Question Video: Finding a Particular Solution for a Separable Differential Equation

Find the particular solution for the following separable differential equation: cos(𝑦) (d𝑦/dπ‘₯) βˆ’ π‘₯ = 0, 𝑦(0) = 0.


Video Transcript

Find the particular solution for the following separable differential equation. Cos of 𝑦 d𝑦 by dπ‘₯ minus π‘₯ equals zero. 𝑦 of zero equals zero.

Let’s start with our differential equation. And what we want to do here is separate the variables. And this means we want functions of 𝑦 and d𝑦 on one side of the equals. And functions of π‘₯ and dπ‘₯ on the other side. So we can start by adding π‘₯ to both sides. Now, although d𝑦 by dπ‘₯ is not a fraction, when we’re separating variables to solve a differential equation, we treat it like a fraction. So this is equivalent to saying cos of 𝑦 d𝑦 equals π‘₯ dπ‘₯. So now, we have the function of 𝑦 and d𝑦 on one side of the equation. And our function of π‘₯ and dπ‘₯ on the other. And now, we have successfully separated the variables. We integrate both sides. On the left, we have the integral of cos of 𝑦 with respect to 𝑦. And at this point, we recall the useful cycle of differentiation and integration of common trig functions. This shows us what each of these functions differentiates to. And if we reverse the cycle, we see what each function integrates to.

So the integral of cos of 𝑦 is sin of 𝑦. And because we don’t have any limits of integration, we need to add a constant of integration, which we can call 𝑐 one. And now, let’s integrate the right-hand side of the equation. To do this, we recall the power rule for integration, which tells us that the integral of π‘₯ to the 𝑛 power is equal to π‘₯ to the power of 𝑛 add one over 𝑛 add one plus a constant of integration 𝑐. And that’s followed for 𝑛 not equal to negative one. So if we think of π‘₯ on its own as π‘₯ to the power of one, we can see that π‘₯ integrates to π‘₯ squared over two. And then, we had a constant of integration, which we can call 𝑐 two this time. So now we’ve integrated both sides of our function. We don’t actually need to have two separate constants here. We can gather them together as one constant, which we can call 𝑐.

And now to find the particular solution, we need to use the condition that we we’re given in the question. 𝑦 of zero equals zero. This tells us that when π‘₯ is equal to zero, 𝑦 is equal to zero. So we substitute π‘₯ equals zero and 𝑦 equals zero into our equation to get sin of zero equals zero squared over two add 𝑐. Now sin of zero equals zero. And zero squared over two equals zero. And so 𝑐 must equal zero. So if we substitute 𝑐 equals zero back into our working, we find that sin of 𝑦 equals π‘₯ squared over two. Finally, let’s write our answer as 𝑦 equals some function of π‘₯. To get 𝑦 on its own, we need to do the reverse of taking the sine of a function. So this is the inverse sine. The inverse sine function works like this. If π‘Ž equals sin of 𝑏, then 𝑏 equals the inverse sin of π‘Ž. Therefore, we can say that 𝑦 is equal to the inverse sin of π‘₯ squared over two.

So to summarize, we started with our differential equation. We separated the variables and integrated each side. And then we used the condition given in the question in order to calculate the constant of integration. And we did some rearranging to get our final answer.

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