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Chapter 18: Problem 8

Compute \(y_{h}(t),\) the general solution to the homogeneous equation, and \(y_{p}(t)\) a particular solution to the nonhomogeneous equation. $$ \begin{array}{lll} \text { a. } y^{\prime \prime}+4 y^{\prime}+3 y=t & \text { b. } y^{\prime \prime}+4 y^{\prime}+3 y= & e^{t} \\ \text { c. } y^{\prime \prime}-4 y^{\prime}+3 y=e^{t} & \text { d. } y^{\prime \prime}+4 y^{\prime}+3 y=\cos t \end{array} $$

Short Answer

Expert verified
a) Combine \(y_h(t) = C_1 e^{-t} + C_2 e^{-3t}\) with \(y_p(t) = \frac{1}{3}t - \frac{4}{9}\). b) Combine \(y_h(t)\) with \(y_p(t) = \frac{1}{8}e^t\). c) Combine \(y_h(t)\) with \(y_p(t) = -\frac{1}{2}te^t\). d) Combine \(y_h(t)\) with \(y_p(t) = \frac{1}{2}\cos t\).

Step by step solution

01

Solve the Homogeneous Equation

Given the equation \(y'' + 4y' + 3y = 0\), we need to find the general solution. The characteristic equation is \(r^2 + 4r + 3 = 0\). Solving this quadratic equation using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), \(c = 3\), we get:\[r = \frac{-4 \pm \sqrt{16 - 12}}{2}\]\[r = \frac{-4 \pm 2}{2}\]This gives roots \(r_1 = -1\) and \(r_2 = -3\). So, the general solution to the homogeneous equation is:\[y_h(t) = C_1 e^{-t} + C_2 e^{-3t}\]
02

Step 2a: Particular Solution for Part (a)

In part (a), the nonhomogeneous term is \(t\). We try a particular solution of the form \(y_p(t) = At + B\). Compute the derivatives: \(y_p' = A\) and \(y_p'' = 0\). Substitute into the equation \(y'' + 4y' + 3y = t\): \[0 + 4A + 3(At + B) = t\]\[4A + 3At + 3B = t\]Equate coefficients:\[3A = 1, \quad 4A + 3B = 0\]Solve for \(A\) and \(B\): \[A = \frac{1}{3},\]\[4\left(\frac{1}{3}\right) + 3B = 0 \implies B = -\frac{4}{9}\]Thus, \(y_p(t) = \frac{1}{3}t - \frac{4}{9}\).
03

Step 3a: General Solution for Part (a)

Combine the homogeneous and particular solutions for part (a):\[y(t) = y_h(t) + y_p(t) = C_1 e^{-t} + C_2 e^{-3t} + \frac{1}{3}t - \frac{4}{9}\]
04

Step 2b: Particular Solution for Part (b)

In part (b), the nonhomogeneous term is \(e^t\). We try a particular solution of the form \(y_p(t) = Ae^t\). Compute derivatives: \(y_p' = Ae^t\) and \(y_p'' = Ae^t\). Substitute into the equation:\[Ae^t + 4Ae^t + 3Ae^t = e^t\]\[8Ae^t = e^t\]Dividing both sides by \(e^t\), we get:\[8A = 1\]\[A = \frac{1}{8}\]Thus, \(y_p(t) = \frac{1}{8}e^t\).
05

Step 3b: General Solution for Part (b)

Combine the homogeneous and particular solutions for part (b):\[y(t) = C_1 e^{-t} + C_2 e^{-3t} + \frac{1}{8}e^t\]
06

Step 2c: Particular Solution for Part (c)

In part (c), the nonhomogeneous term is \(e^t\), similar to part (b). Here, try \(y_p(t) = Ae^t\), compute derivatives: \(y_p' = Ae^t\) and \(y_p'' = Ae^t\). Insert into:\[Ae^t - 4Ae^t + 3Ae^t = e^t\]\[0e^t = e^t\]No particular solution of this form works, indicating a need for a modified guess: \(y_p(t) = Ate^t\). Compute derivatives: \(y_p' = Ae^t + Ate^t\) and \(y_p'' = 2Ae^t + Ate^t\). Substitute:\[(2A + At)e^t - 4(A+At)e^t + 3Ate^t = e^t\]Simplify and equate terms:\[-2A = 1\]\[A = -\frac{1}{2}\]Thus, \[y_p(t) = -\frac{1}{2}te^t\].
07

