Video Transcript
Is the function π¦ equals π to the power of five π₯ minus π to the π₯ a solution to the differential equation π¦ prime equals five π¦ minus four π to the power π₯?
The idea here is that we check the possible solution by substituting it into the differential equation to see if we get a true statement. Our differential equation is π¦ prime equals five π¦ minus four π to the π₯. And weβre going to make substitutions for π¦ prime and π¦. On the right-hand side, we can replace π¦ with the possible solution π to the five π₯ minus π to the π₯, to get five multiplied by π to the five π₯ minus π to the π₯ minus four π to the π₯.
Letβs now distribute the parentheses. And then we can simplify this expression by collecting like terms, to get five π to the five π₯ minus nine π to the π₯. On the left-hand side of the differential equation, we have π¦ prime. Remember, weβre checking if π to the five π₯ minus π to the π₯ is a solution to this. So we need to find π¦ prime for our possible solution in order to make this substitution.
We remember the general rule that the derivative of π raised to π of π₯, a function of π₯, is π prime of π₯ multiplied by π to the power of π of π₯. And so π¦ prime equals five π to the power of five π₯ minus π to the π₯. And so thatβs the left-hand side of our differential equation.
So is this a true statement? Well, it doesnβt work for any value of π₯. So we can conclude that π to the five π₯ minus π to the π₯ is not a solution to this differential equation.