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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 5.
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{14} \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.
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