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Chapter 10: Problem 14

Find the solutions of the following initial-value problems: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{2}+1}{x+2}, \quad x(0)=-2\) (b) \(t(t-1) \frac{\mathrm{d} x}{\mathrm{~d} t}=x(x+1), \quad x(2)=2\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\left(x^{2}-1\right) \cos t, \quad x(0)=2\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\mathrm{e}^{x+t}, \quad x(0)=a\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{4 \ln t}{x^{2}}, \quad x(1)=0\)

Short Answer

Expert verified
Solutions for each are derived by separating variables, integrating, and applying initial conditions.

Step by step solution

01

Solve Part (a) - Separate Variables

To solve \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{t^{2} + 1}{x + 2} \), start by separating variables:\[(x + 2) \mathrm{d} x = (t^{2} + 1) \mathrm{d} t\]
02

Integrate Both Sides - Part (a)

Integrate both sides of the equation:\[\int (x + 2) \, \mathrm{d} x = \int (t^{2} + 1) \, \mathrm{d} t\]The left integral gives \( \frac{x^{2}}{2} + 2x \), and the right integral gives \( \frac{t^{3}}{3} + t \). Add a constant \( C \) on one side:\[\frac{x^{2}}{2} + 2x = \frac{t^{3}}{3} + t + C\]
03

Apply Initial Condition - Part (a)

Use the initial condition \( x(0) = -2 \) to find \( C \):\[\frac{(-2)^{2}}{2} + 2(-2) = \frac{0^{3}}{3} + 0 + C\]\[2 - 4 = C \implies C = -2\]Thus, the equation becomes:\[\frac{x^{2}}{2} + 2x = \frac{t^{3}}{3} + t - 2\]
04

Solve Part (b) - Separate Variables

The equation is \( t(t-1) \frac{\mathrm{d} x}{\mathrm{~d} t} = x(x+1) \). Start by separating variables:\[\frac{1}{x(x+1)} \mathrm{d} x = \frac{1}{t(t-1)} \mathrm{d} t\]
05

Integrate Both Sides - Part (b)

Integrate both sides:\[\int \frac{1}{x} - \frac{1}{x+1} \, \mathrm{d} x = \int \left( \frac{1}{t} - \frac{1}{t-1} \right) \, \mathrm{d} t\]This results in:\[\ln|x| - \ln|x+1| = \ln|t| - \ln|t-1| + C\]
06

Apply Initial Condition - Part (b)

Use the initial condition \( x(2) = 2 \) to find \( C \):\[\ln(2) - \ln(3) = \ln(2) - \ln(1) + C\]\[\ln\left(\frac{2}{3}\right) = \ln2 + C\]Thus, \( C = \ln\left(\frac{2}{3}\right) \). The equation becomes:\[\ln\left(\frac{x}{x+1}\right) = \ln\left( \frac{t}{t-1} \right) + \ln\left(\frac{2}{3}\right)\]
07

Solve Part (c) - Separate Variables

Given \( \frac{\mathrm{d} x}{\mathrm{~d} t} = (x^{2} - 1) \cos t \), separate variables:\[\frac{1}{x^{2} - 1} \mathrm{d} x = \cos t \mathrm{d} t\]
08

Integrate Both Sides - Part (c)

Integrate both sides:\[\int \frac{1}{x-1} - \frac{1}{x+1} \, \mathrm{d} x = \int \cos t \, \mathrm{d} t\]The result is:\[\ln|x-1| - \ln|x+1| = \sin t + C\]
09

Apply Initial Condition - Part (c)

Use the initial condition \( x(0) = 2 \) to solve for \( C \):\[\ln(1) - \ln(3) = \sin 0 + C\]\[-C = \ln(3)\]Thus, \( C = -\ln(3) \) and the solution is:\[\ln\left(\frac{x-1}{x+1}\right) = \sin t - \ln(3)\]
10

Solve Part (d) - Separate Variables

Given \( \frac{\mathrm{d} x}{\mathrm{~d} t} = e^{x+t} \), separate variables:\[e^{-x} \, \mathrm{d} x = e^{t} \, \mathrm{d} t\]
11

Integrate Both Sides - Part (d)

Integrate both sides:\[\int e^{-x} \, \mathrm{d} x = \int e^{t} \, \mathrm{d} t\]This yields:\[-e^{-x} = e^{t} + C\]
12

Apply Initial Condition - Part (d)

Use the initial condition \( x(0) = a \) to find \( C \):\[-e^{-a} = e^{0} + C\]\[C = -e^{-a} - 1\]Thus, the solution is:\[-e^{-x} = e^{t} - e^{-a} - 1\]
13

Solve Part (e) - Separate Variables

For \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{4 \ln t}{x^{2}} \), separate variables:\[x^{2} \, \mathrm{d} x = 4 \ln t \, \mathrm{d} t\]
14

Integrate Both Sides - Part (e)

Integrate both sides:\[\int x^{2} \, \mathrm{d} x = \int 4 \ln t \, \mathrm{d} t\]This results in:\[\frac{x^3}{3} = 4(t \ln t - t) + C\]
15

Apply Initial Condition - Part (e)

Use the initial condition \( x(1) = 0 \) to solve for \( C \):\[\frac{0^3}{3} = 4(1 \cdot 0 - 1) + C\]\[C = 4\]Hence, the solution is:\[\frac{x^3}{3} = 4(t \ln t - t + 1)\]

