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Chapter 4: Problem 17

Consider the equation $$ \frac{d^{2} x}{d t^{2}}-2 \frac{d x}{d t}-3 x=0. $$ (a) Show that \(x_{1}(t)=e^{3 t}\) and \(x_{2}(t)=e^{-t}\) are two solutions. (b) Show that \(x(t)=c_{1} x_{1}(t)+c_{2} x_{2}(t)\) is also a solution.

Short Answer

Expert verified
Both \(x_{1}(t)=e^{3 t}\) and \(x_{2}(t)=e^{-t}\) are solutions, and \(x(t)=c_{1} e^{3 t} + c_{2} e^{-t}\) is a general solution.

Step by step solution

01

Verify the solution for \(x_{1}(t)=e^{3 t}\)

To verify that \(x_{1}(t)=e^{3 t}\) is a solution, substitute \(x(t)=e^{3 t}\) back into the original differential equation. First, find the first and second derivatives of \(x_{1}(t)\): \ \(\frac{d x_{1}}{d t}=3 e^{3 t}\) \ \(\frac{d^{2} x_{1}}{d t^{2}}=9 e^{3 t}\) \ Now substitute these into the differential equation: \ \(\frac{d^{2} x}{d t^{2}}-2 \frac{d x}{d t}-3 x = 9 e^{3 t} - 6 e^{3 t} - 3 e^{3 t} = 0\). \ Since the left-hand side equals the right-hand side, \(x_{1}(t)=e^{3 t}\) is indeed a solution.
02

Verify the solution for \(x_{2}(t)=e^{-t}\)

Next, verify that \(x_{2}(t)=e^{-t}\) is a solution. First, find the first and second derivatives of \(x_{2}(t)\): \ \(\frac{d x_{2}}{d t}=-e^{-t}\) \ \(\frac{d^{2} x_{2}}{d t^{2}}=e^{-t}\) \ Now substitute these into the differential equation: \ \(\frac{d^{2} x}{d t^{2}}-2 \frac{d x}{d t}-3 x = e^{-t} + 2 e^{-t} - 3 e^{-t} = 0\). \ Since the left-hand side equals the right-hand side, \(x_{2}(t)=e^{-t}\) is indeed a solution.
03

Show that \(x(t)=c_{1} x_{1}(t)+c_{2} x_{2}(t)\) is a solution

Now consider the general solution \(x(t)=c_{1} e^{3 t}+c_{2} e^{-t}\). Take the first and second derivatives: \ \(\frac{d x}{d t}=3 c_{1} e^{3 t} - c_{2} e^{-t}\) \ \(\frac{d^{2} x}{d t^{2}}=9 c_{1} e^{3 t} + c_{2} e^{-t}\) \ Substitute these into the original differential equation: \ \(\frac{d^{2} x}{d t^{2}}-2 \frac{d x}{d t}-3 x = \left(9 c_{1} e^{3 t} + c_{2} e^{-t}\right) - 2 \left(3 c_{1} e^{3 t} - c_{2} e^{-t}\right) - 3 \left(c_{1} e^{3 t} + c_{2} e^{-t}\right)\) \ Simplify this expression: \ \(9 c_{1} e^{3 t} + c_{2} e^{-t} - 6 c_{1} e^{3 t} + 2 c_{2} e^{-t} - 3 c_{1} e^{3 t} - 3 c_{2} e^{-t} = 0\) \ Which simplifies to \(0 = 0\), verifying that \(x(t)=c_{1} e^{3 t}+c_{2} e^{-t}\) is indeed a solution.

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Key Concepts

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

Second-order Differential Equations
A differential equation is an equation that relates a function to its derivatives. A second-order differential equation involves the second derivative of a function. In general, the equation is written as \[ \frac{d^2x}{dt^2} + p(t)\frac{dx}{dt} + q(t)x = 0, \] where \(p(t)\) and \(q(t)\) are functions of \(t\). In the specific problem provided, the equation is \( \frac{d^2 x}{d t^{2}}-2 \frac{d x}{d t}-3 x=0 \). This is a classic example of a homogeneous linear second-order differential equation, where the coefficients \(p(t)\) and \(q(t)\) are constants.
General Solution
The general solution to a second-order homogeneous linear differential equation is often a combination of two linearly independent solutions. For the given equation, the solutions \(x_{1}(t)=e^{3 t}\) and \(x_{2}(t)=e^{-t}\) are found by substituting these forms into the differential equation and confirming they satisfy it. The linear combination of these solutions gives the general solution: \[ x(t) = c_{1}e^{3t} + c_{2}e^{-t}, \] where \(c_{1}\) and \(c_{2}\) are arbitrary constants determined by initial conditions. This general solution represents the most comprehensive form of the solution, accommodating all possible specific solutions.
Verification of Solutions
To verify that a function is a solution to the differential equation, we need to substitute the function and its derivatives back into the differential equation and ensure the left-hand side equals the right-hand side. For example, for \(x_{1}(t)=e^{3 t}\), the derivatives are: \[ \frac{d x_{1}}{d t}=3 e^{3 t}, \] \[ \frac{d^{2} x_{1}}{d t^{2}}=9 e^{3 t}. \] Substituting these derivatives back into the original equation confirms that it balances. Similarly, we verify \(x_{2}(t)=e^{-t}\) with the same process. This steps confirm that \(x(t)=c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is also a valid solution by demonstrating that any linear combination of solutions to the homogeneous equation is itself a solution.

