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 is 184. 23 * 8 = 184. 184 * 2 = 368. 368 ÷ 2 = 184
Answer:
x < 0
Step-by-step explanation:
If you were to graph the function, you would get a parabola that opens down with the vertex at (0, 0). So, the graph is increasing from -∞ to 0.
Answer:
False
Step-by-step explanation:
2x + y = 0
Try (1, -2)
2(1) + (-2) = 0
2 + (-2) = 0
0 = 0
The solution works on the first equation.
-x + 2y = 5
-(1) + 2(-2) = 5
-1 - 4 = 5
-5 = 5
The solution does not work in the second equation.
Answer: False
Answer: 100,000.
Explanation: im sorry but thats way too many zeroes for me to write if you want the standard method lol
