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Prove: If \(\alpha\) and \(M\) are constants and \(M \neq 0\) then constant coefficient equation $$ a y^{\prime \prime}+b y^{\prime}+c y=M e^{\alpha x} $$ has a particular solution \(y_{p}=A e^{\alpha x}(A=\) constant \()\) if and only if \(e^{\alpha x}\) isn't a solution of the complementary equation.

Short Answer

Expert verified
Question: Prove that, for a given second-order linear non-homogeneous differential equation in the form \(ay'' + by' + cy = Me^{\alpha x}\) where \(a\), \(b\), \(c\), \(M\), and \(\alpha\) are constants and \(M \neq 0\), there exists a particular solution of the form \(y_p = Ae^{\alpha x}\) if and only if \(e^{\alpha x}\) is not a solution of the complementary homogeneous equation \(ay'' + by' + c = 0\). Answer: We proved that a particular solution of the form \(y_p = Ae^{\alpha x}\) exists if and only if \(e^{\alpha x}\) is not a solution of the complementary equation \(ay'' + by' + c = 0\). This is because, in the case where \(e^{\alpha x}\) is not a solution of the complementary equation, we can find a constant \(A\) to form a valid particular solution, whereas in the case where \(e^{\alpha x}\) is a solution of the complementary equation, it is impossible to obtain a valid particular solution of the given form.

Step by step solution

01

Solve the complementary equation

To find the general solution of the complementary homogeneous equation \(ay'' + by' + c = 0\), we will first find its characteristic equation, which is a quadratic in the form \(ar^2 + br + c = 0\). The general solution of the complementary equation is then given by \(y_c(x) = C_1e^{r_1x} + C_2e^{r_2x}\), where \(r_1\) and \(r_2\) are the roots of the characteristic equation and \(C_1\) and \(C_2\) are arbitrary constants.
02

Substitute the particular solution

Now we will substitute the proposed form for the particular solution, \(y_p = Ae^{\alpha x}\), into the given non-homogeneous equation. That means, we will calculate the first and second derivatives of \(y_p(x) = Ae^{\alpha x}\) and substitute these into the equation: - First derivative: \(y_p'(x) = A\alpha e^{\alpha x}\) - Second derivative: \(y_p''(x) = A\alpha^2 e^{\alpha x}\) Substituting these into the equation \(ay'' + by' + cy = Me^{\alpha x}\) yields: $$(aA\alpha^2 + bA\alpha + cA)e^{\alpha x} = Me^{\alpha x}$$
03

Analyze the coefficients

To satisfy the original equation, the coefficients of the exponential terms must be equal. Therefore, we get the following condition: $$aA\alpha^2 + bA\alpha + cA = M$$ Since \(A\) is a constant, we can divide both sides of the equation by it, obtaining: $$a\alpha^2 + b\alpha + c = \frac{M}{A}$$ Now we analyze the cases:
04

Case 1: \(e^{\alpha x}\) is not a solution of the complementary equation

If \(e^{\alpha x}\) is not a solution of the complementary equation, then \(\alpha\) is not a root of the characteristic equation, which means the equation \(a\alpha^2 + b\alpha + c = 0\) is not satisfied. In this case, we can solve for \(A\) to find the constant that makes the particular solution valid: $$A = \frac{M}{a\alpha^2 + b\alpha + c}$$ This gives us a valid particular solution of the form \(y_p = Ae^{\alpha x}\), proving the first part of the statement.
05

Case 2: \(e^{\alpha x}\) is a solution of the complementary equation

If \(e^{\alpha x}\) is a solution of the complementary equation, then \(\alpha\) is a root of the characteristic equation, which means the equation \(a\alpha^2 + b\alpha + c = 0\) is satisfied. Thus, we have: $$a\alpha^2 + b\alpha + c = 0$$ The equation above implies that \(M/A = 0\). However, since \(M \neq 0\), it is impossible to obtain a valid particular solution of the form \(y_p = Ae^{\alpha x}\), proving the second part of the statement. Finally, we have shown that a particular solution of the form \(y_p = Ae^{\alpha x}\) exists if and only if \(e^{\alpha x}\) is not a solution of the complementary equation, proving the given statement.

