Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
The answer is 80 hope this helps :)
<h3>Answer:</h3>
x = 2
<h3>Explanation:</h3>
The rule for secants is that the product of segment lengths (on the same line) from the point of intersection to the points on the circle is a constant for any given point of intersection. Here, that means ...
... 3×(3+5) = 4×(4+x)
... 6 = 4+x . . . . divide by 4
... 2 = x . . . . . . subtract 4
_____
<em>Comment on this secant relationship</em>
Expressed in this way, the relationship is true whether the point of intersection is inside the circle or outside.
Answer:
a) True
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Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given that the definite integration</em>
<em> </em>
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<em>we know that the trigonometric formula</em>
<em> sin²∝+cos²∝ = 1</em>
<em> cos²∝ = 1-sin²∝</em>
<u><em>step(ii):-</em></u>
<em>Now the integration</em>
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<em> = </em>
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<em>Now, Integrating </em>
<em> </em>
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<em> = sin π - sin 0</em>
<em> = 0-0</em>
<em> = 0</em>
<u><em>Final answer:-</em></u>
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<em></em>
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