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
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
Could I have more details please. I need to review every exercise for my test. thanks
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...
To find the value of x when y = 3, we need to solve the given differential equation. Let's start by separating variables and integrating.
The given differential equation is:
dy/dx = sin^2(πy/4)
Separating variables:
dy/sin^2(πy/4) = dx
Now, let's integrate both sides:
∫ dy/sin^2(πy/4) = ∫ dx
To integrate the left side, we can use the trigonometric identity: 1/sin^2(x) = csc^2(x)
Therefore, the left side becomes:
∫ dy/csc^2(πy/4) = ∫ dx
Next, we can use the substitution u = πy/4. Taking the derivative of both sides with respect to y gives du = π/4 dy.
Applying the substitution, the equation becomes:
∫ csc^2(u)(4/π) du = ∫ dx
Now, integrating the left side:
(4/π) ∫ csc^2(u) du = ∫ dx
Using the integral property, we know that ∫ csc^2(u) du = -cot(u) + C, where C is the constant of integration.
So,
(4/π) (-cot(u) + C) = ∫ dx
Simplifying,
(4/π) (-cot(u) + C) = x + K, where K is the constant of integration.
We can rearrange this equation to solve for x:
x = (4/π) cot(u) + C - K
Now, let's substitute back u = πy/4:
x = (4/π) cot(πy/4) + C - K
To find the constant C, we can utilize the initial condition given in the question where y = 1 when x = 0. Substituting those values into the equation:
1 = (4/π) cot(π(1)/4) + C - K
Since cot(π/4) = 1, we get:
1 = (4/π) + C - K
Now, let's focus on finding the constant K. We are asked to find the value of x when y = 3. Substituting those values into the equation:
x = (4/π) cot(π(3)/4) + C - K
Since cot(3π/4) = -1, we have:
x = -(4/π) + C - K
Comparing the two equations we obtained for x, we can equate the expressions on the right-hand side:
-(4/π) + C - K = (4/π) + C - K
The K term cancels out:
-(4/π) + C = (4/π) + C
Now, we have,
-(4/π) = (4/π)
This is not possible since the left-hand side is negative, and the right-hand side is positive. Therefore, there is no value of x that satisfies y = 3 based on the initial condition given in the question.
So, the correct answer is (d) None of these.
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?