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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section 3.3 Summary & Exercises

Summary of the Key Ideas.

  • Antiderivatives as Solutions to Differential Equations
    • Finding the general antiderivative of a function, \(f(x)\text{,}\) is equivalent to finding the general solution to the differential equation
      \begin{equation*} \frac{dy}{dx} = f(x) \text{.} \end{equation*}
  • Direct Integration
    • Process of integrating both sides of a differential equation to directly solve for the dependent variable, \(y\text{.}\)

Exercises Exercises

General Solution.

Find the general solution for each of the following differential equations. Combine constants where appropriate.
1.
\(\displaystyle \frac{dy}{dx} = 2x - 5 \)
Answer.
\(\displaystyle y = x^2 - 5x + C \)
2.
\(\displaystyle \frac{d}{dx}[y] = e^{2x} \)
Answer.
\(\displaystyle y = \frac{1}{2}e^{2x} + C \)
3.
\(\displaystyle \frac{d}{dx}[x \cdot y] = \cos(x) \)
Answer.
\(\displaystyle y = \frac{\sin(x)}{x} + \frac{C}{x} \)
4.
\(\displaystyle \frac{dy}{dt} = \frac{1}{t^2} + t \)
Answer.
\(\displaystyle y = -\frac{1}{t} + \frac{1}{2}t^2 + C \)
5.
\(\displaystyle \frac{dy}{dx} = x e^x \)
Answer.
\(\displaystyle y = (x - 1)e^x + C \)
6.
\(\displaystyle \left[\sin(x) \cdot y\right]^\prime = \cos^2(x) \)
Answer.
\(\displaystyle y = \frac{\sin(x)}{2} + \frac{C}{\sin(x)} \)
7.
\(\displaystyle y' = \ln(x) + x^2 \)
Answer.
\(\displaystyle y = x \ln(x) - x + \frac{1}{3}x^3 + C \)
8.
\(\displaystyle \frac{d}{dP}\left[e^P \cdot Q\right] = P \)
Answer.
\(\displaystyle Q = e^{-P} \left(\frac{P}{e^P} + C \right) \)

Particular Solution.

Find the particular solution for each of the following differential equations with the given initial condition.
9.
\(\displaystyle \frac{dy}{dx} = 3x^2 + 2, \ \ y(1) = 4 \)
Answer.
\(\displaystyle y = x^3 + 2x + 1 \)
10.
\(\displaystyle \frac{d}{dt}[y] = \sin(t), \ \ y(0) = 2 \)
Answer.
\(\displaystyle y = -\cos(t) + 3 \)
11.
\(\displaystyle \frac{d}{dx}[x \cdot y] = e^x, \ \ y(1) = 0 \)
Answer.
\(\displaystyle y = \frac{e^x}{x} - \frac{e}{x} \)
12.
\(\displaystyle \frac{dy}{dx} = x^3 - 4, \ \ y(2) = 1 \)
Answer.
\(\displaystyle y = \frac{x^4}{4} - 4x + 5 \)
13.
\(\displaystyle \frac{dR}{dh} = \frac{1}{h} + h, \ \ R(1) = 3 \)
Answer.
\(\displaystyle R = \ln|h| + \frac{h^2}{2} + \frac{5}{2} \)
14.
\(\displaystyle \frac{d}{dx}\left[e^x \cdot y\right] = \sin(x), \ \ y(0) = 2 \)
Answer.
\(\displaystyle y = e^{-x} \left(-\cos(x) + 3 \right) \)
15.
\(\displaystyle y' = \cos(x) + x^2, \ \ y(0) = -2 \)
Answer.
\(\displaystyle y = \sin(x) + \frac{x^3}{3} - 2 \)
16.
\(\displaystyle \frac{d}{dx}\left[\tan(x) \cdot y\right] = \sec^2(x), \quad y\left(\frac{\pi}{4}\right) = 1 \)
Answer.
\(\displaystyle y = \frac{x \sec^2(x)}{\tan(x)} + \frac{\pi/4 - 1}{\tan(x)} \)

17.

Attempt to apply direct integration to the differential equation
\begin{equation*} \frac{dy}{dx} = x + y\text{.} \end{equation*}
Get to the point where it becomes clear that you cannot solve for \(y\) directly. What is the obstacle?
Answer.
Integrating both sides gives
\begin{align*} \int \frac{dy}{dx}\ dx \amp = \int\left(x + y\right)\ dx \\ y + C_1 \amp = \int x\ dx + \int y\ dx \\ y + C_1 \amp = \frac12 x^2 + C_2 + \int y\ dx \\ y - \int y\ dx \amp = \frac12 x^2 + C_2 - C_1 \end{align*}
Without knowing \(y\text{,}\) we cannot simplify \(\int y\ dx\text{.}\) So the obstacle is that we are unable to combine the these \(y\) variables into a single \(y\) on the left side.

Exercise Group.

Convert this idea from the integrating factor into some exercises for the direct integration method.
18.
Note 33.
To Readers: This is not an exercise (yet). Ignore it.
In this chapter, we will study a solution technique that will help us solve differential equations that are first order and linear. Consider this (carefully selected) example.
\begin{equation} x^7 \cdot \frac{dy}{dx} + 7x^{6}\cdot y = e^x\tag{3} \end{equation}
Notice that the dependent variable is \(y\) and the independent variable is \(x\text{,}\) so we seek to find a formula for \(y\) in terms of \(x\text{.}\) The DE is also first order and linear.
You have attempted of activities on this page.
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