Master the art of solving differential equations and modeling real-world phenomena
← Back to UnitsUnit 7 focuses on differential equations and their applications. You'll learn how to:
A differential equation is an equation that relates a function with its derivatives. The general form is:
where y' = dy/dx, y'' = d²y/dx², etc.
Differential equations describe how a quantity changes in relation to itself and other variables.
dy/dx = 2x
Solution: y = ∫2x dx = x² + C
This is a family of parabolas, each with a different y-intercept.
Type | Form | Example |
---|---|---|
Separable | dy/dx = f(x)g(y) | dy/dx = xy |
Linear | dy/dx + P(x)y = Q(x) | dy/dx + 2y = x |
Homogeneous | dy/dx = f(y/x) | dy/dx = (x+y)/(x-y) |
A separable differential equation can be written in the form:
We can separate variables and integrate both sides
Solve: dy/dx = xy
Solution:
A slope field (or direction field) is a graphical representation of a differential equation. At each point (x,y), we draw a small line segment with slope equal to dy/dx at that point.
Slope fields help us visualize the family of solutions to a differential equation without solving it analytically.
For dy/dx = x - y:
Draw small line segments at each point with these slopes.
Follow the slope field to sketch approximate solution curves.
Use initial conditions to find specific solution curves.
Given dy/dx = x - y and y(0) = 1, sketch the solution curve.
Solution: Start at (0,1) and follow the slope field to sketch the curve that passes through this point.
Euler's method is a numerical technique for approximating solutions to differential equations:
where h is the step size
Approximate y(1) for dy/dx = x + y, y(0) = 1, using h = 0.5
Solution:
So y(1) ≈ 2.5
The differential equation for exponential growth is:
where k > 0 is the growth constant
A population grows at a rate proportional to its size. If P(0) = 1000 and P(2) = 2000, find P(t).
Solution:
For exponential decay, the equation is the same but k < 0:
where k < 0 and y₀ is the initial amount
A radioactive substance decays at a rate proportional to its mass. If 50% remains after 10 years, find the half-life.
Solution:
Logistic growth models situations where growth is limited by carrying capacity:
where L is the carrying capacity and k > 0
Solve dy/dt = 0.1y(100 - y) with y(0) = 10
Solution:
Solve the separable differential equation: dy/dx = x²y
Solution:
1) Separate variables: dy/y = x² dx
2) Integrate both sides: ∫dy/y = ∫x² dx
3) ln|y| = x³/3 + C
4) y = e^(x³/3 + C) = Ce^(x³/3)
Use Euler's method with h = 0.5 to approximate y(1) for dy/dx = 2x + y, y(0) = 1.
Solution:
Step 1: x₁ = 0.5, y₁ = 1 + 0.5(2(0) + 1) = 1 + 0.5 = 1.5
Step 2: x₂ = 1.0, y₂ = 1.5 + 0.5(2(0.5) + 1.5) = 1.5 + 0.5(2.5) = 2.75
Therefore, y(1) ≈ 2.75
A bacteria culture grows exponentially. If it doubles every 3 hours and starts with 100 bacteria, find the population after 6 hours.
Solution:
1) P(t) = P₀e^(kt) where P₀ = 100
2) P(3) = 200 = 100e^(3k), so e^(3k) = 2
3) k = ln(2)/3
4) P(6) = 100e^(6·ln(2)/3) = 100e^(2ln(2)) = 100·2² = 400
After 6 hours, there are 400 bacteria.
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In Unit 8, you'll learn about applications of integration including area, volume, and arc length.
Go to Unit 8: Applications of Integration →