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Chapter 5: Problem 25

In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}-7 y^{\prime}+12 y=-e^{x}(17-42 x)-e^{3 x} $$

Short Answer

Expert verified
Answer: The general solution is \(y = C_1 e^{4x} + C_2 e^{3x} - xe^{x} -2e^{x} + 2xe^{3x}\), where \(C_1\) and \(C_2\) are constants.

Step by step solution

01

Find the Complementary Solution

To find the complementary solution, we consider the homogeneous version of the given equation. $$ y^{\prime \prime} - 7 y^{\prime} + 12 y = 0 $$ The characteristic equation of this ODE is given by: $$ r^2 - 7r + 12 = 0 $$ Solve this equation for \(r\).
02

Solve the Characteristic Equation

The characteristic equation is a quadratic equation, factoring the quadratic equation, we get: $$ (r-4)(r-3)=0 $$ Solving for \(r\), we get \(r_1=4\) and \(r_2=3\).
03

Write the Complementary Solution

Using the roots of the characteristic equation, we can write the complementary solution, \(y_c\), as: $$ y_c = C_1 e^{4x} + C_2 e^{3x} $$ Where \(C_1\) and \(C_2\) are constants.
04

Write the Particular Solution Form

Now, we need to find the particular solution of the non-homogeneous equation: $$ y^{\prime \prime} - 7 y^{\prime} + 12 y = -e^{x}(17-42x) - e^{3x} $$ For \(-e^{x}(17-42x)\), our guess for the particular solution form is \(Axe^{x} + Be^{x}\), where A and B are constants. For \(-e^{3x}\), since \(e^{3x}\) is part of the complementary solution, our guess for the particular solution form is \(Cxe^{3x}\), where C is a constant. So, the overall particular solution form is: $$ y_p = Axe^{x} + Be^{x} + Cxe^{3x} $$
05

Find the Particular Solution

Take the first and second derivatives of the guessed particular solution form and plug them into the non-homogeneous ODE. Then, solve for constants A, B, and C. First derivative: $$ y_p^{\prime} = (A+B) e^{x} + (A-3C) x e^{x} + 3Ce^{3x} $$ Second derivative: $$ y_p^{\prime\prime} = (A-B+2A-3C) xe^{x} + (3A-10C)e^x + 9Ce^{3x} $$ Plugging the derivatives into the non-homogeneous ODE, we get: $$ (A-B+2A-3C) x e^{x} + (3A-10C)e^x + 9Ce^{3x} -7((A+B)e^{x} + (A-3C) x e^{x} +3Ce^{3x}) + 12(Axe^x + Be^x +Cxe^{3x}) = -e^{x}(17-42x) - e^{3x} $$ After simplifying the equation and grouping similar terms, we get: $$ (-8x + 4)e^x +(42x - 9)e^{3x} = (-42x + 17)e^x - e^{3x} $$ Comparing the coefficients, we get: $$ \begin{cases} -8x + 4 = -42x + 17\\ 42x - 9 = -1 \end{cases} $$ Solving this system of linear equations, we get \(A=-1\), \(B=-2\), and \(C=2\).
06

Write the Particular Solution

Now, substitute the values of A, B, and C in the guessed particular solution form: $$ y_p = -xe^{x} -2e^{x} + 2xe^{3x} $$
07

Combine Complementary and Particular Solutions

Combine the complementary solution and particular solution to find the general solution of the non-homogeneous ODE: $$ y = y_c + y_p = C_1 e^{4x} + C_2 e^{3x} - xe^{x} -2e^{x} + 2xe^{3x} $$ This is the general solution of the given non-homogeneous ODE.

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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 a mathematical expression derived from a linear homogeneous ordinary differential equation (ODE). It serves as a pivotal tool in finding the solution to the ODE.

To obtain the characteristic equation, we replace each derivative of the unknown function with a power of some variable (often denoted as r). For instance, a second-order homogeneous ODE like y'' - 7y' + 12y = 0 leads to a characteristic equation of r2 - 7r + 12 = 0.

By solving this quadratic equation for r, which represents the rate of exponential growth or decay in the solution, we often get two solutions (roots), say r1 and r2. These roots dictate the form of the complementary solution to the homogeneous ODE.
Particular Solution
The particular solution, denoted typically as yp, is a specific solution to a non-homogeneous ODE that includes the function on the right side of the equation.

It is crafted to satisfy the non-homogeneous part of the original ODE exactly. For our given problem, the non-homogeneous ODE has the form y'' - 7y' + 12y = -ex(17-42x) - e3x. To propose a likely form for the particular solution, we employ the method of undetermined coefficients. This method involves guessing a solution form with undefined constants that we then determine by substituting back into the ODE.

After a bit of trial and error and ensuring the proposed particular solution is not a solution of the corresponding homogeneous ODE, we finalize a guess for the particular solution, which we then use to find constants that make it a true solution.
Principle of Superposition
The principle of superposition is a powerful concept that allows the construction of a solution to a linear differential equation from simpler, component solutions.

In the realm of linear ODEs, this principle states that if y1 and y2 are solutions to a homogeneous ODE, their linear combination C1y1 + C2y2 is also a solution, where C1 and C2 are constants.

Additionally, the total solution of a non-homogeneous ODE is the sum of its complementary solution (yc), which solves the corresponding homogeneous equation, and a particular solution (yp) that accounts for the non-homogeneous part. Together, they construct the general solution of the non-homogeneous ODE.
Homogeneous ODE
A homogeneous ordinary differential equation is one where every term is a function of the unknown variable and its derivatives alone, with no external functions or constants added.

