Differential equations
Equations about rates, solved for the function itself.
Pen and paper is fine · no calculator needed why?
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the idea
A differential equation gives a rule for a function’s rate of change and asks for the function. For a first-order equation, one known value, the initial value, fixes the free constant.
Growth in proportion to size, y′ = ky, has the solution y = y(0) × eᵏᵗ. At t = ln m, eᵏᵗ is mᵏ, since e and ln undo each other.
The Laplace transform turns a function of t into a function of s, read from a short table.
techniques
Integrate, then fix the constant
- Integrate term by term: axⁿ becomes axⁿ⁺¹/(n + 1). Add a constant C.
- Put in the known point to find C.
- Evaluate y at the new x.
worked example
y′ = 3x² + 4x and y(1) = 5. What is y(2)?
- Integrate: y = x³ + 2x² + C.
- At x = 1: 1 + 2 + C is 5, so C is 2.
- y(2) = 8 + 8 + 2 = 18.
Answer: 18
Try y = eʳᵗ
- Put in y = eʳᵗ. Each derivative brings down a factor r, so you get r² + br + c = 0.
- Factor it as (r − p)(r − q): p and q multiply to c and add up to −b.
- The roots are p and q.
worked example
y″ − y′ − 6y = 0 has solutions of the form eʳᵗ. One value of r is negative. What is it?
- Putting in eʳᵗ gives r² − r − 6 = 0.
- Two numbers that multiply to −6 and add up to 1: 3 and −2.
- So (r − 3)(r + 2) = 0, and the negative root is −2.
Answer: −2
Read the Laplace table
- 1 becomes 1/s, and eᵃᵗ becomes 1/(s − a).
- tⁿ becomes (1 × 2 × … × n)/sⁿ⁺¹, so t² becomes 2/s³.
- sin bt becomes b/(s² + b²). Then put in s and reduce.
worked example
The Laplace transform of f(t) = t² is F(s). What is F(2)? Give an exact number or fraction.
- t² transforms to 2/s³.
- F(2) = 2/2³ = 2/8 = 1/4.
Answer: 1/4
watch out for
- Forgetting the constant C, or setting C to the starting value when the start is not at x = 0.
- Dropping the rate k. With y′ = 2y, the factor at t = ln 3 is 3², not 3.
- Flipping the signs of the roots. The factor (r − 2) gives r = 2, not −2.
- Using s + a for eᵃᵗ. Its transform is 1/(s − a).
practice
Initial value problem
worked example
y′ = −6x² + 4x − 4 and y(1) = 0. What is y(2)?
Answer: -12
- Integrate: y = −2x³ + 2x² − 4x + C.
- y(1) = −4 + C = 0, so C = 4.
- Then y(2) = −16 + 4 = −12.
y′ = ky
worked example
y′ = y and y(0) = 6. What is y(ln 3)?
Answer: 18
- The solution is y = 6eᵗ.
- At t = ln 3: e^(ln 3) = 3.
- So y(ln 3) = 6 × 3 = 18.
Characteristic roots
worked example
y″ − 9y = 0 has solutions of the form eʳᵗ. What are the two values of r? Enter both, smaller first.
Answer: (-3, 3)
- Substituting y = eʳᵗ gives r² − 9 = 0.
- This factors as (r + 3)(r − 3) = 0.
- So r = −3 or r = 3. Smaller first: −3, 3.
Laplace transforms
worked example
The Laplace transform of f(t) = e⁵ᵗ is F(s). What is F(6)? Give an exact number or fraction.
Answer: 1
- L{eᵃᵗ} = 1/(s − a) for s > a. Here a = 5, so F(s) = 1/(s − 5).
- F(6) = 1/(6 − 5) = 1.