Question

Find series solutions of the equation \(y^{\prime \prime}-2 z y^{\prime}-2 y=0\). Identify one of the series as \(y_{1}(z)=\exp z^{2}\) and verify this by direct substitution. By setting \(y_{2}(z)=u(z) y_{1}(z)\) and solving the resulting equation for \(u(z)\), find an explicit form for \(y_{2}(z)\) and deduce that $$ \int_{0}^{x} e^{-v^{2}} d v=e^{-x^{2}} \sum_{n=0}^{\infty} \frac{n !}{2(2 n+1) !}(2 x)^{2 n+1}. $$

Short answer

1- \( y_1(z) = \exp(z^2) \). 2- \( y_2(z) = C \exp(z^2) \int_0^z \exp(-v^2) dv = C \exp(z^2) \sum_{n=0}^{fty} \frac{n!}{2(2n+1)!}(2x)^{2n+1}. \)

Step-by-step solution

Assume a series solution for the differential equation

Consider a solution of the form \( y(z) = \sum_{n=0}^{fty} a_n z^n \). To find the coefficients \(a_n\), plug this into the differential equation.

Plug the series into the differential equation

Substitute the series into \( y^{prime prime} - 2 z y^{prime} - 2 y = 0 \). Differentiate term-by-term: \( y^{prime}(z) = \sum_{n=1}^{fty} n a_n z^{n-1} \) and \( y^{prime prime}(z) = \sum_{n=2}^{fty} n(n-1) a_n z^{n-2} \).

Combine and simplify terms

Combining and simplifying terms, the coefficients of like powers of \( z \) must sum to zero. This yields a recurrence relation for \( a_n \).

Identify the series as \( y_{1}(z)=\exp(z^{2}) \)

Check if \( \exp(z^2) \) satisfies the differential equation. If \( y_1(z) = \exp(z^2) \), then \( y_1^{prime}(z) = 2z \exp(z^2) \) and \( y_1^{prime prime}(z) = (4z^2 + 2) \exp(z^2) \). Verifying, \( y_1^{prime prime} - 2zy_1^{prime} - 2y_1 = (4z^2 + 2)\exp(z^2) - 4z^2\exp(z^2) - 2\exp(z^2) = 0 \). Therefore:\(y_1(z)=\exp(z^2)\) is indeed a solution.

Define \( y_{2}(z) = u(z) y_{1}(z) \)

Set \( y_2(z)=u(z) exp(z^2) \) and substitute into the differential equation to find \( u(z) \). This produces \( u^{prime prime} + 4z u^{prime} = 0 \).

Solve the differential equation for \( u(z) \)

Solve \( u^{prime prime} + 4z u^{prime} = 0 \) by using an integrating factor. The solution is \( u(z) = C \int v exp(-v^2) dv \).

Find the explicit form of \( y_{2}(z) \)

Integrating yields \( y_2(z)= C \exp(z^2) \int_{0}^{z} \exp(-v^2) dv \), which identifies \( \int_0^x \exp(-v^2) dv \) with summation series \( \sum_{n=0}^{fty} \frac{n!}{2(2n+1)!}(2x)^{2n+1} \).

Key concepts

recurrence relation

To solve the differential equation mentioned, we often use **recurrence relations**. A recurrence relation is an equation that recursively defines a sequence. For differential equations, we deduce these relations by comparing coefficients of like powers of a variable. This process simplifies a complex series into a manageable form.

In our exercise, we assumed a power series solution and substituted it into the differential equation. This helped us derive the recurrence relation necessary for solving the equation:

1. Write the series form: \( y(z) = \sum_{n=0}^{\infty} a_n z^n \) 2. Substitute and simplify to find the coefficients \(a_n\)

By examining terms of the same power, we obtained the necessary recurrence relations that govern the coefficients of the series.

power series

A **power series** is a series of the form \( \sum_{n=0}^{\infty} a_n z^n \) where \(a_n\) are coefficients and \(z\) is a variable. Power series are invaluable in differential equations, providing a method for approximating functions.

In our problem, assuming \( y(z) = \sum_{n=0}^{\infty} a_n z^n \) enabled a structured approach to solve the differential equation.

Steps to use a power series:

• Substitute the power series form into the differential equation.
• Differentiate term-by-term as needed.
• Match coefficients of like powers of \(z\) to form a system of equations.

This method found the series solutions satisfying the differential equation.

exponential functions

**Exponential functions** like \(y_1(z) = \exp(z^2)\) are a class of functions where the variable is in the exponent. These functions grow quickly and appear frequently in solutions to differential equations.

In our case, \(\exp(z^2)\) was identified as a solution.

Steps to verify the solution:

• Compute the first and second derivatives of \(\exp(z^2)\).
• Substitute these into the original differential equation.
• Verify that the resulting equations hold true.

This demonstrated that \(\exp(z^2)\) indeed satisfied the differential equation.

integrating factor method

The **integrating factor method** is a technique to solve first-order linear differential equations of the form \(u^{\prime\prime} + 4z u^{\prime} = 0\). This method involves:
1. Finding an integrating factor, often denoted as \(\mu(z)\).

In practice:

• Multiply the entire differential equation by the integrating factor.
• Transform the equation into an exact differentiation so it can be integrated easily.

For our problem, we solved for \(u(z)\) by finding the integrating factor that simplified solving the differential equation. This led to the explicit form of \(y_2(z)\).