Lessons 27 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
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Problem solving (integrals)
Textbook, Part 2: pp. 13–26
GoalLearn to solve the differential equations y′ = f(x), y′ = ky and y′ = k(y − a) and to apply them to motion, growth and cooling problems.
An equation containing the derivative of an unknown function is called a differential equation. The simplest kind is y′ = f(x): its solution is an antiderivative of f, that is y = F(x) + C, where C is an arbitrary number, so there are infinitely many solutions (the general solution). To pick one solution an initial condition is given, for example y(x₀) = y₀: we substitute it into the general solution and find C. In motion problems s′(t) = v(t) and v′(t) = a(t), so distance is an antiderivative of velocity and velocity of acceleration. If the rate of change of a quantity is proportional to the quantity itself, y′ = ky, and the solution is y = y₀e^(kt) (k > 0 is growth, k < 0 is decay). If the rate is proportional to the difference y − a, then y′ = k(y − a), and the substitution z = y − a gives y = a + (y₀ − a)e^(kt) — this is how Newton’s law of cooling is written.
Worked examples
Find the solution of y′ = 6x² − 4x with y(1) = 5. The general solution is y = 2x³ − 2x² + C. At x = 1 we get 2 − 2 + C = 5, so C = 5. Answer: y = 2x³ − 2x² + 5. Check: y′ = 6x² − 4x and y(1) = 5.
A bacteria count grows in proportion to itself, doubling every 2 hours, starting at 100. Then y = 100e^(kt) with e^(2k) = 2, so after 6 hours y = 100 · (e^(2k))³ = 100 · 8 = 800. Tea is at 80 °C, the room at 20 °C, T′ = k(T − 20); if the difference halves in 10 minutes, T(10) = 20 + (80 − 20) · ½ = 50 °C.
Class activity
“Braking”: a car travels at 20 m/s and brakes with acceleration a = −4 m/s². Write and solve the equations v′ = −4, v(0) = 20 and s′ = v, s(0) = 0. Find the stopping time and the braking distance (v = 20 − 4t, t = 5 s, s = 20t − 2t² = 50 m).
Practice
1
Find the solution of y′ = 4x³ − 3 with y(0) = 2.
y = x⁴ − 3x + 2
2
A point has velocity v(t) = 6 − 2t (m/s). How far does it travel from t = 0 to t = 3 s (v > 0)?
9
3
Given y′ = ky, y(0) = 5 and y(1) = 10, find y(3).
40
4
Why is the solution of y′ = f(x) a whole family of functions rather than one function?
Because an antiderivative is determined only up to a constant term: the derivative of a constant is zero, so F(x) + C has derivative f(x) for every C. An initial condition picks out one solution.