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Question Show that y=sin⁡(t)y=\sin(t) solves the ODE (dydt)2=1−y2(\frac{dy}{dt})^2=1-y^2. Justify your answer in terms of tt.

Studdy Solution

STEP 1

Assumptions
1. We are given the function y=sin⁡(t)y = \sin(t).
2. We need to show that this function is a solution to the differential equation (dydt)2=1−y2\left(\frac{d y}{d t}\right)^{2} = 1 - y^{2}.
3. We will differentiate yy with respect to tt and then verify if the resulting expression satisfies the given differential equation.

STEP 2

First, we need to find the first derivative of y=sin⁡(t)y = \sin(t) with respect to tt.
dydt=ddtsin⁡(t)\frac{d y}{d t} = \frac{d}{d t} \sin(t)

STEP 3

Using the derivative of the sine function, we have:
dydt=cos⁡(t)\frac{d y}{d t} = \cos(t)

STEP 4

Now we need to square the derivative to get (dydt)2\left(\frac{d y}{d t}\right)^{2}.
(dydt)2=cos⁡2(t)\left(\frac{d y}{d t}\right)^{2} = \cos^2(t)

STEP 5

Using the Pythagorean identity for trigonometric functions, we know that sin⁡2(t)+cos⁡2(t)=1\sin^2(t) + \cos^2(t) = 1. We can rearrange this to express cos⁡2(t)\cos^2(t) in terms of sin⁡2(t)\sin^2(t).
cos⁡2(t)=1−sin⁡2(t)\cos^2(t) = 1 - \sin^2(t)

STEP 6

Substitute sin⁡(t)\sin(t) with yy since y=sin⁡(t)y = \sin(t).
cos⁡2(t)=1−y2\cos^2(t) = 1 - y^2

STEP 7

Now we compare the expression from STEP_4 with the given differential equation.
(dydt)2=1−y2\left(\frac{d y}{d t}\right)^{2} = 1 - y^{2}

STEP 8

Since (dydt)2=cos⁡2(t)\left(\frac{d y}{d t}\right)^{2} = \cos^2(t) and cos⁡2(t)=1−y2\cos^2(t) = 1 - y^2, we can see that the given function y=sin⁡(t)y = \sin(t) satisfies the differential equation.
(dydt)2=cos⁡2(t)=1−y2\left(\frac{d y}{d t}\right)^{2} = \cos^2(t) = 1 - y^{2}

STEP 9

Therefore, we have shown that y=sin⁡(t)y = \sin(t) is indeed a solution to the differential equation (dydt)2=1−y2\left(\frac{d y}{d t}\right)^{2} = 1 - y^{2}.
The justification is complete, and the function y=sin⁡(t)y = \sin(t) satisfies the given differential equation.

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