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 for this problem is x=1
Fraction of shampoo left = 
Solution:
Given fraction of full shampoo 
.
Fraction of shampoo Margaret used 
Fraction of shampoo left = Full shampoo – fraction of shampoo used
To make the denominators same do cross multiplication.
Hence of the shampoo bottle is left.
H =8
6 times 6 is 36 and times 8 is 288.
Answer:
an = 115 + (n - 1) (-6)
a25 = - 29
Step-by-step explanation:
We use the definition for the nth term of an arithmetic sequence:
an = a1 + (n - 1) d
a5 = 91 = a1 + (5 - 1) d
91 = a1 + 4 d
a20 = 1 = a1 + (20 - 1) d = a1 + 19 d
1 = a1 + 19 d
now we subtract term by term one expression from the other
90 = 4 d - 19 d
90 = - 15 d
divide both sides by -15 to isolate d
d = 90 / (-15) = -6
Now we can calculate what a1 is using for example:
1 = a1 + 19 d
1 = a1 - 114
add 114 to both sides:
115 = a1
Then our general expression for the sequence is:
an = 115 + (n - 1) (-6)
We can now use it to calculate the value of a25:
a25 = 115 + (24) * (-6) = -29
