Actual Answer:
A (I THINK )
Answer:
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The answer is exponential decay
Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:
![x'=\frac{dx}{dt}](https://tex.z-dn.net/?f=x%27%3D%5Cfrac%7Bdx%7D%7Bdt%7D)
a) x’=t–sin(t), x(0)=1
![dx=(t-sint)dt](https://tex.z-dn.net/?f=dx%3D%28t-sint%29dt)
Apply integral both sides:
![\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k](https://tex.z-dn.net/?f=%5Cint%20%7B%7D%20%5C%2C%20dx%3D%5Cint%20%7B%28t-sint%29%7D%20%5C%2C%20dt%5C%5C%5C%5Cx%3D%5Cfrac%7Bt%5E2%7D%7B2%7D%2Bcost%20%2Bk)
where k is a constant due to integration. With x(0)=1, substitute:
![1=0+cos0+k\\\\1=1+k\\k=0](https://tex.z-dn.net/?f=1%3D0%2Bcos0%2Bk%5C%5C%5C%5C1%3D1%2Bk%5C%5Ck%3D0)
Finally:
![x=\frac{t^2}{2} +cos(t)](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2Bcos%28t%29)
b) x’+2x=4; x(0)=5
![dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\](https://tex.z-dn.net/?f=dx%3D%284-2x%29dt%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7B4-2x%7D%3Ddt%20%5C%5C%5C%5C%5Cint%20%7B%5Cfrac%7Bdx%7D%7B4-2x%7D%7D%3D%20%5Cint%20%7Bdt%7D%5C%5C)
Completing the integral:
![-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cint%7B%5Cfrac%7B%28-2%29dx%7D%7B4-2x%7D%7D%3D%20%5Cint%20%7Bdt%7D)
Solving the operator:
![-\frac{1}{2}ln(4-2x)=t+k](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7Dln%284-2x%29%3Dt%2Bk)
Using algebra, it becomes explicit:
![x=2+ke^{-2t}](https://tex.z-dn.net/?f=x%3D2%2Bke%5E%7B-2t%7D)
With x(0)=5, substitute:
![5=2+ke^{-2(0)}=2+k(1)\\\\k=3](https://tex.z-dn.net/?f=5%3D2%2Bke%5E%7B-2%280%29%7D%3D2%2Bk%281%29%5C%5C%5C%5Ck%3D3)
Finally:
![x=2+3e^{-2t}](https://tex.z-dn.net/?f=x%3D2%2B3e%5E%7B-2t%7D)
c) x’’+4x=0; x(0)=0; x’(0)=1
Let
be the solution for the equation, then:
![x'=me^{mt}\\x''=m^{2}e^{mt}](https://tex.z-dn.net/?f=x%27%3Dme%5E%7Bmt%7D%5C%5Cx%27%27%3Dm%5E%7B2%7De%5E%7Bmt%7D)
Substituting these equations in <em>c)</em>
![m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i](https://tex.z-dn.net/?f=m%5E%7B2%7De%5E%7Bmt%7D%2B4%28e%5E%7Bmt%7D%29%3D0%5C%5C%5C%5Cm%5E%7B2%7D%2B4%3D0%5C%5C%5C%5Cm%5E%7B2%7D%3D-4%5C%5C%5C%5Cm%3D2i)
This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:
![x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}](https://tex.z-dn.net/?f=x%3DAsin%282t%29%2BBcos%282t%29%5C%5C%5C%5Cx%27%3D2Acos%282t%29-2Bsin%282t%29%5C%5C%5C%5C0%3DAsin%282%280%29%29%2BBcos%282%280%29%29%5C%5C%5C%5C0%3D0%2BB%281%29%5C%5C%5C%5CB%3D0%5C%5C%5C%5C1%3D2Acos%282%280%29%29%5C%5C%5C%5C1%3D2A%5C%5C%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D)
Finally:
![x=\frac{1}{2} sin(2t)](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B2%7D%20sin%282t%29)
Answer:
Step-by-step explanation:
a(1) = -7 (given)
a(2) = a(1) + 4 = -7 + 4 = -3
a(3) = a(2) + 4 = -3 + 4 = 1
a(4) = a(3) + 4 = 1 + 4 = 5
and so on.