Question

Consider the initial value problem my''+cy'+ky=F(t), y(0)=0, y'(0)=0, modeling the motion of a spring mass dashpot system initially at rest and subjected to an applied force F(t), where the unit of force is the Newton (N). Assume that m=2 kilograms, c=8 kilograms per second, k=80 Newtons per meter, and F(t)=20 sin(6t) Newtons.

1. Solve the initial value problem. y(t)=?
2. Determine the long term behavior of the system. Is lim as t goes to infinity of y(t)=0? If it is, enter zero. If not, enter a function that approximates y(t) for very large positive values of t. For very large positive values of t, y(t) is approximately.. ?

Answer 1

Hence, the
`y(t)=e^(-2 t)\left((75)/(74) \cos 6 t+(75)/(148) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t))\end{array}` and approximately value of
`y(t)` is
`-0.844`.

Given :

`my''+cy'+ky=F(t), y(0)=0, y'(0)=0,`

Where
`m=2` kilograms

`c=8` kilograms per second

`k=80` Newtons per meter

`F(t)=20\sin (6t)` Newtons

Explanation :

(1)

Solve the initial value problem.
`y(t)`

`my''+cy'+ky=F(t), y(0)=0, y'(0)=0,`

`\Rightarrow 2y''+8y'+80y=20\sin (6t)`

`\Rightarrow y''+4y'+40y=10\sin (6t)`

Auxilary equations :
`F(t)=0`

`\Rightarrow r^2+4r+40=0`

`\Rightarrow r=(-4\pm√(4^2-4* 1* 40))/(2* 1)`

`\Rightarrow r=(-4\pm√(16-160))/(2)`

`\Rightarrow r=(-4\pm√(-144))/(2)`

`\Rightarrow r=(-4\pm12i)/(2)`

`\Rightarrow r=-2\pm6i`

The complementary solution is
`y_c=e^(-2t)\left(c_1\cos 6t+c_2\sin 6t\right)`

The particular Integral,
`y_p=(1)/(f(D))F(t)`

`y_(y) &=(1)/(D^(2)+4 D+40) 25 \sin (6 t) \\\\ y_(y) &=(25)/(-6^(2)+4 D+40) \sin (6 t) \quad\left(D^(2) \text { is replaced with }-6^(2)=-36\right) \\\\y_(y) &=(25)/(4 D+4) \sin (6 t) \\\\y_(y) &=(25)/(4(D+1)) \sin (6 t) \\\\y_(y) &=(25(D-1))/(4(D+1)(D-1)) \sin (6 t) \\\\y_(y) &=(25(D-1))/(4\left(D^(2)-1\right)) \sin (6 t) \\\\y_(y) &=(25(D-1))/(4(-36-1)) \sin (6 t) \\\\y_(y) &=-(25)/(148)(D-1) \sin (6 t) \\y_(y) &=-(25)/(148)\left((d)/(d t) \sin (6 t)-\sin (6`

Hence the general solution is :
`y=y_c+y_p=e^(-2t)(c_1\cos 6t+c_2\sin 6t)-(25)/(148)(6\cos 6t-\sin 6t)`

Now we use given initial condition.

`y(t) &=e^(-2 t)\left(c_(1) \cos 6 t+c_(2) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t)) \\\\\y(0) &=e^(-\alpha 0))\left(c_(1) \cos (0)+c_(2) \sin (0)\right)-(25)/(148)(6 \cos (0)-\sin (0)) \\\\0 &=\left(c_(1)\right)-(25)/(148)(6) \\\\c_(1) &=(75)/(74) \\\\y(t) &=e^(-2 t)\left(c_(1) \cos 6 t+c_(2) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t)) \\\\`

`y^(\prime)(t)=-2 e^(-2 t)\left(c_(1) \cos 6 t+c_(2) \sin 6 t\right)+e^(-2 t)\left(-6 c_(1) \sin 6 t+6 c_(2) \cos 6 t\right)-(25)/(148)(-36 \sin (6 t)-6 \cos (6 t)) \\\\y^(\prime)(0)=-2 e^(0)\left(c_(1) \cos 0+c_(2) \sin 0\right)+e^(0)\left(-6 c_(1) \sin 0+6 c_(2) \cos 0\right)-(25)/(148)(-36 \sin 0-6 \cos 0) \\\\0=-2\left(c_(1)\right)+\left(6 c_(2)\right)-(25)/(148)(-6) \\\\0=-2 c_(1)+6 c_(2)+(75)/(74) \\\\0=-2\left((75)/(74)\right)+6 c_(2)+(75)/(74) \\\\`
`\begin{array}{l}0=-(150)/(74)+6 c_(2)+(75)/(74) \\\\(150)/(74)-(75)/(74)=6 c_(2)\end{array}`

`\begin{array}{l}c_(2)=(25)/(148)\\\\\text { Substitute } c_(1) \text { and } c_(2) \text { in } y(t) \text { . Then }\\\\y(t)=e^(-2 t)\left((75)/(74) \cos 6 t+(75)/(148) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t))\end{array}`

(2)

`y(t)=e^(-2 t)\left((75)/(74) \cos 6 t+(75)/(148) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\left((75)/(74) e^(-2 t) \cos 6 t+(75)/(148) e^(-2 t) \sin 6 t\right)-(25)/(148)(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=(75)/(74)\left(e^(-2 t)-1\right) \cos 6 t+(25)/(148)\left(3 e^(-2 t)+1\right) \sin 6 t \\\\|y(t)| \leq (75)/(74) e^(-2 t)-1|\cos 6 t|+(25)/(148)\left|3 e^(-2 t)+1\right||\sin 6 t| \\\\`

`|y(t)| \leq (75)/(74)\left|e^(-2 t)-1\right|+(25)/(148)\left|3 e^(-2 t)+1\right| \\\\\lim _(t \rightarrow \infty) y(t) \leq \lim _(t \rightarrow \infty)\left\(75)/(74)\left \\\\\lim _(t \rightarrow \infty) y(t)=\left\(75)/(74)\left \\`

`\lim _(t \rightarrow \infty) y(t)=\left\{(75)/(74)(-1)+(25)/(148)(1)\right\}=-(75)/(74)+(25)/(148)=-(-150+25)/(148)=-(125)/(148) \approx-0.844`