Video: Solving a Separable First-Order Differential Equation Involving Using Trigonometric Identities for Integration

Suppose that d𝑦/dπ‘₯ = (3 sinΒ² π‘₯)/(4 cosΒ² 2𝑦) and 𝑦 = πœ‹/4 when π‘₯ = πœ‹/2. Find 𝑦 in terms of π‘₯.

06:19

Video Transcript

Suppose that d𝑦 by dπ‘₯ is equal to three times the sin squared of π‘₯ divided by four times the cos squared of two 𝑦 and 𝑦 is equal to πœ‹ by four when π‘₯ is equal to πœ‹ by two, find 𝑦 in terms of π‘₯.

In this question, we’re given a differential equation and a point on the curve to the solution of this differential equation. We need to use this to find an expression for 𝑦 in terms of π‘₯. So, to do this, we’re going to need to solve our differential equation. We can see d𝑦 by dπ‘₯ is equal to a function in π‘₯ divided by a function in 𝑦. This means we can try doing this by using separable differential equations. We want to separate our variables onto opposite sides of the equation.

First, we’ll multiply both sides of our equation by four times the cos squared of two 𝑦. This gives us four cos squared of two 𝑦 times d𝑦 by dπ‘₯ is equal to three times the sin squared of π‘₯. And it’s worth reiterating at this point we know d𝑦 by dπ‘₯ is not a fraction. However, when we’re solving separable differential equations, we can treat it a little bit like a fraction. Doing this means we can get the equivalent statement in terms of differentials. Four cos squared of two 𝑦 d𝑦 is equal to three sin squared of π‘₯ dπ‘₯. We then want to find the integral of both sides of this equation.

And in fact, we know how to evaluate the integrals of both sides of this equation. The easiest way to do this is by using the double-angle formula. We’ll start with our integral in terms of π‘₯. We need to recall the following version of the double-angle formula for cosine: the cos of two π‘₯ is equivalent to one minus two times the sin squared of π‘₯. We want to use this equivalence to find an expression for three times the sin squared of π‘₯.

We’ll start by rearranging this equivalence to make two sin squared of π‘₯ the subject. This gives us two sin squared of π‘₯ is equivalent to one minus the cos of two π‘₯. Of course, we want an expression for three sin squared of π‘₯. So, we’ll multiply both sides of this equivalence through by three over two. Then, by doing this and simplifying, we get that three sin squared of π‘₯ is equivalent to three minus three cos of two π‘₯ all divided by two. We’ll write this expression into our integral; however, we’ll write this as two separate terms.

Using this, we’ve rewritten our integral on the right-hand side as the integral of three over two minus three cos of two π‘₯ divided by two with respect to π‘₯. And at this point, we could evaluate this integral; however, we’ll also rewrite the integral on the left-hand side of our equation. We’ll do this in the same way. However, this will take a little bit more work. We’ll first recall the following version of the double-angle formula for cosine: the cos of two πœƒ is equivalent to two cos squared of πœƒ minus one.

Remember, we want to use this to rewrite four cos squared of two 𝑦. So, we need an expression for the cos squared of two 𝑦, so we’ll set πœƒ equal to two 𝑦. Doing this gives us the cos of two times two 𝑦 is equivalent to two cos squared of two 𝑦 minus one. And of course, we can simplify this slightly. The cos of two times two 𝑦 is the cos of four 𝑦. Now, we’ll rearrange this equivalence to make two cos squared of two 𝑦 the subject, giving us two cos squared of two 𝑦 is equivalent to the cos of four 𝑦 plus one.

Then, all we need to do is multiply this equivalence through by two, giving us that four cos squared of two 𝑦 is equivalent to two cos of four 𝑦 plus two. Now, we can use this to rewrite our integral. Doing this means we’ve rewritten our integrand on the left-hand side as the integral of two cos of four 𝑦 plus two with respect to 𝑦. And of course, this is still equal to our other integral. We can now evaluate both of these integrals directly.

To evaluate both of these integrals, we need to recall the following trigonometric integral rule: for any real constant π‘Ž not equal to zero, the integral of the cos of π‘Žπœƒ with respect to πœƒ is equal to the sin of π‘Žπœƒ divided by π‘Ž plus the constant of integration 𝐢. We can use this to evaluate both of our integrals. We’ll start with the integral on the left-hand side.

Evaluating this term by term and using our trigonometric integral rule with 𝑦 instead of πœƒ, we get the integral of our first term is two sin of four 𝑦 divided by four. We then need to integrate our second term. The integral of two with respect to 𝑦 is two 𝑦. And of course, we could add a constant of integration; however, we’ll combine this with the constant of integration we’ll get in evaluating our second integral. We’ll now evaluate our second integral term by term.

First, the integral of three over two with respect to π‘₯ is three π‘₯ divided by two. Next, we need to subtract the integral of three times the cos of two π‘₯ over two. We can do this by using our integral rule with π‘Ž equal to two and π‘₯ instead of πœƒ. This means we need to subtract three sin of two π‘₯ divided by two times two. And of course, we need to add a constant of integration we’ll call 𝐢. We could start simplifying this equation, and we’ll do this by multiplying our equation through by four. This gives us two sin of four 𝑦 plus eight 𝑦 is equal to six π‘₯ minus three times the sin of two π‘₯ plus four 𝐢.

And it’s also worth pointing out we could have called our original constant of integration 𝐢 over four. So, when we multiply through by four, we get 𝐢. It doesn’t matter which method you would prefer; it’s all personal preference. We’ll use that method to get our constant of integration now called 𝐢. We can now find the value of 𝐢 by substituting in π‘₯ is equal to πœ‹ by two and 𝑦 is equal to πœ‹ by four. Substituting these values in, we get two times the sin of four times πœ‹ by four plus eight multiplied by πœ‹ by four is equal to six times πœ‹ by two minus three times the sin of two times πœ‹ by two plus our constant of integration 𝐢.

And now, we can start simplifying. First, both of the arguments of the sine simplify to give us πœ‹. But then, we know the sin of πœ‹ is equal to zero. So, both of these terms just simplify to give us zero. Next, eight times πœ‹ by four simplifies to give us two πœ‹. Then, we have six times πœ‹ by two is equal to three πœ‹. So, this equation simplified to give us two πœ‹ is equal to three πœ‹ plus 𝐢. And of course, we can solve this equation for 𝐢. We get that 𝐢 is equal to negative πœ‹. All that’s left to do is substitute this value of 𝐢 into our general solution for the differential equation. And doing this gives us our final answer.

Therefore, we were able to show for the differential equation d𝑦 by dπ‘₯ is equal to three sin squared of π‘₯ divided by four cos squared of two 𝑦, the solution to this differential equation , where when π‘₯ is equal to πœ‹ by two, we have 𝑦 is equal to πœ‹ by four, is given by eight 𝑦 plus two sin of four 𝑦 is equal to six π‘₯ minus three sin of two π‘₯ minus πœ‹.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.