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
It will be facing right...---------->
ok so rearrange things so that it says 6-d=20, then subtract six from each side to get -d=20-6, then subtract 6 from 20 to get -d=-14, then multiply both sides by -1 to get rid of the negative sign in front of the d since a variable cannot be a negative thus making your answer -14 or letter D. :) ❤❤❤❤
hope this helps
Answer:
B
Step-by-step explanation:
To solve -6-(-4) you need to break it down.
First, multiply -(-4) which would equal 4 (negative times negative equals positive)
Then add 4 to -6 (-6+4)
This equation is being expressed through option B
Hope this helps! :)
Answer:
What is the question?
Step-by-step explanation:answer me
