User:Egm4313.s12.team16.r2

Report 2
By: Team 16

R2.1
Given:

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

work:

$$ (-_1)(-_2))=0 $$

$$ ^2-_1-_2+_1_2=0 $$

plug in given values

$$ ^2-3-10=0 $$

Non homogeneous

$${ r(x)=^2-3-10 }$$

Homogeneous solution

$$ {y_n(x)=C_1e^{-2x}+C_2e^{5x}} $$

overall solution

$${ y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x) }$$

its given that this is a case of no excitation therefore

{ r(x)=0}

$$ { y_p(x)=0 } $$

$${ y(0)=1=C_1+C_2}$$

$$ { y'(0)=0=-2C_1+5C_2} $$

therefore C1&C2 are

$$ C_1=5/7 $$

$$C_2=2/7 $$

part B

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

$$ m=2 $$

$$ m(-_1)(-_2))=0 $$

$$ m(^2-_1-_2+_1_2)=0 $$

plug in given values

$$ 2^2-6-20=0 $$

Non homogeneous

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

$$ m=3 $$

$$ m(-_1)(-_2))=0 $$

$$ m(^2-_1-_2+_1_2)=0 $$

plug in given values

$$ 3^2-9-30=0 $$

Non homogeneous

Author: Nyal Jennings

Reviewed by: Brandon Wright

R2.2
For this problem the initial conditions for this equation:

$$ y''+10y'+25=r(x)$$

were given as:

y(0)=1 and y'(0)=0 converting into terms of $$ \lambda $$ this becomes:

$$ \lambda^2+10\lambda+25=0 $$

Factoring this equation becomes $$ (\lambda+5)^2 $$

This means that the solution $$ \lambda = -5 $$ is a real repeated solution.

Because the solution is a double root the general equation is

$$ y_g= C_1e^{-5x}+C_2xe^{-5x} $$

Using the initial conditions y(0)=1 and y'(0)=0 we can solve for $$ C_1, C_2 $$

Plugging x=0,y=1 into the general solution gives

$$ C_1e^{-5(0)}+C_2(0)e^{-5(0)}= 1 $$

The second term in the equation equals 0 making $$ C_1= 1 $$

Next we have to take the derivative of the general equation to solve for $$C_2$$

The derivative of the first term in the equation is simply: $$ -5C_1e^{-5x}$$

Where as the second term derivative is found using the rule of derivatives of two multiplied terms

Using this rule the second term becomes: $$ -5C_2xe^{-5x}+C_2e^{-5x}$$

Making the whole derivative: $$ y'= -5C_1e^{-5x}-5C_2xe^{-5x}+C_2e^{-5x} =0 $$

Now plugging x=0, y=0 and $$ C_1=1 $$ into the derivative we get: $$ -5(1)e^{-5(0)}-5C_2(0)e^{-5(0)}+C_2e^{-5(0)} =0 $$

Solving this equation for $$C_2$$ it is found that $$ C_2=5$$

The final solution to this ODE is

Author: Brandon Wright

Reviewed by: Ryan McDaniel

R2.3
Find the general solution of the ODEs.

R2.3a
Given:

$$ y''+6y'+8.96y=0 \! $$, "Which has the general form of $ y''+ay'+by=0 \! $" Solve for $$ a^2-4b \! $$ where $$ a= 6 \! $$ and $$ b= 8.96 \! $$: $$ 6^2-4(8.96) = 0.16 > 0 \! $$ "Therefore, this is a Case I - 2 distinct roots: $\lambda _1 \! $ and $\lambda _2 \! $"

"with a general solution of $ y=C_1e^{\lambda _1x}+ C_2e^{\lambda _2x} $"

$$ \lambda _1 = \frac{1}{2}\left [ -a+\sqrt{a^2-4b} \right ] $$ "$= \frac{1}{2}\left [ -6+\sqrt{0.16} \right ] = -2.8 $" $$ \lambda _2 = \frac{1}{2}\left [ -a-\sqrt{a^2-4b} \right ]$$ "$= \frac{1}{2}\left [ -6-\sqrt{0.16} \right ] = -3.2 $" Substituting into the general solution where $$ \lambda_1= -2.8 \! $$ and $$ \lambda_2= -3.2 \! $$:

R2.3b
Given:

$$ y''+4y'+(\pi ^2+4)y=0 \! $$,

"Which has a general form of $ y''+ay'+by=0 \! $" Solve for $$ a^2-4b \! $$ where $$ a= 4\! $$ and $$ b=\pi^2+4 \! $$: $$ 4^2-4(\pi^2+4) = -39.4784 < 0 \! $$ "Therefore, this is a Case III - complex conjugate roots: $ \lambda= -\frac{1}{2}a \pm i\omega $"

