Finding Particular Solutions (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Initial conditions

How can I use initial conditions to find the particular solution of a differential equation?

Remember that the general solution of a differential equation represents a family of solutions
- In general there will be an infinite number of possible solutions
- Each of these possible solutions is known as a particular solution
You need additional information to determine a particular solution
- This additional information is known as an initial condition (or sometimes boundary condition)
- For example, you might be given a value of $y$ that corresponds to a particular value of $x$
If you think of the family of solutions as a family of curves on a graph
- then there is only one particular solution that passes through a given point
  - So if you know a point the solution curve goes through
  - then you can determine the unique particular solution
Consider the simple case where $\frac{d y}{d x} = f (x)$
- The general solution can be found by differentiation: $y = \int f (x) d x$
  - This solution will contain a constant of integration, and so represent an infinite number of possible solutions
- But if you know that the solution goes through the point $(a, y_{0})$ , then the particular solution to the equation is
  - $F (x) = y_{0} + \int_{a}^{x} f (t) d t$
    - Note that $F (a) = y_{0} + \int_{a}^{a} f (t) d t = y_{0} + 0 = y_{0}$ , as required
- Note as well that this is an alternative approach to 'finding the constant of integration'

Worked Example

The function $F$ defined by $F (x) = y_{0} + \int_{a}^{x} f (t) d t$ is a particular solution to the differential equation $\frac{d y}{d x} = f (x)$ , satisfying $F (a) = y_{0}$ .

Use this fact to find the particular solution to the differential equation $\frac{d y}{d x} = \frac{3}{x}$ , given that $y = 1$ when $x = e$ .

Answer:

Here $y_{0} = 1$ and $a = e$

Recall that $\ln (e) = 1$

$\begin{array}{rcl} F (x) & = & 1 + \int_{e}^{x} \frac{3}{t} d t \\ = & 1 + 3 \int_{e}^{x} \frac{1}{t} d t \\ = & 1 + 3 {[\ln |t|]}_{e}^{x} \\ = & 1 + 3 (\ln |x| - \ln |e|) \\ = & 1 + 3 (\ln |x| - 1) \\ = & 3 \ln |x| - 2 \end{array}$

$y = 3 \ln |x| - 2$

Finding particular solutions using separation of variables

If you are given an initial condition, then you can find the particular solution to a differential equation solved by separation of variables:
- Substitute the initial condition values into the general solution
- and solve to find the value of the arbitrary constant
E.g. the general solution to $\frac{d y}{d x} = - x y^{3}$ is $y^{2} = \frac{1}{x^{2} + C}$
- If you know that $y = \frac{1}{3}$ when $x = 2$ , then
  - ${(\frac{1}{3})}^{2} = \frac{1}{{(2)}^{2} + C} \Rightarrow \frac{1}{9} = \frac{1}{4 + C} \Rightarrow C + 4 = 9 \Rightarrow C = 5$
- So the particular solution satisfying that initial condition is
  - $y^{2} = \frac{1}{x^{2} + 5}$

Worked Example

Find the solution to the differential equation $(x + 3) \frac{d y}{d x} = \sec y$ , given that the graph of the solution goes through the point $(- 2, \frac{3 π}{2})$ .

Answer:

Separate the variables, getting all the $y$ terms on the $\frac{d y}{d x}$ side and all the $x$ terms on the other side

Recall that $\sec y = \frac{1}{\cos y}$

$(x + 3) \frac{d y}{d x} = \frac{1}{\cos y} \cos y \frac{d y}{d x} = \frac{1}{x + 3}$

Integrate both sides with respect to $x$

$\int \cos y d y = \int \frac{1}{x + 3} d x$

Integrate (and don't forget a constant of integration)

$\sin y = \ln |x + 3| + C$

Don't try to rewrite that as $y = \arcsin (\ln |x + 3| + C)$ ; that would lose parts of the solution

Now bring in the boundary condition, $y = \frac{3 π}{2}$ when $x = - 2$

Recall that $\ln (1) = 0$

$\begin{array}{rcl} \sin (\frac{3 π}{2}) & = & \ln |(- 2) + 3| + C \\ - 1 & = & 0 + C \\ C & = & - 1 \end{array}$

Substitute that value of $C$ into the general solution

$\sin y = \ln |x + 3| - 1$

Last updated: 14 August 2024

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Finding Particular Solutions (College Board AP® Calculus AB)

Study Guide

Initial conditions

How can I use initial conditions to find the particular solution of a differential equation?

Worked Example

Finding particular solutions using separation of variables

Worked Example

You've read 0 of your 5 free study guides this week

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Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Author: Dan Finlay