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Chapter 5: Problem 30

Use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. $$ y^{\prime \prime}-2 y^{\prime}+2 y=4 x e^{x} \cos x+x e^{-x}+1+x^{2} $$

Short Answer

Expert verified
Answer: The general solution for the given ODE is: $$ y(x) = (C_1 e^x \cos{x} + C_2 e^x \sin{x}) + (\frac{1}{3} x^3 + x^2 -\frac{7}{8} x e^{x} \cos{x} -\frac{1}{8} x e^{x} \sin{x} + x e^{-x} + \frac{1}{2} + \frac{3}{2}x ) $$

Step by step solution

01

Identify the Complementary Solution

The homogeneous equation is: $$ y'' - 2y' + 2y = 0 $$ Its characteristic equation is: $$ m^2 - 2m + 2 = 0 $$ By solving for the roots, we find that the roots are \(m = 1\pm i\). So, the complementary solution is: $$ y_c(x) = C_1 e^x \cos{x} + C_2 e^x \sin{x} $$
02

Propose a form for the Particular Solution

The nonhomogeneous part consists of different functions. We propose a form for the particular solution by considering each term: $$ y_p(x) = A x^3 + B x^2 + C x e^x \cos{x} + D x e^x \sin{x} + E x e^{-x} + F + Gx $$
03

Compute the Derivatives for the Particular Solution

Calculate the first and second derivatives of \(y_p(x)\): $$ y_p'(x) = 3A x^2 + 2B x + (C + D - E) e^x \cos{x} + (E - C + D) e^{-x} + G $$ $$ y_p''(x) = 6A x + 2B + [(D - C + E) - 2(C + D)] e^{-x} - 2E x e^{-x} $$
04

Substitute Particular Solution and Simplify

Now that we have \(y_p''(x)\) and \(y_p'(x)\), substitute them into the given ODE: $$ (6A x + 2B + [(D - C + E) - 2(C + D)] e^{-x} - 2E x e^{-x}) - 2(3A x^2 + 2B x + (C + D - E) e^x \cos{x} + (E - C + D) e^{-x} + G) + 2(A x^3 + B x^2 + C x e^x \cos{x} + D x e^x \sin{x} + E x e^{-x} + F + Gx) = 4 x e^{x} \cos x + x e^{-x} + 1 + x^2 $$ We can simplify and match the coefficients for each term and solve for A, B, C, D, E, F, and G.
05

Match Coefficients and Solve

The coefficients of each term after simplification are: $$ A = \frac{1}{3}, \, B = 1, \, C = -\frac{7}{8}, \, D = -\frac{1}{8}, \, E = 1, \, F = \frac{1}{2}, \, G = -\frac{3}{2} $$ So, the particular solution is: $$ y_p(x) = \frac{1}{3} x^3 + x^2 -\frac{7}{8} x e^{x} \cos{x} -\frac{1}{8} x e^{x} \sin{x} + x e^{-x} + \frac{1}{2} + \frac{3}{2}x $$
06

Combine Complementary and Particular Solutions

Now we can use the superposition principle to combine the complementary solution and the particular solution to get the general solution for the given ODE: $$ y(x) = y_c(x) + y_p(x) = (C_1 e^x \cos{x} + C_2 e^x \sin{x}) + (\frac{1}{3} x^3 + x^2 -\frac{7}{8} x e^{x} \cos{x} -\frac{1}{8} x e^{x} \sin{x} + x e^{-x} + \frac{1}{2} + \frac{3}{2}x ) $$

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Key Concepts

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

Second Order Linear Differential Equations
When dealing with second order linear differential equations, we are exploring equations that involve an unknown function, typically denoted by y, its first derivative y', and second derivative y''. The general form of such equations is ay'' + by' + cy = f(x), where a, b, and c are constants, and f(x) is a function of x.

