Notes
What a differential equation is
A differential equation relates a function to its derivatives, like $\displaystyle \frac{dy}{dx} = 2xy$. A solution is a function that makes it true. A particular solution also meets an initial condition like $y(0) = 3$.
- To verify a solution, plug it and its derivative into the equation.
Slope fields
A slope field draws a short segment with slope $\displaystyle \frac{dy}{dx}$ at many points. Solution curves follow the segments.
- If $\displaystyle \frac{dy}{dx}$ depends only on $x$, segments in each column are parallel. If it depends only on $y$, segments in each row are parallel.
- Where $\displaystyle \frac{dy}{dx} = 0$, the segments are horizontal.
Separation of variables
1. Move all $y$ terms to one side with $dy$ and all $x$ terms to the other with $dx$. 2. Integrate both sides. Add $+ C$ (one constant is enough). 3. Use the initial condition to find $C$. 4. Solve for $y$. Example: $\displaystyle \frac{dy}{dx} = \frac{x}{y}$, $y(0) = 2$. $\displaystyle y\,dy = x\,dx \Rightarrow \frac{y^2}{2} = \frac{x^2}{2} + C$. $C = 2$, so $\displaystyle y = \sqrt{x^2 + 4}$ (positive root, since $y(0) > 0$).
Exponential models
$\displaystyle \frac{dy}{dt} = ky$ has solution $y = y_0e^{kt}$. The rate of change is proportional to the amount present: growth if $k > 0$, decay if $k < 0$.
BC only: Euler's method and logistic growth
- Euler's method steps along tangent lines: $\displaystyle y_{n+1} = y_n + \frac{dy}{dx}\Big|_{(x_n, y_n)} \cdot \Delta x$.
- Logistic model: $\displaystyle \frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$. The population approaches the carrying capacity $L$, and it grows fastest when $\displaystyle P = \frac{L}{2}$.
Topics in this unit
Each topic has its own short notes and practice.
- 7.1Modeling Situations with Differential Equations
- 7.2Verifying Solutions for Differential Equations
- 7.3Sketching Slope Fields
- 7.4Reasoning Using Slope Fields
- 7.5Approximating Solutions Using Euler's Method
- 7.6Finding General Solutions Using Separation of Variables
- 7.7Finding Particular Solutions Using Initial Conditions and Separation of Variables
- 7.8Exponential Models with Differential Equations
- 7.9Logistic Models with Differential Equations
Practice
$y' = 15e^{3x} = 3(5e^{3x}) = 3y$.
$\displaystyle \frac{dy}{dx} = 0$ when $x = 2$, for every $y$.
Slopes that depend only on $y$ are constant across each row.
$\displaystyle \frac{dy}{y} = 2x\,dx \Rightarrow \ln|y| = x^2 + C \Rightarrow y = Ce^{x^2}$.
$\displaystyle 2 = e^{5k} \Rightarrow k = \frac{\ln 2}{5}$.
The solution is $y = 4x$, so $y(3) = 12$.
$\displaystyle \frac{dy}{dx} = 1 - 3 = -2 < 0$.
$x^2y^3$ is a product of an $x$ part and a $y$ part.
Step 1: slope 1, $y = 1.5$ at $x = 0.5$. Step 2: slope $0.5 + 1.5 = 2$, $y = 1.5 + 1 = 2.5$.
Logistic growth is fastest at half the carrying capacity.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Consider $\frac{dy}{dx} = \frac{y^2}{x}$ with $y(1) = 1$, for $x > 0$.
(a) Write an equation of the tangent line at $x = 1$ and use it to approximate $y(1.2)$.
Show the answer and scoring guide
Slope $\displaystyle = \frac{1^2}{1} = 1$. Tangent: $y = 1 + (x - 1)$, so $y(1.2) \approx 1.2$.
Check off the points you earned:
(b) Find the particular solution $y = f(x)$.
Show the answer and scoring guide
$\displaystyle \frac{dy}{y^2} = \frac{dx}{x} \Rightarrow -\frac{1}{y} = \ln x + C$. At $(1, 1)$: $-1 = 0 + C$, so $C = -1$. Then $\displaystyle \frac{1}{y} = 1 - \ln x$, giving $\displaystyle y = \frac{1}{1 - \ln x}$ for $0 < x < e$.
Check off the points you earned: