AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/7/7-7)
Unit 7 · Topic 7.7
7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables
An initial condition, such as y(1) = 1, picks out the one curve in a family of solutions that passes through a given point. Use it to find C right after integrating, then solve for y and state any domain restrictions.
Key terms
- particular solution
- initial condition
- separation of variables
- domain restriction
- constant C
From general to particular
A general solution like y²/2 = x³/3 + C is a family of curves. An initial condition, written y(x₀) = y₀ or f(x₀) = y₀, says the curve passes through (x₀, y₀). Substitute that point to find C. The result is the particular solution, the one member of the family through that point.
The full routine
Finding C before solving for y avoids messy algebra with constants inside logs or roots.
- Separate variables.
- Integrate both sides and add C.
- Substitute the initial condition now, while the equation is still simple, and solve for C.
- Solve for y.
- If there's a ± from a square root, choose the sign that matches the initial condition.
- State the domain, if it's restricted.
Domain restrictions
A particular solution is defined on an interval that contains x₀, where the solution is differentiable and satisfies the equation. If the formula breaks down, from dividing by zero, taking a log of a non-positive number or a square root of a negative, the solution's domain stops there.
For example, y = 2/(3 − x²) has vertical asymptotes at x = ±√3. If the initial point is at x = 1, the solution's domain is −√3 < x < √3, the interval between the asymptotes that contains 1. The pieces beyond the asymptotes don't belong to this solution.
Check your answer
Two quick checks catch most errors. First, plug in the initial point: your formula must give y₀ at x₀. Second, differentiate your solution and confirm it satisfies the differential equation (7.2). For y = 2/(3 − x²), the derivative is 4x/(3 − x²)², and xy² = x · 4/(3 − x²)² gives the same thing.
Solutions written as integrals
If dy/dx = f(x) and y(a) = y₀, then the particular solution is y = y₀ + ∫ₐˣ f(t) dt. The Fundamental Theorem (6.4) shows its derivative is f(x), and at x = a the integral is 0, so y(a) = y₀. This form is useful when f has no elementary antiderivative, such as f(x) = sin(x²), and it's the same idea as final amount = starting amount + accumulated change.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Particular solution with a domain
Find the particular solution of dy/dx = xy² with y(1) = 1, and state its domain.
Show the solutionHide the solution
- Step 1: Separate: y⁻² dy = x dx.
- Step 2: Integrate: −1/y = x²/2 + C.
- Step 3: Use (1, 1): −1 = 1/2 + C, so C = −3/2.
- Step 4: Then −1/y = x²/2 − 3/2 = (x² − 3)/2, so y = −2/(x² − 3) = 2/(3 − x²).
- Step 5: Check the point: y(1) = 2/2 = 1. ✓
- Step 6: Domain: the denominator is 0 at x = ±√3. The interval containing x = 1 is −√3 < x < √3.
Answer: y = 2/(3 − x²), for −√3 < x < √3
- Example 2
Trap: choosing the sign of the root
Find the particular solution of dy/dx = cos x/(2y) with y(0) = −3.
Show the solutionHide the solution
- Step 1: Separate: 2y dy = cos x dx.
- Step 2: Integrate: y² = sin x + C.
- Step 3: Use (0, −3): 9 = 0 + C, so C = 9 and y² = sin x + 9.
- Step 4: Take the square root: y = ±√(sin x + 9). Since y(0) = −3 is negative, choose the negative root.
- Step 5: sin x + 9 ≥ 8 for all x, so the solution is defined for all real x.
- Step 6: Common wrong answer: y = √(sin x + 9), which gives y(0) = +3.
Answer: y = −√(sin x + 9), for all real x
- Example 3Calculator allowed
Solution as an accumulation function
Let y = f(x) satisfy dy/dx = sin(x²) with f(2) = 5. Write f(x) using an integral, and use a calculator to find f(3).
Show the solutionHide the solution
- Step 1: sin(x²) has no elementary antiderivative, so write f(x) = 5 + ∫₂ˣ sin(t²) dt.
- Step 2: Check: f′(x) = sin(x²) by the Fundamental Theorem, and f(2) = 5 + 0 = 5.
- Step 3: f(3) = 5 + ∫₂³ sin(t²) dt. The calculator gives ∫₂³ sin(t²) dt ≈ −0.031.
- Step 4: So f(3) ≈ 4.969.
Answer: f(x) = 5 + ∫₂ˣ sin(t²) dt; f(3) ≈ 4.969
Common mistakes
- Solving for y first and then trying to find C inside an exponent or root. Plug in the point right after integrating.
- Keeping ± in the final answer. The initial condition decides the sign.
- Giving a domain that includes an asymptote or crosses it. The domain is one interval, and it must contain the starting x.
- Writing y = ∫₂ˣ sin(t²) dt and forgetting the starting value of 5.
On the exam
- A typical free-response part says: “Find the particular solution y = f(x) … with the initial condition f(1) = 1.” The points usually go to separating, antiderivatives, the constant, using the initial condition and solving for y. Show each step.
- If you can't solve for y, an implicit answer with the correct C can still earn most of the credit.
Connected topics
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Check yourself
4 questions on 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables. Pick an answer to see if you got it, and why.
Let y = f(x) be the solution to the differential equation dy/dx = √(1 + x³) with f(0) = 2. What is f(3)?
Let y = f(x) be the particular solution to the differential equation dy/dx = 2xy² with f(0) = 1. What is f(1/2)?
The function y satisfies dy/dt = 0.3y cos t, with y = 4 when t = 0. What is the value of y when t = 2?
A pan of soup is taken off the stove and left in a kitchen kept at 20°C. Its temperature T, in degrees Celsius, satisfies dT/dt = −k(T − 20), where t is in minutes and k is a positive constant. The soup is 90°C at t = 0 and 60°C at t = 5. At what time is the soup 30°C?
0 of 4 answered