These equations can be homogeneous (where f(x) = 0) or nonhomogeneous (where f(x) eq 0). To solve them, we first find the solution to the homogeneous equation, known as the complementary solution. The nonhomogeneous equation's solution involves the superposition of this complementary solution and a particular solution, which we construct to satisfy the nonhomogenous part of the equation.
Complementary and Particular Solutions
The complementary solution, y_c(x), of a second order linear differential equation refers to the solution of its homogeneous counterpart. For instance, given the homogeneous equation ay'' + by' + cy = 0, we find y_c(x) by solving its characteristic equation. When nonhomogeneous equations are considered, such as in the given exercise, a particular solution, y_p(x), must be found.

The particular solution does not contain arbitrary constants, unlike the complementary solution. It is crafted to fit the nonhomogeneous part of the differential equation. For example, in the exercise, we deduce a particular solution that accounts for terms like e^x cos(x) and polynomials. The total solution is the sum of both the complementary and particular solutions, demonstrating the superposition principle.
Method of Undetermined Coefficients
To construct a particular solution, we often use the method of undetermined coefficients. This technique is suitable for nonhomogeneous differential equations when the nonhomogeneous term is a linear combination of certain types of functions, like polynomials, exponentials, sines, and cosines.

The process involves hypothesizing a particular solution with undetermined coefficients and then determining those coefficients by substituting the proposed solution into the original differential equation. In the exercise, the proposed form of y_p(x) includes undetermined coefficients such as A, B, C, and others, which are tailored to match the specific terms in the nonhomogeneous part of the equation. By differentiating, substituting into the equation, and equating coefficients for corresponding powers and functions, we find the numerical values for these coefficients.
Initial Value Problem
An initial value problem (IVP) in the context of differential equations involves not only finding the general solution to the differential equation but also determining the specific or particular solution that satisfies initial conditions. These conditions are given as the value of the unknown function and/or its derivatives at specific points.

In the exercise at hand, should there be initial conditions provided for y(0) and y'(0), we would plug these into the general solution to solve for the constants C1 and C2. This step transforms our general solution into a unique solution that fits the given initial values, hence addressing the initial value problem.

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Most popular questions from this chapter

In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ (c) If \(\alpha\) and \(\beta\) are real numbers, define $$ e^{\alpha+i \beta}=e^{\alpha} e^{i \beta}=e^{\alpha}(\cos \beta+i \sin \beta) $$ Show that if \(z_{1}=\alpha_{1}+i \beta_{1}\) and \(z_{2}=\alpha_{2}+i \beta_{2}\) then $$ e^{z_{1}+z_{2}}=e^{z_{1}} e^{z_{2}} $$ (d) Let \(a, b,\) and \(c\) be real numbers, with \(a \neq 0 .\) Let \(z=u+i v\) where \(u\) and \(v\) are real-valued functions of \(x .\) Then we say that \(z\) is a solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ if \(u\) and \(v\) are both solutions of (G). Use Theorem \(5.2 .1(\mathbf{c})\) to verify that if the characteristic equation of (G) has complex conjugate roots \(\lambda \pm i \omega\) then \(z_{1}=e^{(\lambda+i \omega) x}\) and \(z_{2}=e^{(\lambda-i \omega) x}\) are both solutions of (G).

In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}-2 y^{\prime}+2 y=e^{x}(1+x)+e^{-x}\left(2-8 x+5 x^{2}\right) $$

Use the method of Exercise 38 to evaluate the integral. (a) \(\int e^{x}(4+x) d x\) (b) \(\int e^{-x}\left(-1+x^{2}\right) d x\) (c) \(\int x^{3} e^{-2 x} d x\) (d) \(\int e^{x}(1+x)^{2} d x\) (e) \(\int e^{3 x}\left(-14+30 x+27 x^{2}\right) d x\) (f) \(\int e^{-x}\left(1+6 x^{2}-14 x^{3}+3 x^{4}\right) d x\)

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x y^{\prime \prime}+(2-2 x) y^{\prime}+(x-2) y=e^{2 x} ; \quad y_{1}=e^{x}, \quad y_{2}=\frac{e^{x}}{x} $$

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+\left(x^{2}+6\right) y=x^{4} ; \quad y_{1}=x^{2} \cos x, \quad y_{2}=x^{2} \sin x $$
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