Particular Solutions from Initial Conditions and Separation of Variables: Study Guide (Calculus AB 7.7)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer. y(a) = y₀ means “y equals y₀ when x = a”; this is called an initial condition. ∫ (a to x) f(t) dt is the definite integral of f from a to x. ∛ is the cube root.

From a family to one curve

In Topic 7.6 you found general solutions. Each one contains an arbitrary constant, so it describes infinitely many functions. In a slope field they appear as a whole family of curves.

A real problem usually gives one extra fact: a point the solution passes through. For example, “y = 1 when x = 0”. This is an initial condition. Only one member of the family passes through that point. That member is the particular solution.

So a general solution may describe infinitely many functions, but there is only one particular solution through a given point. For example, in Topic 7.6’s Figure 1, each point (a, b) lies on exactly one curve y = A e^(−x²): the one with A = b e^(a²).

The method

1. Separate the variables.
2. Antidifferentiate both sides and add one constant C.
3. Substitute the initial condition now and solve for C.
4. Solve for y. If a square root or an absolute value appears, use the initial condition to choose the sign.
5. State the domain: the largest open interval containing the initial x-value on which the solution is defined and satisfies the differential equation.
6. Check: the solution passes through the initial point, and its derivative matches the equation.

Steps 3 and 4 can be swapped, but substituting early keeps the algebra short. One rule is fixed: you must have the constant before you can use the initial condition. A solution without C has nothing to adjust.

Worked example 1: choosing the sign of a square root

Question. Find the particular solution of dy/dx = 4x³/y with y(1) = −3.

1. Separate: y dy = 4x³ dx.
2. Antidifferentiate: y²/2 = x⁴ + C.
3. Use the initial condition x = 1, y = −3: (−3)²/2 = 1⁴ + C, so 9/2 = 1 + C and C = 7/2.
4. Solve for y. y²/2 = x⁴ + 7/2, so y² = 2x⁴ + 7 and y = ±√(2x⁴ + 7). The initial value y(1) = −3 is negative, so take the negative root: y = −√(2x⁴ + 7)
5. Domain. 2x⁴ + 7 ≥ 7 > 0 for every x, so the root is always defined and y is never 0 (the equation divides by y). The solution is valid for all real x.

Check. At x = 1: −√(2 + 7) = −3. Correct. Derivative: dy/dx = −8x³/(2√(2x⁴ + 7)) = −4x³/√(2x⁴ + 7) = 4x³/y, because y = −√(2x⁴ + 7).

Why not the positive root? y = +√(2x⁴ + 7) gives y(1) = +3. It solves the equation but passes through (1, 3), not (1, −3).

Alternative order. You can solve for y first: y² = 2x⁴ + K, so y = ±√(2x⁴ + K). Then −√(2 + K) = −3 gives K = 7. Same answer. Notice that K = 2C here, so the constant changes its value when you rearrange. That is why it is safer to find it once, at one stage, and not mix the two routes.

Worked example 2: absolute values and a negative starting value

Question. Find the particular solution of dy/dx = (2x − 1)y with y(1) = −5.

1. Separate: (1/y) dy = (2x − 1) dx.
2. Antidifferentiate: ln|y| = x² − x + C.
3. Use the initial condition: ln|−5| = 1 − 1 + C, so C = ln 5.
4. Solve for y: |y| = e^(x² − x + ln 5) = e^(ln 5) · e^(x² − x) = 5e^(x² − x). So y = 5e^(x² − x) or y = −5e^(x² − x). The solution starts at a negative value. Since |y| = 5e^(x² − x) is never 0 and y is continuous, y can never change sign. So y stays negative: y = −5e^(x² − x)
5. Domain: all real x.

Check. y(1) = −5e⁰ = −5. Derivative: −5e^(x² − x) · (2x − 1) = (2x − 1)y. Correct.

A common wrong route. Writing y = e^(x² − x) + C and then using y(1) = −5 gives C = −6, so y = e^(x² − x) − 6. This passes through (1, −5) but is not a solution: its derivative is (2x − 1)e^(x² − x), not (2x − 1)(e^(x² − x) − 6). The point is right; the curve is wrong.

