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Consider the following differential equations: a. \(x y^{\prime}=y\); b. \(x^{2} y^{\prime}=y\). In each case, find a power series solution of the form \(\Sigma a_{n} x^{n}\), solve the equation directly, and explain any discrepancies that arise.

Short Answer

Expert verified
For equation a, solutions match; for equation b, both solutions match as well.

Step by step solution

01

Solve equation a directly

Equation a is \(x y' = y\). Rearrange it to \( y' = \frac{y}{x} \), which is a separable equation. Integrate both sides with respect to \(x\) to get \( \ln |y| = \ln |x| + C \), where \(C\) is an integration constant. Simplify to find \( y = Cx \).
02

Power series solution for equation a

Assume a power series solution \( y = \sum_{n=0}^{ft} a_n x^n \). Then \( y' = \sum_{n=1}^{ft} n a_n x^{n-1} \). Substitute these into the equation \( x y' = y \), obtaining \( x \sum_{n=1}^{ft} n a_n x^{n-1} = \sum_{n=0}^{ft} a_n x^n \). Simplify to \( \sum_{n=1}^{ft} n a_n x^n = \sum_{n=0}^{ft} a_n x^n \). Equating coefficients gives \( n a_n = a_{n-1} \), implying \( a_n = 0 \) for \( n\geq 2 \). Thus, \( y = a_0 + a_1 x \).
03

Solve equation b directly

Equation b is \(x^2 y' = y\). Rearrange to \( y' = \frac{y}{x^2} \). This is a separable differential equation. Integrate both sides with respect to \(x\), yielding \( y = -\frac{y}{x} + C \). Simplify to solve for \( y = \frac{C}{x} \).
04

Power series solution for equation b

Assume a power series solution \( y = \sum_{n=0}^{ft} a_n x^n \). Then \( y' = \sum_{n=1}^{ft} n a_n x^{n-1} \). Substitute these into \( x^2 y' = y \), obtaining \( x^2 \sum_{n=1}^{ft} n a_n x^{n-1} = \sum_{n=0}^{ft} a_n x^n \). Simplify to \( \sum_{n=1}^{ft} n a_n x^{n+1} = \sum_{n=0}^{ft} a_n x^n \). Adjust indices and equate coefficients to find \( a_n = 0 \) for \( n>1 \), giving solution \( y = \frac{a_0}{x} \), consistent with direct solution.

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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 for solving differential equations. It's especially useful when traditional methods are difficult to apply. Here, we use a series, typically represented as \[ y = \sum_{n=0}^{\infty} a_n x^n \] to express our solution in terms of powers of \(x\). The series allows us to identify patterns or relationships between coefficients in the series that satisfy the differential equation.For instance, when solving equation (a), \(x y' = y\), we express \( y \) and \( y' \) as power series, where \[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \]. Substituting these into the equation involves aligning terms, so that the powers of \(x\) on both sides match. For equation (a), this process gave us a solution \( y = a_0 + a_1 x \), indicating that higher-order terms are zero.In contrast, equation (b), \(x^2 y' = y\), requires shifting series indexes and equating coefficients carefully. The solution we find, \( y = \frac{a_0}{x} \), reflects the direct solution found by traditional methods. The consistency across methods helps validate the approach.
Separable Differential Equations
Separable differential equations are a type of differential equation that can be rearranged to allow separation of variables. This method involves expressing the equation in terms of one variable on each side.Consider equation (a), \(x y' = y\). This can be rewritten as \[ y' = \frac{y}{x} \], allowing us to separate variables to \[ \frac{dy}{y} = \frac{dx}{x} \]. By integrating both sides, we solve for \(y\) in terms of \(x\). The integration involves adding a constant \(C\), resulting in \[ \ln |y| = \ln |x| + C \]. This simplifies to a solution, \( y = Cx \). Here, dividing and integrating each side is straightforward due to the ability to separate the variables associated with \(y\) and \(x\). The process helps simplify complex equations, providing a clear path to a solution while highlighting the utility of integration constants.
Integration Constants
Integration constants emerge when solving differential equations, especially during integration. As indefinite integrals introduce an arbitrary constant, these constants represent the family of solutions tied to integration.In both equations, the constants represent initial conditions or specific solutions. For example, in equation (a), after integrating \[ \ln |y| = \ln |x| + C \], solving results in \[ y = Cx \]. The constant \(C\) relies on initial conditions or boundary conditions defining the specific solution amongst many potential ones. Similarly, in equation (b), after separating and integrating, we deduce \[ y = \frac{C}{x} \].This constant is essential for illustrating how initial or boundary conditions affect the solution's form. Moreover, it demonstrates the general solution's flexibility to adapt to different scenarios or conditions. Without such constants, solutions would lack context and completeness.

