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Chapter 7: Problem 13

Find the solution of the differential equation that satisfies the given initial condition. $$\frac { d u } { d t } = \frac { 2 t + \sec ^ { 2 } t } { 2 u } , u ( 0 ) = - 5$$

Short Answer

Expert verified
The solution is \( u = -\sqrt{t^2 + \tan(t) + 25} \).

Step by step solution

01

Set Up the Differential Equation

We are given the differential equation \( \frac{du}{dt} = \frac{2t + \sec^2(t)}{2u} \) and the initial condition \( u(0) = -5 \). Our task is to solve this differential equation using the initial condition.
02

Separate Variables

We separate the variables by multiplying both sides by \( 2u \) and \( dt \) to isolate terms involving \( u \) and \( t \) on opposite sides:\[ 2u \frac{du}{dt} = 2t + \sec^2(t) \]\[ 2u \, du = (2t + \sec^2(t)) \, dt \]
03

Integrate Both Sides

Integrate both sides of the equation:\[ \int 2u \, du = \int (2t + \sec^2(t)) \, dt \]The left side integrates to \( u^2 + C_1 \) and requires solving the two parts on the right side:
04

Solve the Two Integrals

Integrate \( 2t \) and \( \sec^2(t) \) separately:\[ \int 2t \, dt = t^2 + C_2 \]\[ \int \sec^2(t) \, dt = \tan(t) + C_3 \]Thus, the right side becomes \( t^2 + \tan(t) + C \) where \( C = C_2 + C_3 \).
05

Combine and Simplify

Write the integrated form as:\[ u^2 = t^2 + \tan(t) + C \]
06

Apply Initial Condition

Using the initial condition \( u(0) = -5 \), substitute \( t = 0 \) and \( u = -5 \) into the equation:\[ (-5)^2 = 0 + \tan(0) + C \]\[ 25 = 0 + C \]Thus, \( C = 25 \).
07

Find the Particular Solution

Substitute \( C = 25 \) into the solution:\[ u^2 = t^2 + \tan(t) + 25 \]Taking the square root gives:\[ u = \pm \sqrt{t^2 + \tan(t) + 25} \]Since \( u(0) = -5 \), choose the negative root:\[ u = -\sqrt{t^2 + \tan(t) + 25} \]

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Key Concepts

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

Separation of Variables
Separation of variables is a powerful technique used to simplify and solve differential equations. It involves rearranging a given equation so that all terms involving the same variable are on one side of the equation. For our exercise, we have the equation:
\[ \frac{du}{dt} = \frac{2t + \sec^2(t)}{2u} \]
To separate the variables, we multiply through by \( 2u \) and \( dt \), rearranging the equation as follows:
  • Multiply both sides by \( 2u \)
  • Reorganize to have all \( u \)s and \( du \)s on one side and \( t \)s and \( dt \)s on the other
This process results in:
\[ 2u \, du = (2t + \sec^2(t)) \, dt \]
By separating variables, we make integrating both sides of the equation possible, enabling us to progress toward a solution.
Integration
Integration is the process of finding the antiderivative or integral of a function. In our context, we've reduced the differential equation to a form that allows us to integrate both sides. This involves finding the integrals of expressions in terms of \( u \) and \( t \):
  • The integral of \( 2u \, du \) calculates to \( u^2 + C_1 \)
  • We need to solve the right side, which consists of two integrals
  • \( \int 2t \, dt = t^2 + C_2 \)
  • The integral of \( \sec^2(t) \, dt \) yields \( \tan(t) + C_3 \)
Combining these, the right side simplifies to \( t^2 + \tan(t) + C \), where \( C = C_2 + C_3 \).
Integration helps us uncover relationships in functions and serves as a foundation for solving differential equations.
Initial Conditions
Initial conditions are vital for finding specific solutions to differential equations, especially when multiple solutions exist. In this case, we use the initial condition \( u(0) = -5 \) to find the constant \( C \). From the general solution:
\[ u^2 = t^2 + \tan(t) + C \]
Substituting \( u = -5 \) and \( t = 0 \), we find:
  • \( (-5)^2 = 0 + \tan(0) + C \)
  • \( 25 = 0 + C \)
  • Thus, \( C = 25 \)
This constant helps us derive the particular solution:
\[ u = \pm \sqrt{t^2 + \tan(t) + 25} \]
Selecting the negative root satisfies \( u(0) = -5 \), leading to:\[ u = -\sqrt{t^2 + \tan(t) + 25} \]
Always remember, initial conditions guide us to the precise form of the solution that fits the given problem.

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Most popular questions from this chapter

Find all equilibria of the autonomous differential equation and construct the phase plot. (a) $$ y^{\prime}=y^{2}-2$$ (b)$$y^{\prime}=\frac{y-3}{y+9}, \quad y \geqslant 0$$ (c) $$y^{\prime}=y(3-y)\left(25-y^{2}\right)$$

Lung ventilation A patient is placed on a ventilator remove \(\mathrm{CO}_{2}\) from the lungs. Suppose that the rate of ventilation is \(100 \mathrm{mL} / \mathrm{s},\) with the percentage of \(\mathrm{CO}_{2}(\mathrm{by}\)volume) in the inflow being zero. Suppose also that air is absorbed by the lungs at a rate of 10 \(\mathrm{mL} / \mathrm{s}\) and gas consisting of 100\(\%\mathrm{CO}_{2}\) is excreted back into the lungs at the same rate. The volume of a typical pair of lungs is around 4000 \(\mathrm{mL} .\) If the patient starts ventilation with 20\(\%\) of lung volume being \(\mathrm{CO}_{2},\) what volume of \(\mathrm{CO}_{2}\) will remain in the lungs after 30 minutes?

A system of differential equations is given. (a) Use a phase plane analysis to determine the values of the constant \(a\) for which the sole equilibrium of the differential equations is locally stable. (b) Obtain an expression for each equilibrium (it may be a function of the constant \(a\) ). \(x^{\prime}=a(x-a), \quad y^{\prime}=4-y-x, \quad a \neq 0\)

A system of differential equations is given. (a) Construct the phase plane, plotting all nullclines, labeling all equilibria, and indicating the direction of motion. (b) Obtain an expression for each equilibrium. \(n^{\prime}=n(1-2 m), \quad m^{\prime}=m(2-2 n-m), \quad n, m \geqslant 0\)

Describe the motion of a particle with position \((x, y)\) as \(t\) varies in the given interval. \(x=5 \sin t, \quad y=2 \cos t, \quad-\pi \leqslant t \leqslant 5 \pi\)
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