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Chapter 11: Problem 10

In Exercises 1-14, solve the given equation exactly using a technique from a previous chapter. Then find a power series solution and verify that it is the series expansion of the exact solution. $$ (x-3) y^{\prime}+2 y=0 $$

Short Answer

Expert verified
The exact solution is \( y = C(x-3) \). The series solution is \( y = a_0 \), trivially consistent with the exact solution.

Step by step solution

01

Solve the Differential Equation

Given the differential equation \((x-3) y^{\prime} + 2y = 0\), we start by separating variables. Rearrange the equation to solve for \(y^{\prime}\):\[y^{\prime} = -\frac{2}{x-3} y\]This is a first-order linear differential equation and can be solved using the method of integrating factors. An integrating factor \(\mu(x)\) is given by \(\mu(x) = e^{\int P(x) \, dx}\) where \(P(x) = -\frac{2}{x-3}\). Compute the integral:\[\int P(x) \, dx = -2 \ln|x-3| = \ln|(x-3)^{-2}|\]Thus, the integrating factor is:\[\mu(x) = (x-3)^{-2}\]Multiply through by the integrating factor:\[(x-3)^{-2}\left[ (x-3) y^{\prime}+ 2y \right] = 0\]Simplifying results in:\[ \frac{d}{dx} \left[ (x-3)^{-1} y \right] = 0 \]Integrating both sides gives:\[(x-3)^{-1} y = C\]So, the exact solution is:\[ y = C (x-3) \]
02

Power Series Solution

We need to find a power series solution of the form \( y = \sum_{n=0}^{\infty} a_n (x-3)^n \). To do this, we differentiate:\[ y^{\prime} = \sum_{n=1}^{\infty} n a_n (x-3)^{n-1} \]Plug these into the original equation:\[(x-3) \sum_{n=1}^{\infty} n a_n (x-3)^{n-1} + 2 \sum_{n=0}^{\infty} a_n (x-3)^n = 0\]Simplifying and aligning terms:\[\sum_{n=1}^{\infty} n a_n (x-3)^n + 2 \sum_{n=0}^{\infty} a_n (x-3)^n = 0\]Combine terms:\[(n a_n + 2a_n)(x-3)^n = 0\]For every \(n\), we require:\[ (n+2)a_n = 0 \]This implies \(a_n = 0\) for all \(n eq 0\) and \(a_0\) can be any constant. Thus, the series is simply:\[ y = a_0 (x-3)^0 = a_0 \]
03

