Video: Solving a Differential Equation Involving Trigonometric Functions

Video Transcript

Find the implicit solution to the following differential equation. The sin of 𝑦 multiplied by d𝑦 by d𝑥 minus the cos of 𝑥 is equal to zero.

The question gives us a differential equation, and it wants us to find the implicit solution of this differential equation. Remember, when we’re asked to find an implicit solution, we don’t need to give our answer as a function of 𝑥. So let’s try and find the solution to our differential equation. The first thing we notice is we’re given a function of 𝑦 and a function of 𝑥 in our differential equation. Since there’s no way of manipulating our equation to remove our function in 𝑦, we should try writing this as a separable differential equation. In fact, in this case, we can just skip the step where we write this as a separable differential equation. We’ll just add the cos of 𝑥 to both sides of the equation.

Doing this, we get the sin of 𝑦 multiplied by d𝑦 by d𝑥 is equal to the cos of 𝑥. And now we want to separate our variables. We have our function of 𝑦 on the left-hand side of our equation and our function of 𝑥 on the right-hand side of our equation. And to solve this, we need to use a trick which we use when we’re solving separable differential equations. We know d𝑦 by d𝑥 is not a fraction. However, we can treat it a little bit like a fraction when we’re solving separable differential equations. This gives us the equivalent statement the sin of 𝑦 d𝑦 is equal to the cos of 𝑥 d𝑥. And we can then solve this by integrating both sides of our equation. We get the integral of the sin of 𝑦 with respect to 𝑦 is equal to the integral of the cos of 𝑥 with respect to 𝑥.

We’re now ready to evaluate both of these integrals. First, to integrate the sin of 𝑦 with respect to 𝑦, we recall the integral of the sin of 𝜃 with respect to 𝜃 is equal to negative the cos of 𝜃 plus a constant of integration. So by calling our variable 𝑦, we get that this integral was equal to negative the cos of 𝑦 plus a constant of integration we’ll call 𝐶 one. Now, to evaluate the integral of the cos of 𝑥 with respect to 𝑥, we recall the integral of the cos of 𝜃 with respect to 𝜃 is equal to the sin of 𝜃 plus a constant of integration. This time, since we’re integrating the cos of 𝑥 with respect to 𝑥, we’ll label our variable 𝑥. This gives us the sin of 𝑥 plus a constant of integration we’ve called 𝐶 two.

At this point, we try and write this in the form 𝑦 is equal to some function of 𝑥. But the question is telling us to find an implicit solution, so we don’t need to do this step. Instead, we can rearrange our solution into any form which we want. We’ll start by combining our constants 𝐶 one and 𝐶 two into a new constant on the left-hand side of our equation we’ll call 𝐶. Next, we’ll add the cos of 𝑦 to both sides of this equation. This gives us 𝐶 is equal to the sin of 𝑥 plus the cos of 𝑦. And we’ll rearrange this equation one more time. We’ll rearrange this to get the cos of 𝑦 plus the sin of 𝑥 is equal to 𝐶.

Therefore, given the differential equation the sin of 𝑦 times d𝑦 by d𝑥 minus the cos of 𝑥 is equal to zero, we were able to find the implicit solution the cos of 𝑦 plus the sin of 𝑥 is equal to 𝐶.