Answer:
njblhgkc dggh,bn bhjvbhj,n hg cfxzedsetxdrycfuvgibkhjnm
Step-by-step explanation:
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
24
Step-by-step explanation:
Well 6 times 8 is 48 so
8 times 3 is 24
Answer:
Step-by-step expla3
x
−
2
y
<
10
Solve for y
.
Tap for more steps...
y
>
−
5+
3
x
2
Find the slope and the y-intercept for the boundary line.
Tap for more steps...
Slope:
3
2
y-intercept:
(
0
,
−
5
)
Graph a dashed line, then shade the area above the boundary line since
y
is greater than
−
5
+
3
x
2
.
y
>
−
5
+3
x2
