Introduction to Differential Equations

Warmup At each point (x 0,y 0 ) on a function y = f(x), the tangent line has the equation y =2x 0 x - y 0. Find the equation for f(x). Solution: The equation for the tangent line to y = f(x) at (x 0,y 0 ) is y = y 0 + f’(x 0 )(x – x 0 ) OR y = f’(x 0 )x + (y 0 – x 0 f’(x 0 )) But we are given that the equation of the tangent is y =2x 0 x – y 0. Hence, f’(x 0 ) = 2x 0 and y 0 = x 0 f’(x 0 ) – y 0 which gives y 0 = x 0 2. Thus, f(x) = x 2.

Since we are given the derivative and we must find the original function, the process we must follow must be go ‘backwards’ from taking the derivative or to take the ‘anti-derivative’. Because the derivative of a sum is the sum of the derivatives, we must look at each term in F’(x) in order to find an antiderivative. But what is the antiderivative of 0? We need a constant, but we can’t know what it is without additional information. Solution: Don’t forget the C when taking a general antiderivative

We begin with two examples. Find any function whose derivative is – a)f(x) = 9x 2 +4x b) f(x)=3cos2x These are the antiderivatives An antiderivative of the function f(x) is a function F(x) that satisfies F’(x) = f(x) How many possible antiderivatives are there for each function? - There are an infinite number of solutions! - They can be represented as F(x) = 3x 3 +2x 2 +C F(x) = 1.5sin2x+C F(x) = 3x 3 +2x 2 + C F(x) = 1.5sin2x + C

Solve:with the initial condition s = 3 when t = 0. Solution: S(t) = t 2 + C is the most general antiderivative of 2t. At s = 3 and t = 0 3 = C Hence C = 3 Therefore S(t) = t Solve:when y = 1 and t = 2. Solution: y(t) = at y = 1 and t = 2 C = Therefore y(t) =

Determine the equation of the curve y = f(x) that passes through (0, 1) and satisfies Solution: y = 2x 2 – x + C at (0, 1), C = 1 Hence y = 2x 2 – x + 1 Note: (1) the graphs are parallel since the constant distance between points with the same x-coordinate (2) the set of all solutions of the form y = F(x) + C is a one-parameter family of solution curves, with C being the parameter

Find the antiderivative of each: a) f(x) = sin πx b) f(x) = (2x+5) 4 c) Solutions:

Many of the general laws of nature find their most useful form in equations that involve rates of change. These equations are called differential equations because they contain functions and their differential quotients. Some examples of differential equations are: P’ = 3P We have begun by working with equations of the form y’ = f(x), the solutions of which are called antiderivatives. Recall: an antiderivative of a function f(x) is a function F(x) where F’(x) = f(x). This is a simple case of a d.e. More generally, a differential equation is any equation that involves an unknown Function and its derivatives. For example, Where k is a constant is a common form of a d.e., with y denoting the unknown function. The process of finding the unknown function is referred to as solving the d.e.. Any function that when substituted for the unknown function, reduces the d.e. to an identity, is said to be a solution of the d.e..