Topic 7.7 Notes – Finding Particular Solutions Using Initial Conditions and Separation of Variables

General solution vs particular solution

When you solve a differential equation, you usually get a general solution.

Example:
If
$$\frac{dy}{dx} = 4x,$$
then integrating gives
$$y = 2x^2 + C.$$

That $+C$ means:

• There are infinitely many solution curves.
• Each value of $C$ gives a different graph.
• All of them satisfy the differential equation.

On a slope field, this shows up as many different curves following the same pattern of line segments.

Slope field and several solutions to $\frac{dy}{dx} = 4x$

Each parabola shown corresponds to a different value of $C$, but they all match the same slope field.

Now suppose you’re given an initial condition, like $y(1) = 5$.

Plug into the general solution:

$$5 = 2(1)^2 + C \Rightarrow C = 3.$$

So the particular solution is
$$y = 2x^2 + 3.$$

Key idea:

• A general solution describes a whole family.
• A particular solution is the one curve that passes through a specific point.
• For a given point $(a, y_0)$, there is exactly one solution curve through that point.

That uniqueness is something the AP expects you to understand conceptually.

Writing a particular solution with an integral

When the differential equation is

$$\frac{dy}{dx} = f(x),$$

you can write the particular solution satisfying $y(a)=y_0$ as

$$F(x) = y_0 + \int_a^x f(t)\,dt$$

Why this works:

• By the Fundamental Theorem of Calculus,
  $F'(x) = f(x)$.
• When $x = a$, the integral is 0, so
  $F(a) = y_0$.

This form:

• Automatically builds in the initial condition.
• Avoids solving for $C$.
• Often appears on FRQs that say “write an expression for the solution.”

Notice the variable inside the integral is $t$, not $x$. That prevents confusion with the upper limit.

Separation of variables with an initial condition

This is the most common skill tested here.

Suppose you’re given

$$\frac{dy}{dx} = x y.$$

Step-by-step logic

1. Separate variables
   $$\frac{1}{y} dy = x\, dx$$

2. Integrate both sides
   $$\int \frac{1}{y} dy = \int x\, dx$$

   $$\ln |y| = \frac{x^2}{2} + C$$

3. Solve for $y$

   $$y = Ce^{x^2/2}$$

   (The constant absorbs the ± from exponentiating.)

That’s the general solution.

Now use an initial condition, say $y(0)=4$:

$$4 = Ce^{0} \Rightarrow C=4.$$

So the particular solution is

$$y = 4e^{x^2/2}.$$

Algebra matters a lot here. Common trouble spots:

• Forgetting absolute values in $\ln |y|$.
• Losing the constant when exponentiating.
• Plugging in the initial condition before integrating.

On FRQs, even correct separation can earn credit, so show that step clearly.

Domain restrictions of solutions

Not every solution works for all $x$.

Restrictions happen because of:

• Division by zero (denominators).
• Logarithms (argument must be positive).
• Even roots (radicand ≥ 0).
• Context (time can’t be negative in many models).

Example:
If your solution ends up as
$$y = \frac{1}{x-3},$$
then $x \neq 3$.

If an initial condition is given, the solution is valid on the interval containing that point without crossing a discontinuity.

This matters on the AP. A correct formula with the wrong domain can cost you.