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe many physical phenomena, from physics to economics. In essence, these equations relate a function with changes in its rates.
For instance, consider an equation like \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{t^2 + 1}{x + 2} \). Here, it says that the derivative of \( x \) with respect to \( t \), i.e., \( \frac{\mathrm{d} x}{\mathrm{~d} t} \), is a ratio of a function of \( t \) and \( x \).
  • A common task is to find the function \( x(t) \) that satisfies the equation and any initial conditions provided.
  • Initial-value problems supply extra information that enable finding one specific solution among many possible ones.
Understanding and solving differential equations is crucial as they model real-world systems by describing how quantities change over time.
Separation of Variables
Separation of variables is a method for solving differential equations. It's particularly useful when both sides of an equation can be rewritten as a function of one variable.
This technique involves rearranging the equation so that each variable is on a different side, such as transforming \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{t^2 + 1}{x + 2} \) into \( (x + 2) \mathrm{d} x = (t^2 + 1) \mathrm{d} t \).
  • Separating variables sets the stage for the next step: integration.
  • This technique is highly applicable to first-order differential equations, allowing them to become integrable equations.
  • After the separation step, integration on both sides can derive a general solution.
Successful use of this method depends on recognizing when an equation can be rearranged suitably for variable separation.
Integration of Functions
Integration is a cornerstone in solving differential equations once variables have been separated. The goal is to find the antiderivative or the integral of both sides of the equation.
Consider an equation like \( \int (x + 2) \mathrm{d} x = \int (t^2 + 1) \mathrm{d} t \). Integrating both sides, you get the solution in terms of indefinite integrals: \( \frac{x^2}{2} + 2x \) and \( \frac{t^3}{3} + t \), respectively.
  • Integrating involves finding functions whose derivatives give back the original functions.
  • Each integration step gives a function plus a constant of integration \( C \), which is determined by any given initial conditions.
  • These integrations transform the relationship between \( \,\mathrm{d} x \) and \( \,\mathrm{d} t \) into an expression for variables directly.
Applying the initial conditions helps to determine a constant, narrowing down to a particular solution within possibly infinite family of solutions.

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Most popular questions from this chapter

Transform the following initial-value problems into sets of first-order differential equations with appropriate initial conditions: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4\left(x^{2}-t^{2}\right)^{1 / 2}=0\) $$ x(1)=2, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(1)=0.5 $$ (b) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+t \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+6 \mathrm{e}^{t} \frac{\mathrm{d} x}{\mathrm{~d} t}-x^{2} t=\mathrm{e}^{2 t}\) $$ x(0)=1, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(0)=2, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(0)=0 $$ (c) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+t \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+x^{2}=\sin t\) $$ x(1)=1, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(1)=0, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(1)=-2 $$ (d) \(\left(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}\right)^{1 / 2}+t \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+x^{2} t^{2}=0\) \(x(2)=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(2)=0, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(2)=2\) (e) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}+x \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+x^{2}=\ln t, \quad x(0)=0\) \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0)=0, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(0)=4, \quad \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}(0)=-3\) (f) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}+\left(\frac{\mathrm{d} x}{\mathrm{~d} t}-1\right) \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+\frac{\mathrm{d} x}{\mathrm{~d} t}-(x t)^{1 / 2}\) \(=t^{2}+4 t-5\) \(x(0)=a, \frac{\mathrm{d} x}{\mathrm{~d} t}(0)=0, \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(0)=b, \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}(0)=0\)

Classify the following differential equations as linear homogeneous, linear nonhomogeneous or nonlinear differential equations, state their order and name the dependent and independent variables. (a) \(\frac{\mathrm{d}^{2} p}{\mathrm{~d} z^{2}} \frac{\mathrm{d} p}{\mathrm{~d} z}+(\sin z) p=\ln z\) (b) \(\frac{\mathrm{d}^{2} s}{\mathrm{~d} t^{2}}+(\sin t) \frac{\mathrm{d} s}{\mathrm{~d} t}+(t+\cos t) s=\mathrm{e}^{t}\) (c) \(\left(\frac{\mathrm{d}^{3} p}{\mathrm{~d} y^{3}}\right)^{1 / 2}+4 \frac{\mathrm{d}^{2} p}{\mathrm{~d} y^{2}}-6 \frac{\mathrm{d} p}{\mathrm{~d} y}+8 p=0\) (d) \(\frac{\mathrm{d} r}{\mathrm{~d} z}+z^{2}=0\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=f(t) x\) (f) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=f(t) x+g(t)\) (g) \(\frac{\mathrm{d}^{3} p}{\mathrm{~d} q^{3}}+\frac{\mathrm{d}^{2} p}{\mathrm{~d} q^{2}} p+4 q^{2}=0\) (h) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} y^{2}}=\frac{y}{x^{2}-1}\) (i) \((\sin z) \frac{\mathrm{d} y}{\mathrm{~d} z}+\frac{\cos z}{z} y=0\)

State the order of each of the following differential equations and name the dependent and independent variables. Classify each equation as linear homogeneous, linear nonhomogeneous or nonlinear differential equations. (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=0\) (c) \(\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}+x=0\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=t^{2}\) (e) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=\cos t+\mathrm{e}^{t}\)

Find the solutions of the following initial-value problems: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x^{3}-x t^{2}}{t^{3}}, \quad x(1)=2\) (b) \(x t \frac{\mathrm{d} x}{\mathrm{~d} t}=2\left(x^{2}+t^{2}\right), \quad x(2)=-1\) (c) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}=t \mathrm{e}^{-x / t}+x, \quad x(2)=4\) (d) \(x t \frac{\mathrm{d} x}{\mathrm{~d} t}=t^{2} \mathrm{e}^{-x^{2} / t^{2}}+x^{2}, \quad x(1)=2\) (e) \(t^{2} \frac{\mathrm{d} x}{\mathrm{~d} t}=x^{2}+2 x t, \quad x(1)=4\)

Find the solution of the following differential equations: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=2\) (b) \(\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=t\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=\mathrm{e}^{-4 t}\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+t x=-2 t\)
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