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
Answer:
i use slope calculators for these type of problems :)
Step-by-step explanation:
When t = 0 the point on the graph is (0,36). Your equation is just a line and always has the same tangent line so your equation is the same s = 36 - t.
Answer: SECOND OPTION.
Step-by-step explanation:
It is important to remember that, by definition, the "Resultant vector " is the vector obtained by adding two or more vectors.
For this exercise you need to use the formula for calculate the magnitude of the resultant vector obtained by adding the vectors "a" and "b" shown in the picture attached:
Where "c" is the magnitude of the resultant vector asked in the exercise.
You need to analyze the graph given in the exercise.
You can identify that:
Then the next step, knowing which are "a" and "b", is to substitute values into the formula given at the beginning of this explanation, as following:
Finally, evaluating, you get that "c" is:
