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
Alternate exterior angles(AEA).
Given their relationship the angles are congruent.
8x-71=5x+7
3x-71=7
3x=78
X=26
Answer:
m= 1/4
Step-by-step explanation:
1) Take two points. Let's use the first and second. Subtract one y coordinate from the other. 6 - 5 = 1
2) Do the same process with the x coordinates. 0 - -4 = 4 (0 + 4)
3) Make the answer from the y coordinates the numerator and the answer from the x coordinates the denominator, and you have your final answer: 1/4
Answer:
10/33 there 33 chocolates in the bag but only 10 are dark chocolate
Step-by-step explanation:
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