An example of a homogeneous ODE is y'' - 7y' + 12y = 0. These types of equations are significant because their solutions exhibit specific patterns, and they form the complementary solution (yc) part of the general solution for a non-homogeneous ODE.

The homogeneous version forms the backbone of the ODE solution as it establishes the fundamental part of the response to the system that the differential equation models, irrespective of any external forces or inputs.
Non-homogeneous ODE
A non-homogeneous ODE includes additional functions or constants beyond the unknown variable and its derivatives. The equation y'' - 7y' + 12y = -ex(17-42x) - e3x is an example of such an equation.

These extra terms represent external forces or inputs to the system being modeled by the ODE. The solution to a non-homogeneous ODE requires the determination of a particular solution that specifically addresses the non-homogeneous part of the equation.

The method of undetermined coefficients or variation of parameters are techniques used to find the particular solution, which, when added to the homogeneous (complementary) solution, yields the complete response of the system described by the non-homogeneous differential equation.

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Most popular questions from this chapter

In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}-8 y^{\prime}+16 y=6 x e^{4 x}+2+16 x+16 x^{2} $$

Let \(a, b, c,\) and \(\omega\) be constants, with \(a \neq 0\) and \(\omega>0\), and let $$ P(x)=p_{0}+p_{1} x+\cdots+p_{k} x^{k} \quad \text { and } \quad Q(x)=q_{0}+q_{1} x+\cdots+q_{k} x^{k} $$ where at least one of the coefficients \(p_{k}, q_{k}\) is nonzero, so \(k\) is the larger of the degrees of \(P\) and \(Q\). (a) Show that if \(\cos \omega x\) and \(\sin \omega x\) are not solutions of the complementary equation $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ then there are polynomials $$ A(x)=A_{0}+A_{1} x+\cdots+A_{k} x^{k} \quad \text { and } \quad B(x)=B_{0}+B_{1} x+\cdots+B_{k} x^{k} $$ such that $$ \begin{aligned} \left(c-a \omega^{2}\right) A+b \omega B+2 a \omega B^{\prime}+b A^{\prime}+a A^{\prime \prime} &=P \\ -b \omega A+\left(c-a \omega^{2}\right) B-2 a \omega A^{\prime}+b B^{\prime}+a B^{\prime \prime} &=Q \end{aligned} $$ where \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right), \ldots,\left(A_{0}, B_{0}\right)\) can be computed successively by solving the systems $$ \begin{aligned} \left(c-a \omega^{2}\right) A_{k}+b \omega B_{k} &=p_{k} \\ -b \omega A_{k}+\left(c-a \omega^{2}\right) B_{k} &=q_{k} \end{aligned} $$ and, if \(1 \leq r \leq k,\) $$ \begin{aligned} \left(c-a \omega^{2}\right) A_{k-r}+b \omega B_{k-r} &=p_{k-r}+\cdots \\ -b \omega A_{k-r}+\left(c-a \omega^{2}\right) B_{k-r} &=q_{k-r}+\cdots \end{aligned} $$ where the terms indicated by "..." depend upon the previously computed coefficients with subscripts greater than \(k-r\). Conclude from this and Exercise \(36(\mathbf{b})\) that $$ y_{p}=A(x) \cos \omega x+B(x) \sin \omega x $$ is a particular solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=P(x) \cos \omega x+Q(x) \sin \omega x . $$ (b) Conclude from Exercise \(36(\mathbf{c})\) that the equation $$ a\left(y^{\prime \prime}+\omega^{2} y\right)=P(x) \cos \omega x+Q(x) \sin \omega x $$ does not have a solution of the form (B) with \(A\) and \(B\) as in (A). Then show that there are polynomials $$ A(x)=A_{0} x+A_{1} x^{2}+\cdots+A_{k} x^{k+1} \quad \text { and } \quad B(x)=B_{0} x+B_{1} x^{2}+\cdots+B_{k} x^{k+1} $$ such that $$ \begin{aligned} a\left(A^{\prime \prime}+2 \omega B^{\prime}\right) &=P \\ a\left(B^{\prime \prime}-2 \omega A^{\prime}\right) &=Q \end{aligned} $$ where the pairs \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right), \ldots,\left(A_{0}, B_{0}\right)\) can be computed successively as follows: $$ \begin{aligned} A_{k} &=-\frac{q_{k}}{2 a \omega(k+1)} \\ B_{k} &=\frac{p_{k}}{2 a \omega(k+1)}, \end{aligned} $$ and, if \(k \geq 1\), $$ A_{k-j}=-\frac{1}{2 \omega}\left[\frac{q_{k-j}}{a(k-j+1)}-(k-j+2) B_{k-j+1}\right] $$ $$ B_{k-j}=\frac{1}{2 \omega}\left[\frac{p_{k-j}}{a(k-j+1)}-(k-j+2) A_{k-j+1}\right] $$ for \(1 \leq j \leq k\). Conclude that (B) with this choice of the polynomials \(A\) and \(B\) is a particular solution of \((\mathrm{C})\)

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

Find a particular solution. $$ y^{\prime \prime}+y=(-4+8 x) \cos x+(8-4 x) \sin x $$

Use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. $$ y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 x}+25 \sin x, \quad y(0)=5, y^{\prime}(0)=3 $$
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