differential equation v/s algebraic equations
Okay, let's begin by considering a very basic differential equation: y prime equals cosine of x, and let's solve this equation. To solve means to find the unknown function y, and we can do this by integrating both sides of the equation with respect to x. The integral of y prime dx equals the integral of the cosine of x dx, and that implies y is equal to the integral of cosine, which is the sine of x plus C.
Here, the constant of integration will be very important. So let's see if you can remember why we need it. For example, if C equals 1 then y equals sine of x plus 1, and the derivative of y is the cosine of x plus the derivative of one, which is zero because one is a constant. Therefore, y prime is just cosine of x. If C is a different number such as two thirds, then y equals sine of x plus two thirds, and again, the derivative y prime equals cosine of x. This is what we want: a function y satisfying y prime equals cosine of x, and these two functions are two different solutions. Since C can be any number, they are infinitely many solutions altogether.
What's different, then, about the solutions of differential equations and the solutions of algebraic equations. Let's consider the example from the last slide, y prime equals cosine of x, and an equation from algebra such as x squared plus five x plus six equals zero. With the differential equation, we saw that the solution was y equals sine of x plus C, and for each value of C, the solution is a function. On the other hand, if we solve this algebraic equation, which we can do by factoring, we get that x is either negative 2 or negative 3 which are numbers, as opposed to the solutions of the differential equation, which are functions.
Now that we've solved a basic differential equation and have considered a key difference between differential equations and the more familiar algebraic equations, we'll set up and solve an important differential equation in the next video. See you then!
Post a Comment