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

General vs. Particular Solutions

A general solution is a family of functions that all satisfy a differential equation.
It includes an arbitrary constant $C$.

Example forms:

• $y = Ce^{2x}$
• $y = \ln|x| + C$

Each different value of $C$ gives a different curve.

A particular solution is the one specific function that:

1. Satisfies the differential equation
2. Passes through a given point like $y(1) = 3$

That initial condition picks out one curve from the family.

Here’s what that looks like on a slope field.

Slope field with a family of solution curves and one particular solution

The light blue curves represent the general solution, one for each value of $C$. The highlighted curve is the particular solution that passes through $(1,3)$.

Important idea:
For the kinds of equations you solve in AP Calculus AB, one initial condition → exactly one solution curve (on an interval where the function is defined).

Separation of Variables with an Initial Condition

This applies when the differential equation can be written in the form

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

The process

1. Separate the variables

   Move all $y$-terms with $dy$, all $x$-terms with $dx$.

   $$\frac{1}{g(y)}\,dy = f(x)\,dx$$

   If variables aren’t fully separated, you won’t get credit on an FRQ.

2. Integrate both sides

   $$\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$$

   Add + C (just one constant, since constants combine).

   Now you have the general solution.

3. Solve for $y$ (if possible)

   You might need:

   • Exponentials to undo logs
   • Log rules
   • Care with absolute values

   Most algebra mistakes happen here, not in the calculus.

4. Apply the initial condition

   Plug in the given point to solve for $C$.

5. Write the particular solution

   Substitute your value of $C$ back in.

Quick example (structure only)

Suppose
$$\frac{dy}{dx} = 3xy$$

Separate:
$$\frac{1}{y}dy = 3x\,dx$$

Integrate:
$$\ln|y| = \frac{3}{2}x^2 + C$$

Solve:
$$y = Ce^{\frac{3}{2}x^2}$$

If given $y(0)=4$, then $4 = Ce^0$, so $C=4$.

Particular solution:
$$y = 4e^{\frac{3}{2}x^2}$$

That’s the full pipeline.

Writing a Particular Solution as an Integral

Sometimes you’re given

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

with an initial condition $y(a) = y_0$.

A particular solution can be written as

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

This guarantees:

• $F'(x) = f(x, y(x))$
• $F(a) = y_0$

If the equation is just $\frac{dy}{dx} = f(x)$, then this becomes

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

On free response, when they say “write an expression for the particular solution,” this format is often exactly what they want. Include the initial value in the expression or you lose the point.

Domain Restrictions

Solutions don’t always work for all $x$.

1. Algebra restrictions

• Denominator ≠ 0
• Inside a log must be positive
• Even roots need nonnegative input

Example:
If your solution contains $\ln(x-5)$, then $x>5$.

2. Restrictions from separation

If you divided by something like $y$, you assumed $y \neq 0$.
Constant solutions (like $y=0$) might need to be checked separately.

3. Interval containing the initial condition

Your solution is valid only on the interval:

• Where the formula makes sense
• That includes the given initial value

You cannot cross a vertical asymptote.