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Most popular questions from this chapter

In this question we deal with a number of variants of the chemostat. (a) How would you expect the model to differ if there were two growth-limiting nutrients? (b) Suppose that at high densities bacteria start secreting a chemical that inhibits their own growth. How would you model this situation? (c) In certain cases two (or more) bacterial species are kept in the same chemostat and compete for a common nutrient. Suggest a model for such competition experiments.

Problems 15 through 24 deal with ODEs and techniques discussed in Section \(4.8 .\) The differential equation $$ \frac{d^{2} x}{d t^{2}}+3 \frac{d x}{d t}+2 x=0 $$ has the general solution $$ x(t)=c_{1} e^{-t}+c_{2} e^{-2 t} \text {. } $$ If we are told that, when \(t=0, x(0)=1\) and its derivative \(x^{\prime}(0)=1\), we can determine \(c_{1}\) and \(c_{2}\) by solving the equations $$ \begin{aligned} &x(0)=c_{1} e^{-0}+c_{2} e^{(-2)}=c_{1}(1)+c_{2}(1)=1, \\ &x^{\prime}(0)=c_{1}(-1) e^{-6}+c_{2}(-2) e^{-0}=-c_{1}-2 c_{2}=1 . \end{aligned} $$ (a) Find the values of \(c_{1}\) and \(c_{2}\) by solving the above. For questions (b) through (e) find the solution of the differential equation subject to the specified initial condition. (b) \(y^{\prime}=10 y ; \quad y(0)=0.001\). (c) \(y^{\prime \prime}-3 y^{\prime}-4 y=0 ; \quad y(0)=0, y^{\prime}(0)=1\). (d) \(y^{\prime \prime}-9 y=0 ; \quad y(0)=5, y^{\prime}(0)=0\). (e) \(y^{\prime \prime}-5 y^{\prime}=0 ; \quad y(0)=1, y^{\prime}(0)=2\).

Show that for a decaying population $$ d N / d t=-K N \quad(K>0) $$ the time at which only half of the original population remains (the half-life) is $$ \tau_{1 / 2}=\frac{\ln 2}{K} . $$

A model due to Landahl and Grodsky (1982) for insulin secretion is based on the assumption that there are separate storage and labile compartments, a provisionary factor, and a signal for release. They define the following variables: \(G=\) glucose concentration, \(X=\) moiety formed from glucose in presence of calcium ions \(\mathrm{Ca}^{++}\), \(P=\) provisionary quantity required for insulin production, \(Q=\) total amount of insulin available for release \(=C V\), where \(C\) is its concentration and \(V\) is the volume, \(S=\) secretion rate of insulin, \(I=\) concentration of an inhibitory quantity. (a) The following equation describes the amount of insulin available for release: $$ V \frac{d C}{d t}=k_{+} C_{2}-k_{-} C+\gamma V P-S $$ where \(k_{*}, k\). and \(\gamma\) are constant and \(C_{s}\) is the concentration in the storage compartment (of volume \(V_{L}\) ), assumed constant. Explain the terms and assumptions made in deriving the equation. (b) Show that another way of expressing part (a) is $$ \frac{d Q}{d t}=H\left(Q_{0}-Q\right)+\gamma^{\prime} P-S \text {. } $$ How do \(H, \gamma^{\prime}\), and \(Q_{0}\) relate to parameters which appear above? (c) An equation for the provisional factor \(P\) is given as follows: $$ \frac{d P}{d t}=\alpha\left[P_{\infty}(G)-P\right] $$ where \(P_{-}(G)\) is just some function of \(G\left(e, g, P_{w}(G)=G\right)\). Explain what has been assumed about \(P\). (d) The inhibitory entity \(I\) is assumed to be produced at the rate $$ \frac{d l}{d t}=B(N X-I) \text {, } $$ where \(B\) is a rate constant and \(N\) is a proportionality constant. Explain this equation. (e) Secretion of insulin \(S\) is assumed to be determined by two processes, and governed by the equation $$ S=\left[M_{1} Y(G)+M_{2}(X-I)\right] Q . $$ where \(M_{1}\) is constant, \(Y\) is a function of the glucose concentration, and \(M_{2}\) is a step function; that is, $$ M_{2}(X-I)=\left\\{\begin{array}{cc} 0 & \text { if } X

Problems 15 through 24 deal with ODEs and techniques discussed in Section \(4.8 .\) In the following exercises find the general solution to the system of equations \(d x / d t=A x\), where the matrix \(A\) is as follows: (a) \(\mathbf{A}=\left(\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right)\). (d) \(\mathbf{A}=\left(\begin{array}{ll}-1 & 4 \\ -2 & 5\end{array}\right)\). (b) \(A=\left(\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right)\). (e) \(A=\left(\begin{array}{ll}2 & -3 \\ 1 & -2\end{array}\right)\). (c) \(\mathrm{A}=\left(\begin{array}{rr}-2 & 7 \\ 2 & 3\end{array}\right)\). (f) \(A=\left(\begin{array}{rr}-4 & 1 \\ 3 & 0\end{array}\right)\).
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