The Initial Value Problem for Differential Equations

1. The Initial Value Problem for Differential Equations 1.1. Summary on differential equations. A differential equation is an equation, where the unknown is a function, and both the function and its derivatives appear in the equation. Examples of differential equations for the unknown function y are given below: dy = −2 y, dx dy = −2 y + 3 y 7 , dx ( dy )2 y ln(x) tan(y) = . dx (1 + x2 ) A solution to a differential equation is a function y that satifies the equation. Example 1.1: Verify that the functions y(x) = c e−2x , for all constants c ∈ R, are solutions dy = −2 y. to the differential equation dx Solution: We need to compute the left-hand side and the right-hand side of the differential equation, and check that we obtain the same result. This is the case, since, dy = −2c e−2x dx and − 2y(x) = −2ce−2x ⇒ dy = −2 y. dx (1) (2) (3)

▹ In the example above we saw that there are infinitely many solutions y of the differential equation, one for each value of the constant c. It is often the case in physics that one is interested in only one solution among all solutions to a differential equation. One way to select one particular solution of a differential equation is to require that y(x0 ) = y0 . That is, the solution at a given point x0 takes the given value y0 . The point x0 is usually called the initial point, and the condition y(x0 ) = y0 is called the initial condition. An initial value problem for Eq. (1) is the following: Given the constants x0 , y0 , find all functions y solutions of dy = −2 y, y(x0 ) = y0 . dx It is not difficult to see that the initial value problem above has a unique solution. A method to find such solution is to find the general solution, involving a constant c, and then fix the constant using the initial condition. Example 1.2: Find the solution to the initial value problem dy = −2 y, dx y(0) = 3.

Solution: We know that all solutions of the differential equation are y(x) = c e−2x . The constant c is determined by the initial condition. Indeed, 3 =...