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Key Concepts

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

Characteristic Equation
The characteristic equation is at the heart of understanding constant coefficient differential equations. It's a tool we use to find solutions to linear homogeneous differential equations, which appear frequently in fields like physics and engineering. Simply put, for an equation with constant coefficients such as
\[ ay'' + by' + c = 0,\]
we substitute \( y = e^{rx} \) (where \( r \) is a number we want to find) into the equation. This gives us a new equation, \( ar^2 + br + c = 0 \), where \( a \), \( b \), and \( c \) are given constants. This polynomial equation is what we call the characteristic equation.

Finding the roots of the characteristic equation, which may be real or complex, helps us establish the complementary solution to the homogeneous equation. These roots represent the exponents in the exponential solutions that make up the complementary function, often denoted as \( y_c(x) \).
Particular Solution
Working with non-homogeneous differential equations involves finding a particular solution, \( y_p \), that satisfies the non-homogeneous part of the equation. This solution typically reflects the external force or influence not accounted for by the complementary homogeneous equation.

In the context of our exercise, where the non-homogeneous term is \( Me^{\alpha x} \), we look for a particular solution of the same form, \( y_p = Ae^{\alpha x} \). We assume this form based on the method of undetermined coefficients. After substituting \( y_p \) into the original differential equation, we end up with an equation that allows us to solve for the constant \( A \), thus finding our particular solution.
To determine \( A \), we need to ensure that the exponential term \( e^{\alpha x} \) is not a solution to the complementary equation. If it is not, we can directly solve for \( A \), but if it is, the particular solution must be modified, and a simple \( Ae^{\alpha x} \) will not suffice.
Complementary Homogeneous Equation
Associated with every non-homogeneous differential equation is a corresponding complementary homogeneous equation. It represents the system's response devoid of any external forces or sources. For the equation
\[ ay'' + by' + c = Me^{\alpha x},\]
the complementary homogeneous equation would be
\[ ay'' + by' + c = 0.\]
The solutions to this equation form a general solution called the complementary function, usually denoted as \( y_c(x) \), which is constructed from the linear combination of the solutions to the characteristic equation.

The complementary solution is crucial because it gives us the basis of the system's natural behavior. When solving the entire non-homogeneous equation, we combine this complementary solution with the particular solution to form the general solution to the original differential equation.
Non-Homogeneous Differential Equation
A non-homogeneous differential equation, unlike its homogeneous counterpart, includes a non-zero term that accounts for an external input or force acting on the system. In our exercise example, the equation
\[ ay'' + by' + cy = Me^{\alpha x} \]
features a non-homogeneous term represented by \( Me^{\alpha x} \), where \( M \) and \( \alpha \) are constants. The non-homogeneous term can take various forms, like trigonometric functions, polynomials, or exponential functions, reflecting different kinds of external influences.

The entire solution to a non-homogeneous differential equation consists of two parts: the complementary solution, \( y_c(x) \), which we get from the homogeneous equation, and the particular solution, \( y_p(x) \), which fits the non-homogeneous part. The full general solution is then the sum of these two parts.
In some cases, the particular solution can affect the form of the complementary solution—especially if the non-homogeneous term resembles any part of the complementary solution, leading to a necessary adjustment in our approach to finding the particular solution.

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Most popular questions from this chapter

Find the general solution, given that \(y_{1}\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation. \(x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=4 x^{4} ; \quad y_{1}=x^{2}\)

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

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).

Solve the initial value problem, given that \(y_{1}\) satisfies the complementary equation. \(x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=4 x^{4}, \quad y(-1)=7, \quad y^{\prime}(-1)=-8 ; \quad y_{1}=x^{2}\)

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-2 x y^{\prime}+\left(x^{2}+2\right) y=x^{3} \cos x ; \quad y_{1}=x \cos x, \quad y_{2}=x \sin x $$

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