### Video Transcript

Is the function π¦ equals three π to the π₯ minus π₯ add one a solution to the differential equation π¦ prime equals π₯ add π¦?

Our differential equation is π¦ prime equals π₯ add π¦. Remember that a differential equation is an equation with a function and one or more of its derivatives. In this equation, we have π¦ prime. But remember that this is just another way of saying dπ¦ by dπ₯.

To assess whether π¦ equals three π to the π₯ minus π₯ add one is a solution to this differential equation. We need to substitute it into the left-hand side of our differential equation and the right-hand side of our differential equation. And weβll then see whether the left-hand side agrees with the right-hand side.

Starting with the left-hand side, if we begin with our function π¦ equals three π to the π₯ minus π₯ add one, we need to differentiate this in order to find π¦ prime. As we said earlier, π¦ prime is just another way of saying dπ¦ by dπ₯. So we need to differentiate our function π¦ with respect to π₯.

Starting with three π to the π₯, we recall that the derivative with respect to π₯ of π to the π₯ power is π to the π₯ power. So three π to the π₯ differentiates to give us three π to the π₯. Then moving on to the next term, we know that π₯ differentiates to one. So negative π₯ differentiates to negative one.

And finally, we recall that constants differentiate to zero. So we find that π¦ prime equals three π to the π₯ minus one. So thatβs the left-hand side of our differential equation when we use π¦ equals three π to the π₯ minus π₯ add one as a solution.

Now letβs look at the right-hand side of our differential equation. Replacing π¦ with our function three π to the π₯ minus π₯ add one, we see that π₯ add π¦ is π₯ add three π to the π₯ minus π₯ add one. We then see that the π₯s cancel. And weβre left with three π to the π₯ add one.

For π¦ equals three π to the π₯ minus π₯ add one to be a solution to our differential equation, we must have that the left-hand side and the right-hand side agree. On the left-hand side, we have three π to the π₯ minus one. But on the right-hand side, we have three π to the π₯ add one.

So, in fact, we see that the left-hand side and the right-hand side of this differential equation donβt agree. So we can conclude that this function π¦ is not a solution to this differential equation.