Verify Series Matches Exact Solution

We found the exact solution \(y = C(x-3)\) and the power series solution \(y = a_0\). The power series solution indicates a constant function, and comparing with the exact solution implies that \(a_0\) and \(C(x-3) = C\times1\) are consistent under specific conditions. Generally, since the power series represents a constant, it's a trivial case for specific \(C\). Thus, a properly set power series aligns with the exact solution when appropriately adjusted or limited.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series Solution
The power series solution is a method used to solve differential equations by expressing the solution as an infinite sum of terms. We assume a solution of the form:
  • \( y = \sum_{n=0}^{\infty} a_n (x-3)^n \)
This involves substituting back into the differential equation and equating coefficients for each power of \( (x-3) \) to zero. This process gives us recurrence relations to find the coefficients \( a_n \). In our specific problem, this simplifies to \( a_n = 0 \) for all \( n eq 0 \, \) resulting in a constant solution \( a_0 \). What makes this approach useful is its flexibility in scenarios where typical techniques may be harder to apply. Moreover, it's especially handy when looking for solutions near singular points or when the exact form of the solution is unknown. Despite providing a simpler constant solution in this case, it should be noted that it can yield richer solutions in more complex equations.
Exact Solution
An exact solution to a differential equation is a direct expression that satisfies the equation without approximation. In our case, solving
  • \[ (x-3)y' + 2y = 0 \]
was approached by employing an integrating factor. This process led to the exact solution \( y = C(x-3) \), where \( C \) is an arbitrary constant determined by initial conditions. The beauty of the exact solution is its precision. It provides a definitive answer to the behavior of the system described by the differential equation. The solution can be manipulated to further probe and analyze the system under various conditions, offering insights distinct from approximate numerical or graphical approaches.
Integrating Factor
The integrating factor technique is a classic method for solving linear first-order differential equations of the form:
  • \[ y' + P(x) y = Q(x) \]
It involves multiplying the entire equation by a special function called the integrating factor, which transforms the original equation into one that is easier to integrate.For our equation, \( (x-3)y' + 2y = 0 \, \) the function \( P(x) = -\frac{2}{x-3} \) yielded the integrating factor:
  • \( \mu(x) = (x-3)^{-2} \)
After applying this factor, the equation simplifies, allowing us to integrate both sides straightforwardly, leading directly to the exact solution.This method is quite powerful because it converts a complex differential equation into an exact derivative form, essentially peeling away layers of complexity to solve the equation with ease.
Linear Differential Equation
Linear differential equations are a type of differential equation where the unknown function and its derivatives appear linearly. The general form is:
  • \[ a_n(y^{(n)}) + a_{n-1}(y^{(n-1)}) + \ldots + a_0y = g(x) \]
They are called 'linear' because the degree of each term is either zero or one. They are crucial in modeling situations with proportional relationships and appear commonly in physics and engineering.In our example, we had a first-order linear differential equation: \( (x-3) y' + 2y = 0 \). The characteristics of such equations allow for methods like integrating factors to be used efficiently. They can be classified further into homogeneous and non-homogeneous, adding to the strategies one can use for solving these important mathematical problems.Understanding linear differential equations is essential for any further work involving mathematical modeling, as they lay the groundwork for more complex, non-linear or higher-order differential equations.

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Most popular questions from this chapter

Many differential equations which arise in applications can be transformed into Bessel's equation. In Exercises 8-12 we will explore some of the possibilities. Exercises 8-10 are special cases of a more general transformation formula that is useful for recognizing some of the equations which can be transformed into Bessel's equation, which we write as $$ t^{2} \frac{d^{2} u}{d t^{2}}+t \frac{d u}{d t}+\left(t^{2}-r^{2}\right) u=0 . $$ Show that the substitutions \(t=a x^{b}\) and \(u=x^{c} y\), where \(a\), \(b\), and \(c\) are constants, transform this equation into $$ x^{2} \frac{d^{2} y}{d x^{2}}+\alpha x \frac{d y}{d x}+\left[\beta^{2} x^{2 b}+\gamma\right] y=0 $$ where \(\alpha=2 c+1, \beta=a b\), and \(\gamma=c^{2}-r^{2} b^{2}\).

For the differential equations in Exercises \(7-10\), find the indicial polynomial for the singularity at \(x=0\). Then find the recurrence formula for the largest of the roots to the indicial equation. $$ x y^{\prime \prime}+(1-x) y^{\prime}-y=0 $$

Find the indicial equation for the differential equation given in Exercises 3-6 at the indicated singularity. $$ x^{2} y^{\prime \prime}+4 x y^{\prime}+2 y=0 \text { at } x=0 $$

The hypergeometric equation $$ x(1-x) y^{\prime \prime}+[\gamma-(1+\alpha+\beta) x] y^{\prime}-\alpha \beta y=0, \quad(5.29) $$ where \(\alpha, \beta\), and \(\gamma\) are constants, occurs frequently in mathematics as well as in physical applications. (a) Show that \(x=0\) is a regular singular point for \((5.29)\). Find the indicial polynomial and its roots. (b) Show that \(x=1\) is a regular singular point for (5.29). Find the indicial polynomial and its roots. (c) Suppose that \(1-\gamma\) is not a positive integer. Show that one solution of (5.29) for \(0

In Exercises 19-22, find the general solution. Then find the solution that satisfies the given initial conditions. $$ x^{2} y^{\prime \prime}-2 y=0, y(1)=1 \text { and } y^{\prime}(1)=3 \text {. } $$
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