### Video Transcript

Is the function π¦ is equal to two plus the natural logarithm of two π₯ a solution to the differential equation π₯ times π¦ prime is equal to two?

The question gives us a differential equation and a function. It wants us to determine whether this function is a solution to our differential equation.

To start, letβs take a closer look at our differential equation. We can see that the only part of this differential equation which contains π¦ is the expression π¦ prime. So, to check that this function is a solution to this differential equation, weβre going to need to find an expression for π¦ prime. Weβll then substitute this into our differential equation and see if it satisfies it.

Itβs also worth pointing out since π¦ is a function of π₯, π¦ prime means the derivative of π¦ with respect to π₯. So, letβs find an expression for π¦ prime. Thatβs the derivative of two plus the natural logarithm of two π₯ with respect to π₯. We want to evaluate this derivative term by term. Remember, for any nonzero constant π, the derivative of the natural logarithm of ππ₯ with respect to π₯ is equal to one over π₯.

This means we can now evaluate this derivative term by term. First, the derivative of the constant two is equal to zero. Next, the derivative of the natural logarithm of two π₯ equal to one over π₯. So, in this case, π¦ prime is equal to zero plus one over π₯. And, of course, we can simplify this to give us π¦ prime is equal to one over π₯. Now, we want to substitute this expression for π¦ prime into our differential equation.

So, we want to substitute π¦ prime is equal to one over π₯ into the equation π₯ times π¦ prime is equal to two. Doing this, we get π₯ times one over π₯ is equal to two. And we can simplify the left-hand side of this equation. We can cancel the shared factor of π₯ in our numerator and our dominator. And this means weβve simplified our equation to give us that one is equal to two. This is, of course, not true.

And because this is not true, this means that our function did not satisfy our differential equation, so it canβt have been a solution to the differential equation. Therefore, we were able to show that the function π¦ is equal to two plus the natural logarithm of two π₯ is not a solution of the differential equation π₯ times π¦ prime is equal to two.