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:
do you still need an answer?
Step-by-step explanation:
?
Answer:
y= -4x + 54
Step-by-step explanation:
4x + y =54
-4x -4x
y = -4x +54
Answer:
C
Step-by-step explanation:
We want the equation of the line that passes through (3, 6) and is perpendicular to:
First, convert the second equation into slope-intercept form:
So, we can see that the slope of the line is 3/4.
The slopes of perpendicular lines are negative reciprocals of each other.
Therefore, the slope of the new line is -4/3.
It passes through the point (3, 6).
We can use the point-slope form:
Substitute:
Distribute:
Therefore:
The answer is C.
It’s that last one 8 square root 10^3x
