \(y = e^{rx}\) into the given differential equation \(x^2 y'' - 3x y' + 3y = 0\). This gives:\(x^2 r^2 e^{rx} - 3x r e^{rx} + 3 e^{rx} = 0\). Simplifying, we get:\( r^2 - \frac{3}{x} r + \frac{3}{x^2} = 0 \).

- Find the characteristic equation solutions

Since the equation in Step 1 needs to be transformed further, express it as a polynomial in terms of \(r\), by multiplying through by \(x^2\):\(x^2 r^2 - 3xr + 3 = 0\). Now solve for \(r\) using the quadratic formula:\(r = \frac{3 \, \pm \, \sqrt{9-12}}{2} = \frac{3 \, \pm \, \sqrt{-3}}{2}\). Hence, \(r\) can be written as:\(r = \frac{3}{2} \pm \frac{\sqrt{3}}{2}i\). So, the general solution is:\(y = e^{\frac{3}{2}t}(C_1 \cos(\frac{\sqrt{3}}{2} t) + C_2 \sin(\frac{\sqrt{3}}{2} t))\).

- Assume a power series solution for the differential equation

Consider a power series solution \(y = \sum_{n=0}^{\infty} a_n x^n\). Then we can compute the series for its derivatives:\(y' = \sum_{n=1}^{\infty} n a_n x^{n-1}\) and \(y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\). Substituting this back into the original differential equation yields:\(x^2\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 3x \sum_{n=1}^{\infty} n a_n x^{n-1} + 3 \sum_{n=0}^{\infty} a_n x^n = 0\).

- Equate powers of x

Shift indices to combine series for the same powers of \(x\) and set the coefficients of like powers to zero. This results in the recurrence relation \(a_n = \frac{n(n-1)}{(n+2)(n+1)} a_{n-2}\). Solving this recurrence gives:\(a_2 = \frac{1}{6} a_0\),\(a_4 = \frac{1}{20} a_2 = \frac{1}{120} a_0\), etc....

- Formulate the power series solution

Substitute the values of \(a_n\) found from the recurrence relation back into the series representation to obtain:\(y = a_0 (1 + \frac{1}{6} x^2 + \frac{1}{120} x^4 ...) + a_1 (x + \frac{1}{12} x^3 + ...)\).

- Verify the series solution with the elementary solution

To verify, recall the exponential series \(e^{rx} = \sum_{n=0}^{\infty} \frac{(rx)^n}{n!}\) and observe whether the two solutions agree term by term. For \(y = e^{\frac{3}{2}x}(C_1 \cos(\frac{\sqrt{3}}{2} x) + C_2 \sin(\frac{\sqrt{3}}{2} x))\) to match series terms derived in Step 5.

Key concepts

Power Series Solution

A power series solution is a method for solving differential equations by expressing the solution as an infinite sum of terms, each involving a power of the independent variable. In our case, the differential equation is \[ x^2 y'' - 3x y' + 3y = 0 \] We assume a solution of the form: \[ y = \sum_{n=0}^{\infty} a_n x^n \] This means we write \(y\) as a series \(y = a_0 + a_1 x + a_2 x^2 + \cdots\). Next, we need to find the derivatives and substitute them back into the differential equation to form new series:

• \[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \]
• \[ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]

Insert these into the original equation to equate powers of \(x\). Combining and shifting the series indices, we create a recurrence relation. This approach transforms the problem into an algebraic one, allowing us to solve for the coefficients \(a_n\). This process validates our solution term by term.

Characteristic Equation

A characteristic equation is used to find solutions to linear differential equations. We form it by substituting a trial solution of the form \(y = e^{rx}\). For our differential equation \[ x^2 y'' - 3x y' + 3y = 0 \] By trying \( y = e^{rx} \), we substitute and obtain: \[ x^2 r^2 e^{rx} - 3x r e^{rx} + 3 e^{rx} = 0 \] Simplifying this gives the characteristic equation: \[ r^2 - \frac{3}{x} r + \frac{3}{x^2} = 0 \] Multiplying through by \( x^2\), we convert it into a standard quadratic form: \[ x^2 r^2 - 3xr + 3 = 0 \] Solving this using the quadratic formula \( r = \frac{3 \pm \sqrt{9-12}}{2} = \frac{3 \pm \sqrt{-3}}{2}\), we find our characteristic roots to be: \[ r = \frac{3}{2} \pm \frac{\sqrt{3}}{2} i \] This results in a general solution involving complex exponentials and trigonometric functions.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence, each term relying on the preceding terms. For our power series method, upon substituting the series solution into our differential equation, we equate powers of \(x\) and derive the recurrence relation: \[ a_n = \frac{n(n-1)}{(n+2)(n+1)} a_{n-2} \] This allows us to find coefficients \(a_n\) based on previous terms. Starting with initial conditions like \(a_0\) and \(a_1\), we find:

• \[ a_2 = \frac{1}{6} a_0 \]
• \[ a_4 = \frac{1}{20} a_2 = \frac{1}{120} a_0 \]

This pattern establishes how each \(a_n\) follows from earlier coefficients, ultimately constructing the series solution. These relations simplify solving the differential equation through algebraic methods, ensuring accurate and computationally simpler solutions.