Answer:
y = sin(t) is a solution to the differential equation
(dy/dt)² = 1 - y²
Step-by-step explanation:
Given (dy/dt)² = 1 - y²
Suppose y = sin(t) is a solution, then it satisfies the differential equation.
That is
[d(sin(t))]² = 1 - y²
Let y = sin(t)
dy/dt = d(sin(t)) = cos(t)
(dy/dt)² = cos²t
But cos²t + sin²t = 1
=> 1 - sin²t = cos²t
So
(dy/dt)² = 1 - sin²t
Since sin²t = (sint)² = y²,
we have
(dy/dt)² = 1 - y²
as required.
The differential equation becomes [tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
Given the function;
Take the differential of the function
Square both sides of the equation to have:
[tex](\frac{dy}{dx} )^2 = (cost)^2[/tex]
Recall from trigonometry identity that [tex]sin^2t + cos^2t = 1[/tex]
Hence, [tex]cos^2t = 1- sin^2t[/tex]
Replace into the differential expression to have:
[tex](\frac{dy}{dx} )^2 = 1-sin^2t[/tex]
Recall that y = sin(t). On replacing, the differential equation becomes:
[tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
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