Find the particular solution to y ' = sin(x) given the general solution is y = C - cos(x) and the initial condition y of pi over 2 equals 1

Question

a) -cos(x)

b) 2 - cos(x)

c) -1 - cos(x)

d) 1 - cos(x)

Answer 1

Yes, the general solution to y' = sin (x) is

y = C - cos(x)

Initial condition:

y(π/2) = 1 is same as x=π/2, y=1

Common mistake: The phrase "y of pi over 2" doesn't mean "y times pi over 2". It means "y as a function of pi over 2". We were taught that the word "of" means multiplication and that is correct. However, it's not always like that and this problem is a good example.

substitute: 1 = C - cos (π/2)

cos (π/2) = 0

∴ 1 = C - 0

∴ C = 1

Therefore, the particular solution is:

y = 1 - cos(x)

Answer 2

Hi Sreeram!

The initial condition, y(π/2) = 1, tells us that when we plug π/2 in for x, we get 1 for y.

So, we should plug in π/2 in for x and 1 for y and solve for c!

Doing so, we get the following:

1 = C - cos(π/2)

Now, we wanna solve for C. First, note that the cosine of π/2 is 0.

1 = C - 0

∴ C = 1

Plugging this in to y = C - cos(x), we find that the particular solution is y = 1 - cos(x)!

I hope this answer was clear! If not, feel free to comment questions and I'll do my best to answer.