"with a general solution of $ y=e^{-ax/2}\left ( A\cos(\omega x) + B\sin (\omega x) \right )\! $"

Also, $$ \omega^2=b-\frac{1}{4}a^2 \! $$

"$ =(\pi^2+4)-\frac{1}{4}(4^2) \! $"

"$ \omega ^2= \pi^2 \! $"

"$ \omega = \pi \! $"

Substituting into the general solution where $$ a= 4 \! $$ and $$ \omega = \pi \! $$:

Author: Erin Standish Reviewed by:Brandon Wright

R2.4
Find the general solution of the ODE.

R2.4a
Given:

$$ y''+2 \pi y'+\pi^2y=0 \! $$

"which has the general form of $ y''+ay'+by=0 \! $." Solve for $$ a^2-4b \! $$ where $$a=2 \pi \! $$ and $$ b=\pi \! $$: $$ (2\pi)^2-4(\pi)^2 = 0 = 0 \! $$ "Since 0 = 0, this is a Case II - real double root: $\lambda = -a/2 \!$"

"with a general solution of $ y=(C_1+C_2x)e^{-ax/2} \!$." Substituting into the general solution where $$ a= 2 \pi \!$$ :

R2.4b
Given:

$$ 10y''-32y'+25.6y=0 \! $$

"We must get into the form $y''+ay'+by=0 \! $ by diving through by 10:" $$ y''-3.2y'+2.56y=0 \! $$ Solve for $$ a^2-4b \! $$ where $$ a= 3.2 \! $$ and $$ b= 2.56 \! $$: $$ 3.2^2-4(2.56) = 0 = 0 \! $$ "Since 0 = 0, this is a Case II - real double root: $ \lambda = -a/2 \! $"

"with a general solution of $ y=(C_1+C_2x)e^{-ax/2} \! $." Substituting into the general solution where $$ a = 3.2 \! $$:

Author: Erin Standish Reviewed by: Nyal Jennings

R2.5
Find an ODE with form $$ y''+ay'+by=0 \! $$ for the given basis.

R2.5a
Given: $$ e^{2.6x} \! $$, $$ e^{-4.3x} \! $$

This basis implies that we are dealing with an ODE of Case I - 2 distinct roots: "$ \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6$ and    $\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3$" "Also, $ a^2-4b>0 \! $."

The given basis of an ODE of Case I is: $$e^{\lambda_1x}$$ and $$e^{\lambda_2x}$$.

Substituting $$ c = \sqrt{a^2-4b} $$ yields: "$ \lambda_1=\frac{1}{2}(-a+c)=2.6$ and    $\lambda_2=\frac{1}{2}(-a-c)=-4.3 $"

This yields two equations & two unknowns:

$$ \lambda_1=\frac{1}{2}(-a+c)=2.6$$......(1)

$$\lambda_2=\frac{1}{2}(-a-c)=-4.3$$.....(2)

Solving for $$ a \! $$ by adding (1) to (2) yields: "$ (5.2=-a+c) \! $" "$+(-8.6=-a-c) \! $" "$ -3.4=-2a \! $" "$ 1.7=a \! $"

Substituting $$ a \! $$ into (1) and solving for $$ c \! $$ yields: "$ \frac{1}{2}(-1.7+c)=2.6 $" "$ 5.2=-1.7+c \! $" "$ 6.9=c \! $"

Substituting back into $$ c = \sqrt{a^2-4b} $$ and solving for $$ b \! $$ yields: "$6.9=\sqrt{1.7^2-4b}$" "$47.61=2.89-4b \! $" "$ -11.18=b \! $"

Substituting into the general form:

R2.5b
Given: $$e^{ \sqrt{5} x} $$, $$ xe^{ -\sqrt{5} x} $$

This basis implies that we are dealing with an ODE of Case II - real double root: "$ \lambda =-\frac{1}{2}a $." "Also, $ a^2-4b=0 \! $."

The given basis of an ODE of Case II is: $$e^{\frac{-ax}{2}}$$.

