# Question Video: Solving a First-Order Separable Differential Equation given in the Normal Form

Suppose that d𝑦/d𝑥 = (4𝑥 − 4 cos 2𝑥)/(4 sin 𝑦 + 9) and 𝑦 = 0 when 𝑥 = 0. Find 𝑦 in terms of 𝑥.

03:29

### Video Transcript

Suppose that d𝑦 by d𝑥 is equal to four 𝑥 minus four times the cos of two 𝑥 all divided by four times the sin of 𝑦 plus nine and 𝑦 is equal to zero when 𝑥 is equal to zero. Find 𝑦 in terms of 𝑥.

The question gives us a differential equation, and it wants us to find an equation for 𝑦 in terms of 𝑥, given that when 𝑦 is equal to zero, 𝑥 is equal to zero. We can see that d𝑦 by d𝑥 is the product of a function in 𝑥 and a function in 𝑦. And we call first-order differential equations, which have this property, separable. This is because we can separate the 𝑥-variables and the 𝑦-variables onto opposite sides of our equation. So to solve this, we’ll start by multiplying both sides for our equation by four times the sin of 𝑦 plus nine. This gives us that four sin 𝑦 plus nine multiplied by d𝑦 by d𝑥 is equal to four 𝑥 minus four times the cos of two 𝑥.

And at this point, it’s worth reiterating, d𝑦 by d𝑥 is definitely not a fraction. However, when we’re solving separable differential equations, we can treat it a little bit like a fraction. Doing this gives us the equivalent statement four sin 𝑦 plus nine d𝑦 is equal to four 𝑥 minus four times the cos of two 𝑥 d𝑥. Then we just take the integral of both sides of this equation. And we see that these integrals are in a form which we can solve. We know the integral of the sin of 𝜃 with respect to 𝜃 is equal to negative the cos of 𝜃 plus a constant of integration 𝑐. So integrating four sin of 𝑦 gives us negative four times the cos of 𝑦. And we know the integral of nine is just nine 𝑦. Then we just add our constant of integration we will call 𝑐 one.

Similarly, we know for constants 𝑎 and 𝑛, where 𝑛 is not equal to zero, the integral of 𝑎 times the cos of 𝑛𝜃 with respect to 𝜃 is equal to 𝑎 times the sin of 𝑛𝜃 divided by 𝑛 plus the constant of integration 𝑐. We integrate four 𝑥 by using the power rule for integration. We add one to the exponent and divide by this new exponent. This gives us four 𝑥 squared over two, which is just two 𝑥 squared. We apply our integral rule to integrate negative four times the cos of two 𝑥. This gives us negative four times the sin of two 𝑥 divided by two, which is just negative two times the sin of two 𝑥.

And finally, we add a constant of integration we will call 𝑐 two. We can simplify this. We can combine the constants 𝑐 one and 𝑐 two into a new constant we will just call 𝑐. And we can find the value of 𝑐, since we’re told that when 𝑦 is equal to zero, 𝑥 is equal to zero. Substituting 𝑥 is equal to zero and 𝑦 is equal to zero gives us negative four times the cos of zero plus nine times zero is equal to two times zero squared minus two times the sin of two times zero plus 𝑐. By using the fact that the cos of zero is equal to one and the sin of zero is equal to zero, we can simplify this expression to see that 𝑐 is equal to negative four.

So by using 𝑐 is equal to negative four, we must have that nine 𝑦 minus four times the cos of 𝑦 is equal to two 𝑥 squared minus two times the sin of two 𝑥 minus four. Therefore, we’ve shown if d𝑦 by d𝑥 is equal to four 𝑥 minus four times the cos of two 𝑥 divided by four sin 𝑦 plus nine and 𝑦 is equal to zero when 𝑥 is equal to zero. Then we must have that nine 𝑦 minus four cos of 𝑦 is equal to two 𝑥 squared minus two times the sin of two 𝑥 minus four.