Linear First-Order Equations

When separation fails, multiply by something clever.

12 problems · about 24 minutes · F-IF.B.6, A-CED.A.1

What this lesson teaches

The student recognises linear first-order equations and solves them with an integrating factor.

A linear first-order equation has the form dy/dx + P(x)y = Q(x).
Multiplying by e to the integral of P makes the left side a product-rule derivative.
The solution splits into a fading part carrying the constant and a persisting part.

Straightforward practice. Get the method working first.

5 problems

dy/dx + 6y = 18. What constant value does y settle toward?
Answer 3
Why 3.
What does the integrating factor accomplish?
Answer It turns the left side into a single product-rule derivative.
Why The product rule in reverse.
dy/dx + 4y = 8. What is P?
Answer 4
Why The coefficient of y.
dy/dx + 4y = 8. What constant value does y settle toward?
Answer 2
Why Set dy/dx to zero and solve.
dy/dx + 3y = 12. What constant value does y settle toward?
Answer 4
Why 12 divided by 3.

The same ideas with more to keep track of.

3 problems

Is dy/dx + 2y = x linear? 1 yes, 0 no.
Answer 1
Why y appears to the first power only.
Is dy/dx + y² = x linear? 1 yes, 0 no.
Answer 0
Why A squared y.
y = 7 + Ce^(−3x) with y = 10 at x = 0. What is C?
Answer 3
Why 10 − 7.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Order the steps of the integrating factor method.
Answer 1. Write the equation in standard form 2. Compute the integrating factor from P 3. Multiply through and recognise the product rule 4. Integrate both sides and divide the factor back out
Why Standard form must come first, or P is not visible.
y = 5 + Ce^(−x). What value does y approach as x grows large?
Answer 5
Why The exponential fades.
The Factor: dy/dx + 5y = 1. What is the coefficient P?
Answer 5
Why 5.
The Long Run: y = 3 + Ce^(−2x). What value does y approach as x grows large?
Answer 3
Why 3.