Step 3c: General Solution for Part (c)

Combine the homogeneous and particular solutions for part (c):\[y(t) = C_1 e^{2t} + C_2 e^t - \frac{1}{2}te^t\]
08

Step 2d: Particular Solution for Part (d)

In part (d), the nonhomogeneous term is \(\cos t\). Try \(y_p(t) = A\cos t + B\sin t\). Compute derivatives: \(y_p' = -A\sin t + B\cos t\), \(y_p'' = -A\cos t - B\sin t\). Substitute:\[-A\cos t - B\sin t + 4(-A\sin t + B\cos t) + 3(A\cos t + B\sin t) = \cos t\]Simplify the equation:\[2A\cos t + 10B\sin t = \cos t\]Equate coefficients:\[2A = 1\] and \[10B = 0\]Solve for \(A\) and \(B\):\[A = \frac{1}{2}, B = 0\]Thus, \(y_p(t) = \frac{1}{2}\cos t\).
09

Step 3d: General Solution for Part (d)

Combine the homogeneous and particular solutions for part (d):\[y(t) = C_1 e^{-t} + C_2 e^{-3t} + \frac{1}{2}\cos t\]

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Key Concepts

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

Homogeneous Equations
A homogeneous differential equation is one where the right-hand side of the equation is zero. This means there are no external forces or inputs affecting the system described by the equation. For example, in the exercise, the homogeneous part is given by:\[ y'' + 4y' + 3y = 0 \]The key to solving such equations is finding the characteristic equation, which is obtained by replacing the derivatives with powers of a variable, typically \( r \). In our case, we derive the characteristic equation as:\[ r^2 + 4r + 3 = 0 \]This is a quadratic equation. Using the quadratic formula, we solve for \( r \) and find two roots, \( r_1 = -1 \) and \( r_2 = -3 \). These roots lead directly to the general solution:\[ y_h(t) = C_1 e^{-t} + C_2 e^{-3t} \]where \( C_1 \) and \( C_2 \) are arbitrary constants. They take different values depending on initial conditions.
Particular Solutions
To solve a nonhomogeneous differential equation, one must find a particular solution for the given nonhomogeneous term, which is the part of the equation that distinguishes it from being homogeneous. In the exercise, this entails discovering a solution to equations like:- \( y'' + 4y' + 3y = t \)- \( y'' + 4y' + 3y = e^t \)- \( y'' + 4y' + 3y = \cos t \)Each of these requires a guess for the particular solution form, based on the type of nonhomogeneous term:- For polynomial terms, try a polynomial solution.- For exponential terms, try an exponential solution.- For trigonometric terms, try combinations of sine and cosine.In part (a), with a nonhomogeneous term \( t \), we use:\[ y_p(t) = At + B \]In part (b), the particular solution for \( e^t \) involves:\[ y_p(t) = Ae^t \]We substitute these forms into the original equation, derive and adjust coefficients to match the nonhomogeneous term, effectively determining values for constants like \( A \) and \( B \). This approach is unique to each type of term, making these particular solutions tailored responses to the specific input affecting the system.
Nonhomogeneous Equations
Nonhomogeneous differential equations include an additional non-zero term that represents external influences or inputs. For example, the general form for the exercise is:\[ y'' + 4y' + 3y = f(t) \]where \( f(t) \) is not zero and could be functions like a polynomial, an exponential, or a trigonometric function.The solution to a nonhomogeneous equation is a combination of two parts:- The general solution to the corresponding homogeneous equation, which represents the natural behavior of the system without external influences.- A particular solution that takes into account the non-zero right-hand side.The full solution to the nonhomogeneous equation is then:\[ y(t) = y_h(t) + y_p(t) \]This represents the overall behavior of the system, integrating both natural dynamics and external inputs. To effectively solve such equations, understanding the impact of \( f(t) \) on the system and choosing appropriate techniques for finding \( y_p(t) \) is crucial. This process illustrates the principle of superposition, where the response of the system is the sum of its responses to internal and external factors.