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Most popular questions from this chapter

Find the indicial equation and its roots for each of the following differ-. ential equations: a. \(x^{3} y^{\prime}+(\cos 2 x-1) y^{\prime}+2 x y=0\) b. \(4 x^{2} y^{\prime \prime}+\left(2 x^{4}-5 x\right) y^{\prime}+\left(3 x^{2}+2\right) y=0\).

The equation \(y^{\prime \prime}+\left(p+\frac{1}{2}-\frac{1}{4} x^{2}\right) y=0\), where \(p\) is a constant, certainly has a series solution of the form \(y=\Sigma a_{n} x^{n}\). a. Show that the coefficients \(a_{n}\) are related by the threc-term recursion formula $$ (n+1)(n+2) a_{n+2}+\left(p+\frac{1}{2}\right) a_{n}-\frac{1}{4} a_{n-2}=0 $$ b. If the dependent variable is changed from \(y\) to \(w\) by means of \(y=\) \(w e^{-x^{2} / 4}\), show that the equation is transformed into \(w^{*}-x w^{\prime}+\) \(p w=0\) c. Verify that the equation in (b) has a two-term recursion formula and find its general solution.

a. Show that the series for \(\cos x\), $$ y=1-\frac{x^{2}}{1 \cdot 2}+\frac{x^{4}}{1 \cdot 2 \cdot 3 \cdot 4}-\frac{x^{6}}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}+\cdots $$ has the property that \(y^{\prime \prime}=-y\), and is therefore a solution of equation (1). b. Show that the series $$ y=1-\frac{x^{2}}{2^{2}}+\frac{x^{4}}{2^{2} \cdot 4^{2}}-\frac{x^{6}}{2^{2} \cdot 4^{2} \cdot 6^{2}}+\cdots $$ converges for all \(x\), and verify that it is a solution of equation (2). [Observe that this series can be obtained from the one in (a) by replacing each odd factor in the denominators by the next greater even number. The sum of this series is a useful special function denoted by \(J_{0}(x)\) and called the Bessel function of order 0 ; it will be studied in detail in the next chapter.]

Consider the differential equation $$ y^{\prime \prime}+\frac{p}{x^{b}} y^{\prime}+\frac{q}{x^{4}} y=0 $$ where \(p\) and \(q\) are nonzero real numbers and \(b\) and \(c\) are positive integers. It is clear that \(x=0\) is an irregular singular point if \(b>1\) or \(c>2\). a. If \(b=2\) and \(c=3\), show that there is only one possible value of \(m\) for which there might exist a Frobenius series solution. b. Show similarly that \(m\) satisfies a quadratic equation -and hence we can hope for two Frobenius series solutions, corresponding to the roots of this equation-if and only if \(b=1\) and \(c \leq 2\). Observe that these are exactly the conditions that characterize \(x=0\) as a "weak" or regular singular point as opposed to a "strong" or irregular singular point.

Consider the following differential equations: a. \(y^{\prime}=2 x y\) : b. \(y^{\prime}+y=1\). In each case, find a power series solution of the form \(\Sigma a_{n} x^{n}\), try to recognize the resulting series as the expansion of a familiar function, and verify your conclusion by solving the equation directly.

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