17–20. Verifying solutions of initial value problems Verify th...

Question

17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.

y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5

Accepted Answer

Welcome back, everyone. Determine whether Y of S equals 35 minus 4 solves the initial value problems of minus 5 Y S equals 20, where Y 1 equals -1. A says yes and B says no. For this problem, we're going to identify the terms required for the differential equation. Considering the left hand side, we're going to begin with Y S. Which is the derivative of 3 to 5 minus 4. Applying the power rule, we get 15s to the 4th, right? That's our derivative Y S. Now, let's identify as multiplied by Y S minus 5 Y S. and we're going to check if it is equal to 20. We get S multiplied by 15s. To the 4th minus 5 multiplied by. 3 to 5th minus 4. Simplifying, we get 15s to the 5th minus 15s to the 5th plus 5 multiplied by 4. That's 20, right? We can notice that our terms cancel each other out and we get 20 on the left hand side and we indeed have 20 on the right hand side. So we have satisfied the first condition and we want to check. The initial value, right? It says that Y of 1 must be equal to -1. So we're going to use our original function. And sets equals 1. And check if S equals 1 produces -1 for Y, right? So we get 3 multiplied by 1 to the 5th minus 4, which is 3 minus 4, and that's indeed -1, right? We get -1 for Y when S is equal to 1. So we also satisfy the initial value problem. Therefore, the correct answer for this problem is A. Yes, Y of S is a solution to the initial value problem. Let's label it as our final answer, and thank you for watching.

17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5

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17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.
y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5