Differential equations and initial conditions
Objective. Recover functions from rates and known points.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
A differential equation describes a rate of change, and solving dy/dx = f(x) by integration gives a general solution y = F(x) + c containing an unknown constant that must be fixed using a given initial condition, such as a known point the curve passes through. After integrating to find the general solution, substitute the given x and y values to solve for c, giving one specific function rather than a whole family of curves. For example, if dy/dx = 2x and the curve passes through (1, 4), then y = x² + c, and substituting gives 4 = 1 + c, so c = 3 and y = x² + 3. This technique recovers an exact function from information about its rate of change plus one known point.
Worked example
A curve satisfies dy/dx = 3x² − 4 and passes through the point (2, 5). Use this initial condition to find y in terms of x.
- 1Integrate: y = x³ − 4x + c.
- 2Substitute the point (2, 5): 5 = 8 − 8 + c.
- 3Solve: 5 = 0 + c, so c = 5.
- 4Write the specific solution: y = x³ − 4x + 5.
Answer: y = x³ − 4x + 5
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Practice
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Start practisingCommon mistake. Students sometimes solve for c using the wrong coordinate, substituting the x-value in for y or vice versa. Always substitute the x-coordinate into x and the y-coordinate into y of the general solution before solving for c.
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