AP Calculus AB · Lesson 5 of 5
Differential equations
Read a differential equation as a statement about a rate, and use an initial condition to select one solution.
What the equation says
dy/dx = 2x says the slope at each x is 2x. Solutions are functions, not a single number. Integrating both sides gives y = x² + C, a family of parabolas with the same derivative.
An initial condition such as y(0) = 3 selects C. Here 3 = 0 + C, so y = x² + 3. A solution must satisfy both the differential equation and the initial condition.
Separable equations and verification
If dy/dx = ky, separating variables leads to an exponential solution y = Ae^(kx), provided y is not zero. The constant A is chosen from the initial amount. This is the model for continuous proportional growth or decay.
To verify a proposed solution, differentiate it and substitute it back into the differential equation. Then check the initial condition. Passing only one of those tests is not enough.
Worked example
Solve dy/dx = 4y with y(0) = 2.
- Separate: dy/y = 4 dx, for y ≠ 0.
- Integrate: ln|y| = 4x + C, so y = Ae^(4x).
- y(0) = 2 gives A = 2, so y = 2e^(4x). Differentiating returns 8e^(4x), which equals 4y.
Why this works. The differential equation produces the family. The initial condition produces the particular function.
Check your understanding
- Interpret dy/dx as a slope or a rate before integrating.
- Use the initial condition to find the constant.
- Verify by differentiating and by checking the initial value.