Introduction to Differential Equations

Warmup

At each point (x0,y0) on a function y = f(x), the tangent line has the equationy =2x0x - y0. Find the equation for f(x).

Solution:

The equation for the tangent line to y = f(x) at (x0,y0) is

y = y0 + f’(x0)(x – x0)

OR

y = f’(x0)x + (y0 – x0f’(x0))

But we are given that the equation of the tangent is y =2x0x – y0.

Hence, f’(x0) = 2x0 and y0 = x0f’(x0) – y0 which gives y0 = x02.

Thus, f(x) = x2.

⇨

If F’(x) = 4-6+11 on the interval (-Determine the most general expression for F(x).

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 derivativeor to take the ‘anti-derivative’.

Because the derivative of a sum is the sum of the derivatives, we must look ateach term in F’(x) in order to find an antiderivative.

Consider: is the antiderivative of 4 -2 is the antiderivative of-6 11x is the antiderivative of 11

But what is the antiderivative of 0?

We need a constant, but we can’t know whatit is without additional information.

Solution:

Therefore F(x) = +11x + C

Don’t forget the C when taking a general antiderivative

⇨

We begin with two examples.

Find any function whose derivative is –

a) f(x) = 9x2+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) = 3x3 +2x2+C F(x) = 1.5sin2x+C

F(x) = 3x3 +2x2 + C F(x) = 1.5sin2x + C

⇨

Solve: tdt

ds2 with the initial condition s = 3 when t = 0.

Solution:

S(t) = t2 + C is the most general antiderivative of 2t.

At s = 3 and t = 0 3 = 02 + C Hence C = 3

Therefore S(t) = t2 + 3

Solve: 2)23( tdt

dy when y = 1 and t = 2.

Solution:y(t) = ct

3)23(6

1 at y = 1 and t = 2

C)43(61

1 3

C = 65

Therefore y(t) = 65

)t23(61 3

⇨

Determine the equation of the curve y = f(x) that passes through (0, 1) and satisfies

14 xdx

dy

Solution:y = 2x2 – x + C

at (0, 1), C = 1

Hence y = 2x2 – x + 1

x

y

x

y

x

y

Note: (1) the graphs are parallel since the constant distance between pointswith the same x-coordinate(2) the set of all solutions of the form y = F(x) + C is a one-parameter familyof solution curves, with C being the parameter ⇨

Find the antiderivative of each:

a) f(x) = sin πx

b) f(x) = (2x+5)4

c) 2x3

1

dx

dy

⇨

Solutions:

a) F(x) = + C

b) F(x) = + C

c) y = ln(3x-2) + C

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:

xdx

dy2 8.9

dt

dv 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 unknownFunction and its derivatives. For example,

kydt

dy

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..

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