User:Egm4313.s12.team4.jkl

= Problem 5: Homogeneous Linear ODEs with Constant Coefficients =

Problem Statement
Problems found in the textbook on page 59.

Solution (2.5)
The following table, taken from page 58 of the textbook. will help with problems 16 and 17.

16. Find an ODE given the basis.


 * {| style="width:100%" border="0"

$$\displaystyle e^{2.6x}, e^{-4.3x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.1)
 * }

This falls under Case I with the two roots being:


 * {| style="width:100%" border="0"

$$\displaystyle \lambda_1=2.6$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.2)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \lambda_2=-4.3$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.3)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.4)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.5)
 * }

Plugging $$\lambda_1$$ into the characteristic equation(2.5.5):


 * {| style="width:100%" border="0"

$$\displaystyle (2.6)^2+a(2.6)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.6)
 * }

Simplifying:


 * {| style="width:100%" border="0"

$$\displaystyle 6.76+2.6a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.7)
 * }

Plugging $$\lambda_2$$ into the characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle (-4.3)^2+a(-4.3)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.8)
 * }

Simplifying:


 * {| style="width:100%" border="0"

$$\displaystyle 18.49-4.3a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.9)
 * }

Subtracting equation 2.5.9 from equation 2.5.7:


 * {| style="width:100%" border="0"

$$\displaystyle -11.73+6.9a=0$$ Solving for a:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.10)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle a=1.7$$ Plugging a into equation 1:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.11)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle 6.76+2.6(1.7)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.12)
 * }

Solving for b:


 * {| style="width:100%" border="0"

$$\displaystyle b=-11.18$$ Plugging a and b to complete the ODE:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.13)
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (2.5.14)
 * }

17. Find an ODE given the basis.


 * {| style="width:100%" border="0"

$$\displaystyle e^{-\sqrt5x}, xe^{-\sqrt5x}$$ This falls under Case II with the root being:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.15)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \lambda=-\frac{1}{2}a=-\sqrt5$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.16)
 * }

An ODE of this form is desired:


 * {| style="width:100%" border="0"

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.17)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.18)
 * }

Multiplying both sides of the equation by -2 yields:


 * {| style="width:100%" border="0"

$$\displaystyle a=2\sqrt5$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.19)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.20)
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.21)
 * }

Plugging λ and a into the characteristic equation:


 * {| style="width:100%" border="0"

$$\displaystyle (-\sqrt5)^2+(2\sqrt5)(-\sqrt5)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.22)
 * }

Simplifying:


 * {| style="width:100%" border="0"

$$\displaystyle 5-10+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.23)
 * }

Solving:
 * {| style="width:100%" border="0"

$$\displaystyle b=5$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.24)
 * }

Plugging a and b to complete the ODE:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.25)
 * }

Author
Solved and typed by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC)

In order to derive an equation of motion, all of the forces must be analyzed. In the spring-mass-dashpot model there are forces from the spring, inertia, and a damping force caused by the dashpot.

The force of a spring oriented in the y-direction is given by Hooke's Law.

$$F=-ky$$

The damping force of a dashpot is proportional to the mass' velocity.

Relating velocity to position and time

$$V=\frac{dy}{dt}$$

Proportionality is related by a dashpot coefficient 'c'

$$F_{d}=c\frac{dy}{dt}$$

According to Newton's Second Law of Motion

Σ$$F_{y}=ma_{y}$$

The force of inertia comes from this law

Relating acceleration to position and time

$$a_{y}=\frac{\mathrm{d^2}y}{\mathrm{d} t^2}$$

Summing spring and damping forces and substituting new acceleration term

$$m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}=-c\frac{dy}{dt}-ky$$

(Both forces are negative beacuse they resist the motion of the spring)

Moving all terms to one side

$$m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky=0$$

Setting the equation equal to r(t)

$$r(t)=m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky$$