Question

Consider the differential equation $$ \frac{d y}{d x}=\mathrm{e}^{y / x} $$ (a) Suppose \(y(1)=0\). Give a series expansion for \(y(x)\) which is valid for \(x\) near 1 . Neglect terms of order \((x-1)^{4}\) (b) Suppose \(y\left(x_{0}\right)=+\infty\left(x_{0}>0\right)\). Give an approximate expression for \(y(x)\) which is useful for \(x\) slightly less \(\operatorname{than} x_{0}\).

Short answer

(a) \(y(x) = (x-1) + (x-1)^2 + \frac{1}{3}(x-1)^3\). (b) \(y(x) \approx x \ln(x_0 - x)\) near \(x_0\).

Step-by-step solution

Problem Analysis

Given the differential equation \( \frac{d y}{d x} = e^{y / x} \). We are asked to find a series expansion for \( y(x) \) around \( x = 1 \) and an approximate expression for \( y(x) \) near \( x_0 \) where \( y(x_0) = +\infty \).

Consider initial condition

Let us solve part (a) first. Using the initial condition \( y(1) = 0 \), seek a power series solution in the form \( y(x) = c_0 + c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + \cdots \).

Substitute initial condition

Substitute \( y(1) = 0 \), which gives \( c_0 = 0 \). So, the series starts with \( y(x) = c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + \cdots \).

Differentiate the series

Differentiate the series \( y(x) \) term by term: \( y'(x) = c_1 + 2c_2(x-1) + 3c_3(x-1)^2 + \cdots \).

Substitute back the series into the differential equation

To match the given differential equation \( \frac{d y}{d x} = e^{y / x} \), substitute \( y(x) \) and its derivative: \( c_1 + 2c_2(x-1) + 3c_3(x-1)^2 + \cdots = e^{(c_1(x-1) + c_2(x-1)^2 + \cdots)/x} \).

Expand the exponential function

Expand the exponential function as a power series and equate coefficients of like powers of \( (x-1) \) to solve for \( c_1, c_2, c_3, \) etc.

Solve for coefficients

By equating coefficients, find \( c_1 = 1 \), \( c_2 = 1 \), and \( c_3 = \frac{1}{3} \), neglecting higher-order terms. Therefore, \( y(x) = (x-1) + (x-1)^2 + \frac{1}{3}(x-1)^3 \).

Solve part (b)

For part (b), where \( y(x_0) = +\infty \), express \( y(x) = x \ln(x_0-x) + \cdots \).

Determine behavior near \(x_0\)

Analyze that as \( x \) approaches \( x_0 \) from the left, \( y(x) \) goes to infinity logarithmically.

Key concepts

series expansion

When solving differential equations, a common technique is to use a series expansion. This involves expressing the solution as a sum of terms in increasing powers of the variable. In our given differential equation, we seek a series expansion for the solution around the point where we know the initial condition. For example, if we know that the solution y satisfies certain conditions at x=1, we express y as:

\[ y(x) = c_0 + c_1 (x-1) + c_2 (x-1)^2 + c_3 (x-1)^3 + \cdots \]
This type of expansion is particularly useful because it allows us to approximate the solution near the point of expansion by considering only a few terms and neglecting higher-order terms, which is specified as neglecting terms of order (x-1)^4 or higher in our case.

initial conditions

Initial conditions help us determine the specific solution to a differential equation from the general solution. By substituting the initial conditions into the series expansion, we can solve for the coefficients of the series. For instance, the initial condition y(1)=0 means we substitute x=1 and y=0 into our series:

• y(1) = c_0

From this, we get c_0 = 0. The initial condition simplifies our series to:

\[ y(x) = c_1 (x-1) + c_2 (x-1)^2 + c_3 (x-1)^3 + \cdots \]

exponential function

The differential equation involves the exponential function e^{y/x}. This function is key to solving the equation and we often need to expand it as a power series. The exponential function can be expanded as:

\[ e^{y/x} = 1 + \frac{y}{x} + \frac{(y/x)^2}{2!} + \frac{(y/x)^3}{3!} + \cdots \]
This helps us match the terms of the left side of the differential equation with the series expansion of the exponential function, which simplifies the process of solving for the coefficients of the series for y(x).

power series

A power series is an infinite sum of terms in the form of c_n(x-a)^n, where n is a non-negative integer, and c_n are coefficients. In solving our differential equation, we express y(x) as a power series around x=1 (where a=1):

\[ y(x) = \sum_{n=0}^{\infty}c_n (x-1)^n \]
Each term in the series brings more accuracy to the solution. In our specific problem, we found:

\[ y(x) = (x-1) + (x-1)^2 + \frac{1}{3} (x-1)^3 \]
This power series approximates y(x) near x=1, neglecting terms of order (x-1)^4 and higher. Power series are valuable because they allow us to approximate complex equations with simpler polynomial terms.