Finding Particular Solutions (College Board AP® Calculus AB): Revision Note

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

  • The general solution can be found by differentiation: $y=\int f\left(x\right)dx$

    • 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 $\left(a,y_0\right)$, then the particular solution to the equation is

    • $F\left(x\right)=y_0+{\int }_a^{x}f\left(t\right)dt$

      • Note that $F\left(a\right)=y_0+{\int }_a^{a}f\left(t\right)dt=y_0+0=y_0$, as required

  • Note as well that this is an alternative approach to 'finding the constant of integration'

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 $\frac{dy}{dx}=-x{y}^3$is $y^2=\frac{1}{x^2+C}$

  • If you know that $y=\frac{1}{3}$ when $x=2$, then

    • ${\left(\frac{1}{3}\right)}^2=\frac{1}{{\left(2\right)}^2+C}\Rightarrow \frac{1}{9}=\frac{1}{4+C}\Rightarrow C+4=9\Rightarrow C=5$

  • So the particular solution satisfying that initial condition is

    • $y^2=\frac{1}{x^2+5}$