### Video Transcript

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

The question gives us a differential equation and a function. It wants us to find out if this function is a solution to our differential equation. To start, letβs take a look at our differential equation. Itβs in the form π¦ prime is equal to some function of π₯. And, of course, since π¦ is a function of π₯, we have π¦ prime is equal to the first derivative of π¦ with respect to π₯. Remember, we call π¦ is equal to some function of π₯ a solution to a differential equation if it satisfies that differential equation.

What this means in practice is we want to find the derivatives of π¦ with respect to π₯ from our solution and substitute these into our differential equation. In our case, our differential equation only contains dπ¦ by dπ₯. So letβs find that expression for dπ¦ by dπ₯. For the solution given to us, thatβs the derivative of five minus three π₯ plus three π₯ times the natural logarithm of π₯ with respect to π₯. We want to evaluate this term by term. We see we can differentiate the constant five. We can differentiate negative three π₯. However, weβre going to need to use the product rule to differentiate three π₯ times the natural logarithm of π₯, since itβs the product of two functions.

We recall the product rule tells us the derivative of π’ of π₯ times π£ of π₯ with respect to π₯ is equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯. So to differentiate three π₯ times the natural logarithm of π₯, weβll set our function π’ to be three π₯ and our function π£ to be the natural logarithm of π₯. To use the product rule, we need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of the linear function three π₯ with respect to π₯. Since this is a linear function, its derivative is just the coefficient of π₯, which we know is three.

Next, we want to differentiate π£ of π₯ with respect to π₯. Thatβs the derivative of the natural logarithm of π₯ with respect to π₯. We know this is equal to one over π₯. Weβre now ready to evaluate our derivative term by term. First, the derivative of the constant five is equal to zero. Next, to differentiate negative three π₯ with respect to π₯, we remember this is a linear function. So its derivative is just the coefficient of π₯, which we know is negative three. And remember, to differentiate our third term, weβre using the product rule. Itβs equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯.

So letβs substitute in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯. This gives us that π¦ prime is equal to negative three plus three times the natural logarithm of π₯ plus one over π₯ times three π₯. We can then simplify this expression. In our third term, we have one over π₯ multiplied by three π₯ is just equal to three. But then we see we have negative three plus three. So we can cancel these to get zero.

So this simplifies to give us that π¦ prime is equal to three times the natural logarithm of π₯. And we can see this is exactly the same as the differential equation given to us in the question. And because this is exactly the same as the differential equation given to us in the question, this means our function was a solution to this differential equation.

Therefore, we were able to show that π¦ is equal to five minus three π₯ plus three π₯ times the natural logarithm of π₯ is indeed a solution to the differential equation. π¦ prime is equal to three times the natural logarithm of π₯.