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:
[tex]x'=\frac{dx}{dt}[/tex]
a) x’=t–sin(t), x(0)=1
[tex]dx=(t-sint)dt[/tex]
Apply integral both sides:
[tex]\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k[/tex]
where k is a constant due to integration. With x(0)=1, substitute:
[tex]1=0+cos0+k\\\\1=1+k\\k=0[/tex]
Finally:
[tex]x=\frac{t^2}{2} +cos(t)[/tex]
b) x’+2x=4; x(0)=5
[tex]dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\[/tex]
Completing the integral:
[tex]-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}[/tex]
Solving the operator:
[tex]-\frac{1}{2}ln(4-2x)=t+k[/tex]
Using algebra, it becomes explicit:
[tex]x=2+ke^{-2t}[/tex]
With x(0)=5, substitute:
[tex]5=2+ke^{-2(0)}=2+k(1)\\\\k=3[/tex]
Finally:
[tex]x=2+3e^{-2t}[/tex]
c) x’’+4x=0; x(0)=0; x’(0)=1
Let [tex]x=e^{mt}[/tex] be the solution for the equation, then:
[tex]x'=me^{mt}\\x''=m^{2}e^{mt}[/tex]
Substituting these equations in c)
[tex]m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i[/tex]
This becomes the solution m=α±βi where α=0 and β=2
[tex]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)[/tex]
Where A and B are constants. With x(0)=0; x’(0)=1:
[tex]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}[/tex]
Finally:
[tex]x=\frac{1}{2} sin(2t)[/tex]
