4.1.5
Speaker: So now we know that example 4 and 5 are both initial value problems, we're going to do two things. For example 4, we're going to verify that this function y equals 2e to the negative 2t plus e to the t is a solution to this given initial value problem, which means we actually need to do two things for that. First, we need to do as we did earlier verify this equation or that that function satisfies the differential equation and also determine whether y of 0 is in fact 3, and then next we're going to solve this other problem. And so rather than verify, we will have one to solve.
So let's leave it there. So first, we want to take the derivative of y. And say this is y prime. And y prime is negative 4e to the negative 2t plus e to the t. Yes? OK. Derivative of an exponential is an exponential.
Now, taking that paired with the function y, we want to show that y prime-- let's do this. You want to show that y prime plus 2y equals 3e to the t. Now, that will be negative 4e to the negative 2t plus e to the t plus 2, OK, plus 2 times the original function 2e to the negative 2t plus e to the t. You want to show that that is equal to 3 to the t.
Now, distributing, this is 4e to the negative 2t plus 2e to the t. Bring these down. Negative 4e to the negative 2t. These terms eliminate. They're a zero pair. These two combine to be 3e to the t, which is in fact equal to 3e to the t. So that function is a solution to the noninitial value problem version of this. It is a solution to that differential equation. Now you also want to verify that y of 0 equals 3.
Now, to do that, we will go to the function y equals 2e to the negative 2t plus e to the t and evaluate that at 0. 2e to the-- well, negative 2 times 0 is 0 plus e to the 0. That's 2 plus 1 or 3. So y of 0 is in fact 3, and this is a solution to this initial value problem.
[ENDS AT: 2:51]