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Most popular questions from this chapter

Consider now the case that there is resistance \(r>0\) in the spring-mass equation with a harmonic forcing function, $$ m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=\cos \omega t $$ a. Show that in the real root case, \(\mu_{1}\) and \(\mu_{2}\) are negative. b. Show that in the repeated root case, the value of \(\mu\) is negative. c. Show that in the and complex roots case, the value of \(\mu\) is negative. d. Show that in all of the cases, $$ \lim _{t \rightarrow \infty} y_{h}(t)=0 $$ The next few exercises examine the importance of \(y_{p},\) the particular solution of Equation \(18.20 .\) We will need $$ \begin{array}{ll} y_{p}=A \cos 0.3 t+B \sin 0.3 t & \text { and one of } \\ y_{h}=C_{1} e^{\mu_{1} t}+C_{2} e^{\mu_{2} t}, & \text { real roots, } \\ y_{h}=C_{1} e^{\mu t}+C_{2} t e^{\mu t}, & \text { repeated root, and } \\ y_{h}=e^{\mu t}\left(C_{1} \cos (\omega t)+C_{2} \sin (\omega t)\right) & \text { complex roots. } \end{array} $$

Symbiotic relationships are common and persist for long periods. It is curious that there are no known or very few symbiotic relationships between mammals. There are, however, many symbiotic relationships between mammals and other organisms, Escherichia coli, for example. Analysis of the equations for symbiosis, Equations 18.63: $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ a. Show that the equilibrium point without zeros is $$ x_{1}=\frac{b+d}{a d-b c}, \quad y_{1}=\frac{a+c}{a d-b c} $$ $$ \text { if } a d-b c \neq 0 $$ b. \(x_{1}\) and \(y_{1}\) are positive only if \(a d-b c>0 .\) This is a surprise to us. Set \(b=c=d=1\) and examine the equilibrium point for \(a>1\) and \(a<1\). c. Assume \(a d-b c>0\) so that \(x_{1}\) and \(y_{1}\) are positive. Either work it out (no!) or accept our analysis that the Jacobian at \(\left(x_{1}, y_{1}\right)\) is $$ J=\left[\begin{array}{cc} -a \frac{b+d}{a d-b c} & b \frac{b+d}{a d-b c} \\ c \frac{a+c}{a d-b c} & -d \frac{a+c}{a d-b c} \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} -a(b+d) & b(b+d) \\ c(a+c) & -d(a+c) \end{array}\right] $$ Argue that if \(a d-b c>0,\left(x_{1}, y_{1}\right)\) is an asymptotically stable equilibrium. d. With persistence you might show that the characteristic roots are not complex. It requires showing that the discrimiant $$ \begin{array}{l} (a(b+d)+d(a+c))^{2}-4(a(b+d) d(a+c)-c(a+c) d(b+d))= \\ (a(b+d)-d(a+c))^{2}+4 c(a+c) d(b+d) \geq 0 \end{array} $$

Show by substitution that if \(y_{1}(t)\) and \(y_{2}(t)\) are solutions to \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and each of \(C_{1}\) and \(C_{2}\) is a number then $$ y(t)=C_{1} y_{1}(t)+C_{2} y_{2}(t) $$ is a solution to \(y^{\prime \prime}(t)-p y^{\prime}(t)+q y(t)=0\).

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

Find the local linear approximation to the system $$ \begin{array}{l} x^{\prime}=x-x^{2}-x y \\ y^{\prime}=y-0.5 x y-2 y^{2} \end{array} $$ a. At the equilibrium point (0,0) . b. At the equilibrium point (0,0.5) . c. At the equilibrium point (1,0) . d. At the equilibrium point \((2 / 3,1 / 3)\). For each of the local linear approximations, determine whether it is stable.
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