Question

If dy/dx= sin^2(πy/4) and y=1 when x = 0, then find the value of x when y = 3.

a) 0
b) 8/π
c) -π/8
d) None of these

Answer 1

y' = sin^2(πy/4)
dy/sin^2(πy/4) = dx
csc^2(πy/4) dy = dx
Think you can take it from here?
You know what ∫(csc^2 u) du is, right?

Answer 2

Could I have more details please. I need to review every exercise for my test. thanks

Answer 3

recall that
∫(csc^2 u) du = -cot u
So, if u = π/4 y, then du = π/4 dy
But, you only have dy = 4/π du
Thus,
∫csc^2(πy/4) dy = dx
4/π ∫csc^2(πy/4) π/4 dy = dx
4/π (-cot π/4 y) = x+c
cot π/4 y = -4/π x + C
y(0) = 1
? huh? cot(0) is undefined. Maybe you can spot an error in my math...

Answer 4

Oops. Let's start again at this point
4/π (-cot π/4 y) = x+c
y = -4/π cot^-1(π/4 (x+c))
Now, y(0) = 1, so
-4/π cot^-1(π/4 c) = 1
cot^-1(π/4 c) = -π/4
since cot^-1(-1) = -π/4,
π/4 c = -1
c = -4/π
y = -4/π cot^-1(π/4 (x-4/π))
y = -4/π cot^-1(π/4 x - 1)
So, when y=3,
-4/π cot^-1(π/4 x - 1) = 3
cot^-1(π/4 x - 1) = -3π/4
cot -3π/4 = 1
so,
π/4 x - 1 = 1
π/4 x = 2
x = 8/π