Families of Solutions, Geometric Interpretation MATH 374 Lecture 2 Families of Solutions, Geometric Interpretation

1.3: Families of Solutions Consider the following differential equations: To solve (1), we can integrate (hopefully), and find where C is any real number.

One-Parameter Families In fact, since all antiderivatives of f(x) on an interval differ by a constant, all solutions of (1) are given by (3). We say that (1) has a one-parameter family of solutions.

One-Parameter Families It turns out that all solutions of (2) are given by the one-parameter family: y = C ek x, (4) where C is any real number. In general, what we find is that solutions to an nth order ODE (if they can be found) will occur as an n-parameter family of solutions. To find a particular solution to a given differential equation, we will need to have more information.

Example 1 Find the solution to that passes through (x,y) = (1,5).

Example 2 We will find that all solutions to y’’ + y = 0 (6) are of the form y = c1 cos x + c2 sin x. (7) To find a particular solution to (6), we need to specify two pieces of information, such as y(0) = 1 and y’(0) = -4. (8)

Example 2 Putting (8) into (7), we can find c1 and c2: 1 = y(0) = c1 cos(0) +c2 sin(0) = c1 -4 = y’(0) = -c1 sin(0) + c2 cos(0) = c2 Therefore, y = cos x – 4 sin x.

Elimination of Constants Given a family of curves with parameters, we can often find an associated differential equation. The way to do this is to eliminate the constants via differentiation or other mathematical techniques.

Example 3 Eliminate the constant a from (x-a)2 + y2 = a2. (9) Solution: Thinking of y as a function of x and differentiating, we find 2(x-a) + 2 y dy/dx = 0. Solving for a gives a = x + y dy/dx. (10) Solution (continued): Substituting (10) into (9) for a gives: Equation (11) can be simplified to the differential equation: y2 = x2 + 2xy dy/dx. (12)

Notes Another way to get (12) is to solve (9) for 2a: (x2 + y2)/x = 2a and differentiate both sides. (Try this!) It is much “easier” to start with a relation with constants and end up with a differential equation than it is to start with a differential equation and end up with a relation (or family of solutions)!

1.4: Geometric Interpretation To study the behavior of solutions of a differential equation, it is often useful to graph several members of a given family of solutions. Example 4: Sketch some solutions of (5), Note that through each point in the plane there passes one and only one member of the family of solutions! (Proof later …)

Example 5 Repeat with some solutions of (6), y’’ + y =0. Solution: Try y1 = cos x + sin x (c1 = c2 = 1) y2 = sin x (c1 = 0, c2 = 1) y3 = -4 sin x (c1 = 0, c2 = -4) Plotting these functions, we find that solutions cross many times! For example, y2 and y3 cross at multiples of . Therefore, the property of solutions we saw in Example 4 doesn’t hold here. It turns out a different uniqueness property holds here – through each point in the plane passes one and only one member of the family of solutions that has a given slope! (Proof later …)

Example 6 Sketch a few members of the one-parameter family of curves given by (9), (x-a)2 + y2 = a2. Solution: This family gives implicit solutions to differential equation (12): For example, the solution through (2,2) is valid for 0<x<4.

Example 6 (continued) Note that at any point (x,y) such that xy  0, (12) gives the slope of the line tangent to the curve through (x,y). Also, any point (x,y) where xy  0 has a unique member of the family passing through it! Question: What’s “wrong” with xy = 0? Answer: Rewriting the differential equation (12) gives: Therefore, infinite slope at points where xy =0.