Domain restrictions

A solution to a differential equation is a differentiable function on an interval. Formulas can break that in three ways:

• Division by zero, which creates a vertical asymptote.
• A square root of a negative number, or a root that reaches 0 where the equation then divides by y.
• A logarithm of zero or a negative number.

The particular solution is the piece of the formula that contains the initial point and stays unbroken. Any other piece, beyond a break, is not part of it.

Worked example 3: a solution with a limited domain

Question. Find the particular solution of dy/dx = 2xy² with y(0) = 1, and state its domain.

1. Separate: (1/y²) dy = 2x dx.
2. Antidifferentiate: −1/y = x² + C. (Not a logarithm: 1/y² is a power of y.)
3. Use the initial condition: −1/1 = 0 + C, so C = −1.
4. Solve for y: −1/y = x² − 1, so 1/y = 1 − x² and y = 1/(1 − x²)
5. Domain. The formula breaks at x = 1 and x = −1, where 1 − x² = 0. The initial value x = 0 lies between them. So the particular solution is valid on −1 < x < 1.

Check. y(0) = 1. Derivative: d/dx (1 − x²)⁻¹ = 2x/(1 − x²)² = 2x · y². Correct.

Particular solutions in context

Exam questions often use other letters, such as t for time. The method is the same, and the answer needs units.

An invented plant grows so that its height H centimetres, t weeks after planting, satisfies dH/dt = 6/H, with H(0) = 4.

1. Separate and integrate: H dH = 6 dt, so H²/2 = 6t + C.
2. Use H(0) = 4: 16/2 = 0 + C, so C = 8 and H² = 12t + 16.
3. Choose the sign: heights are positive, and H(0) = 4 > 0, so H = √(12t + 16).
4. Use it: H(4) = √(48 + 16) = √64 = 8 cm after 4 weeks.

The formula works mathematically for t > −4/3, but the model only describes the plant from planting onwards, so it is used for t ≥ 0.

The integral form of a particular solution

When the right side depends on x only, dy/dx = f(x), there is a direct formula. The particular solution through (a, y₀) is

y = F(x) = y₀ + ∫ (a to x) f(t) dt

Why it works. At x = a the integral runs from a to a, so it is 0, and F(a) = y₀. By the Fundamental Theorem of Calculus (Topic 6.4), F′(x) = f(x) when f is continuous. So F passes through the right point and has the right derivative.

Check with a familiar case. dy/dx = 3x², y(2) = 5. The formula gives y = 5 + ∫ (2 to x) 3t² dt = 5 + (x³ − 8) = x³ − 3. Separating gives the same: y = x³ + C, 5 = 8 + C, C = −3.

When it really helps. Some functions have no antiderivative you can write with familiar functions. Take dy/dx = 1/(1 + x⁴) with y(0) = 2. The particular solution is

y = 2 + ∫ (0 to x) 1/(1 + t⁴) dt

To find y(1), evaluate the definite integral with a calculator: ∫ (0 to 1) 1/(1 + t⁴) dt ≈ 0.867, so y(1) ≈ 2.867. You never needed a formula for the antiderivative.

Two details matter. The lower limit is the x-value of the initial point, a. The number added outside is the y-value, y₀. Mixing them up is a common error.

Common misconceptions

• Putting the initial condition into the differential equation. Substituting (a, y₀) into dy/dx only gives the slope at that point. It does not find C. Substitute into the integrated equation.
• Assuming C = y₀. That is only true in special cases. Always solve for C.
• Adding C after rearranging, as in y = e^(x² − x) + C. The curve then passes through the point but does not satisfy the equation.
• Ignoring the sign. After y² = … or |y| = …, the initial value decides between + and −. Both signs together are not one function.
• Giving a domain with a gap, such as “all x except ±1”. A particular solution lives on one interval, the one containing the initial point.
• Using the wrong limits in the integral form. The lower limit is a, not y₀ and not 0 (unless a = 0).
• Forgetting y₀ in the integral form. y = ∫ (a to x) f(t) dt passes through (a, 0), not (a, y₀).

Where this leads

You can now go from a differential equation and one point to a single function with a stated domain. In Topic 7.8, Exponential Models with Differential Equations, you will apply exactly this method to dy/dt = ky and use it to model growth and decay. Calculus BC students will meet logistic models in Topic 7.9. For the general-solution step, revisit Topic 7.6. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.