# Question Video: Checking If a Function Is a Solution to a Differential Equation Mathematics • Higher Education

Is the function 𝑦 = 5 − 3𝑥 + 3𝑥 ln 𝑥 a solution to the differential equation 𝑦′ = 3 ln 𝑥?

03:20

### Video Transcript

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

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, let’s take a look at our differential equation. It’s in the form 𝑦 prime is equal to some function of 𝑥. And, of course, since 𝑦 is a function of 𝑥, we have 𝑦 prime is equal to the first derivative of 𝑦 with respect to 𝑥. Remember, we call 𝑦 is equal to some function of 𝑥 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 𝑦 with respect to 𝑥 from our solution and substitute these into our differential equation. In our case, our differential equation only contains d𝑦 by d𝑥. So let’s find that expression for d𝑦 by d𝑥. For the solution given to us, that’s the derivative of five minus three 𝑥 plus three 𝑥 times the natural logarithm of 𝑥 with respect to 𝑥. We want to evaluate this term by term. We see we can differentiate the constant five. We can differentiate negative three 𝑥. However, we’re going to need to use the product rule to differentiate three 𝑥 times the natural logarithm of 𝑥, since it’s the product of two functions.

We recall the product rule tells us the derivative of 𝑢 of 𝑥 times 𝑣 of 𝑥 with respect to 𝑥 is equal to 𝑢 prime of 𝑥 times 𝑣 of 𝑥 plus 𝑣 prime of 𝑥 times 𝑢 of 𝑥. So to differentiate three 𝑥 times the natural logarithm of 𝑥, we’ll set our function 𝑢 to be three 𝑥 and our function 𝑣 to be the natural logarithm of 𝑥. To use the product rule, we need to find expressions for 𝑢 prime of 𝑥 and 𝑣 prime of 𝑥. Let’s start with 𝑢 prime of 𝑥. That’s the derivative of the linear function three 𝑥 with respect to 𝑥. Since this is a linear function, its derivative is just the coefficient of 𝑥, which we know is three.

Next, we want to differentiate 𝑣 of 𝑥 with respect to 𝑥. That’s the derivative of the natural logarithm of 𝑥 with respect to 𝑥. We know this is equal to one over 𝑥. We’re 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 𝑥 with respect to 𝑥, we remember this is a linear function. So its derivative is just the coefficient of 𝑥, which we know is negative three. And remember, to differentiate our third term, we’re using the product rule. It’s equal to 𝑢 prime of 𝑥 times 𝑣 of 𝑥 plus 𝑣 prime of 𝑥 times 𝑢 of 𝑥.

So let’s substitute in our expressions for 𝑢 of 𝑥, 𝑣 of 𝑥, 𝑢 prime of 𝑥, and 𝑣 prime of 𝑥. This gives us that 𝑦 prime is equal to negative three plus three times the natural logarithm of 𝑥 plus one over 𝑥 times three 𝑥. We can then simplify this expression. In our third term, we have one over 𝑥 multiplied by three 𝑥 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 𝑦 prime is equal to three times the natural logarithm of 𝑥. 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 𝑦 is equal to five minus three 𝑥 plus three 𝑥 times the natural logarithm of 𝑥 is indeed a solution to the differential equation. 𝑦 prime is equal to three times the natural logarithm of 𝑥.

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