dy/dx= (4-y^2)^(1/2) y=1 when x=TT/6?

Question

I just need the equation that makes this true. I don't know how to find it.

Answer 1

This is a separable differential equation. Divide by √(4-y²):

y'/√(4-y²) = 1

Integrate both sides with respect to x:

∫1/√(4-y²) dy/dx dx = ∫1 dx

Cancel the dx, and evaluate the integral on the right:

∫1/√(4-y²) dy = x+C

Make the substitution θ = arcsin (y/2), so y = 2 sin θ, dy = 2 cos θ dθ. Then we have:

∫1/√(4-4 sin² θ) 2 cos θ dθ = x+C

Factor the denominator:

∫2 cos θ/(2√(1-sin² θ)) dθ = x+C

Simplify:

∫cos θ/√cos² θ dθ = x+C
∫1 dθ = x+C
θ = x+C

But as you recall, θ = arcsin (y/2), so:

arcsin (y/2) = x+C

Taking the sine of both sides:

y/2 = sin (x+C)

Multiplying by 2:

y = 2 sin (x+C)

Now, to find the value of C. Note that y=1 when x=π/6, so:

1=2 sin (π/6 + C)
sin (π/6+C) = 1/2
π/6+C = arcsin (1/2) = π/6
C=0

so in fact y=2 sin x, and we are done.

Answer 2

discover the final answer with the help of isolating the variables then integrating: dy / dx = y² / x? dy / y² = dx / x? ? a million / y² dy = ? a million / x? dx -a million / y = -a million / (3x³) + C -a million / y = C - a million / (3x³) a million / y = C + a million / (3x³) a million / y = (Cx³ + a million) / (3x³) y = 3x³ / (Cx³ + a million) discover the particular answer with the help of fixing for the consistent: whilst x = a million, y = a million 3 / (C + a million) = a million a million / (C + a million) = ? C - a million = 2 C = 4 y = 3x³ / (2x³ + a million)

Answer 3

dy/dx means take the derivative.

The derivative of (4-y^2)^(1/2) is:

(1/2)(4 - y^2)(-2y)

You are given y=1.
Plug in 1 into y in your equation.

(1/2)(4 - 1^2)(-2*1)

You get:

-3

Answer 4

hey, you probably need to clarify your question
how does the derivitive of function y equal that
long equation?

because if y did equal one, than the
derivative must be 0
also, x isn't stated anywhere in the f(x) or
f(y) equation... >_>

Answer 5

easy, type hello upside down on your ti83