Solving for a: "$e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $|undefined" "$+\frac{a}{2}=+ \sqrt{5} $" "$a=2\sqrt{5}$"

Solving for b using $$ a^2-4b=0 \! $$: "$(2\sqrt{5})^2-4b=0$" "$-4b=-20 \! $" "$b=5 \! $"

Substituting into the general form:

Author: Erin Standish Reviewed by: Martin Dolgiej

R2.6
Given

From (3) p.1-5 the equation of motion of the spring-mass-dashpot system

$$ my_k'' + \frac{mk}{c}y_k' + ky =f(t)...(1a) $$

Solution

Using the double real root λ=-3 the characteristic equation can be developed

$$ (\lambda+3)^2=\lambda^2 + 6\lambda + 9 = 0 \! $$

Relating this to the equation of motion in (1a)

$$ m=1 \! $$

$$ k=9 \! $$

$$ c=\frac{3}{2} $$

Author:James Welch

Reviewed by: Erin Standish

R2.7
Equation for Maclaurin Series

$$ f(x) = \sum_{n=0}^\infty a_0 (x - x_0)^n $$

where

$$ a_0 = \frac{1}{n!}f^{(n)}(x_0) \! $$

and

$$ x_0 = 0 \! $$

PART 1

$$ f(t) = e^t = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} \! $$

PART 2

$$ f(t) = cos(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} + \frac{1}{24}f^{(4)}(0)t^{(4)} \! $$

PART 3

$$ f(t) = sin(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} + \frac{1}{24}f^{(4)}(0)t^{(4)} \! $$

Author:Ryan McDaniel

Reviewed By: Martin Dolgiej

R2.8
Page 59

8. Put the equation in general form $$y''+y'+3.25y=0 \! $$ We can start with $$y = e^{st} \! $$ and from that derive $$y' = se^{st \! }$$ and $$y'' = s^{2}e^{st} \! $$ where $$s = i\omega \! $$. Replacing into the original equation, $$s^{2}e^{st}+se^{st}+3.25e^{st}=0 \! $$ and factoring out the exponential term, $$e^{st}(s^{2}+s+3.25)=0 \! $$ Since $$e^{st}\neq 0 $$, we have our characteristic equation as $$s^2+s+3.25=0 \! $$ using the quadratic formula to solve, the roots are found at $$ \frac{1}{2}\pm i\sqrt{3} \! $$ and substituting into standard form, the equation becomes $$y = e^{\frac{-x}{2}}(Acos\sqrt3 x+Bsin\sqrt3 x)\! $$ Page 59 15. Put the equation in general form $$y''+0.54y'+(0.0729+\pi)y=0 \! $$ We can start with $$y = e^{st} \! $$ and from that derive $$y' = se^{st} \! $$ and $$y'' = s^{2}e^{st} \! $$ where $$s = i\omega \! $$. Replacing into the original equation, $$s^{2}e^{st}+0.54se^{st}+(0.0729+\pi)e^{st}=0 \! $$ and factoring out the exponential term, $$e^{st}(s^{2}+0.54s+(0.0729+\pi))=0 \! $$ Since $$e^{st}\neq 0$$, we have our characteristic equation as $$s^2+0.54s+(0.0729+\pi)=0 \! $$ using the quadratic formula to solve, the roots are found at $$-0.27\pm \sqrt{\pi}i \! $$ and substituting into standard form, the equation becomes $$y = e^{-0.27x}(Acos\sqrt{\pi} x+Bsin\sqrt{\pi} x)\! $$ Author: Matt Hilsenrath Reviewed by: Martin Dolgiej

Given:
This is the characteristic equation from the ODE: Given the following initial conditions: $$ y(0) = 1, y'(0) = 0 \! $$ $$ r(x) = 0 \! $$

Solution:
Solving the characteristic equation (Eq. 1) for $$ \lambda \! $$: $$ \lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \! $$ $$ \lambda = \frac{-4 \pm \sqrt{4^2-4(1)(13)}}{2(1)} \! $$ $$ \lambda = \frac{-4 \pm 6i}{2} \! $$ $$ \lambda = -2 \pm 3i \! $$ Therefore, the solution will take the form of: $$ y(x) = e^{-ax}(C_1\cos({\omega x}) + C_2\sin({\omega x})) \! $$ Where the a and $$ \omega $$ are: $$ a = -2, \omega = 3 \! $$ Substituting the values and taking the derivative of $$ y(x) $$: $$ y(x) = e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ $$ y(x)' = e^{-2x}(-3C_1\sin({3x}) + 3C_2\cos({3x})) - 2e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ Using the initial condition states above and solving for $$ C_1, C_2 $$: $$ y(0)= 1 = e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ y(0)' = 0 = e^{-2(0)}(-3C_1\sin({3(0)}) + 3C_2\cos({3(0)})) - 2e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ C_1 = 1, C_2 = \frac{2}{3} $$ Final solution: Equations for Graph: Author: Martin Dolgiej Reviewed by: Ryan McDaniel