User:Egm4313.s12.team13.r2

=Report 2=

Problem Statement
Find the solution for the general excitation for each of the two systems listed below:

No excitation when $$ r(x)=0\!$$. Solution will then be plotted.

Since the roots of the differential equations have already been given to us, it is simply a matter of transforming it into a general solution from here.

$$ \mathbf{Part\ a}\!$$

$$ \lambda_1=3,\!\qquad\lambda_2= -6\!$$

$$ y(0)=-5,\qquad\! y'(0)=2$$

The roots of the differential equation can be written as

(2.0) $$\color{Red} y(x)=c_1 e^{3x} + c_2 e^{-6x}\!$$

To find the constants, we shall now apply the initial conditions. Since the there are two unknowns in this expression, it will be necessary to take the derivative of equation 2.0 to be able to apply the second initial condition.

(2.1 $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= 3c_1 e^{3x} - 6c_2 e^{-6x}\!$$

When applying the first initial condition, it results in

(2.2) $$ -5=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}-5=c_1+c_2\!$$

Applying the second initial conditions yields

(2.3) $$ 2=3c_1 e^{0} + -6c_2 e^{0}\longrightarrow\!\color{Red}2=3c_1-6c_2\!$$

By solving the system of equations, the constants come out to be

$$\color{Blue}c_1= \frac{-28}{9}\,\qquad c_2=\frac{-17}{9}\,$$

Solution Part a
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-28}{9} e^{3x} + \frac{-17}{9}e^{-6x} }$$



$$ \mathbf{Part\ b}\!$$

$$ \lambda_1=-2,\!\qquad\!\lambda_2=+5\!$$

$$ y(0)=1,\!\qquad y'(0)=0$$

Following the exact same procedure in part b as in part a, the general solution is shown below:

(2.4) $$ \color{Red} y(x)=c_1 e^{-2x} + c_2 e^{5x}\!$$

The derivative of equation 2.4 is

(2.3) $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= -2c_1 e^{-2x} +5c_2 e^{5x}\!$$

Applying the first initial condition yields

(2.5) $$ 1=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}1=c_1+c_2\!$$

Applying the second initial condition yields

(2.6) $$ 0=-2c_1 e^{0} + 5c_2 e^{0}\longrightarrow\!\color{Red}0=-2c_1+5c_2\!$$

Solving the system of equations, the constants are figured out to be

$$\color{Blue}c_1= \frac{-5}{2}\,\qquad c_2=\frac{2}{7}\,$$

Solution Part b
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-5}{2} e^{-2x} + \frac{2}{7}e^{5x} }$$



Section 3 Notes

Author: Frank Portillo Reviewed : Frank Portillo

Problem Statement
Find and plot the solution for $$ y''-10y'+25=0 \!$$.

Initial Conditions: $$ y(0)=1, y'(0)=0 \!$$

No excitation: $$ r(x)=0 \!$$

Background Information
Finding Homogeneous Solutions in cases where $$ \Delta=a^2-4b=0 \!$$

The two roots are a double root:

$$ \lambda_1=\lambda_2=\lambda=-a/2\!$$ (R2.1)

The first homogeneous solution is:

$$ y_{h,1}(x)= e^{yx} \!$$ (R2.2)

To find the 2nd homogeneous solution we use a trial solution:

$$ y_{h,2}=u(x)y_{h,1}(x) \!$$where u(x) is an unknown function of x (R2.3)

$$ y'_{h,2}=uy'_{h,1}+u'y_{h,1} \!$$ (R2.4)

$$ y_{h,2}=uy_{h,1}+2u'y'_{h,1}+u''y_{h,1}\!$$ (R2.5)

Adding up equations (R2.3), (R2.4), and (R2.5) we get:

$$ 0=u[y_{h,1}+ay'_{h,1}+by_{h,1}]+u'[2y'_{h,1}+ay_{h,1}]+uy_{h,1} \!$$ (R2.6)

We can simplify equation (R2.6) by applying known equations:

$$ 0=u''y_{h,1} \!$$ (R2.7)

Hence:

$$ u''=0 \!$$ (R2.8)

Integrating with respect to x twice yields:

$$ u(x)=xc_3 \!$$<p style="text-align:right"> (R2.9)

So $$ y_{h,2}=(x+c_3)y_{h,1}\!$$<p style="text-align:right"> (R2.10)

The final homogeneous solution is:

$$ y_h=c_1y_{h,1}+c_2xy_{h,1} \!$$<p style="text-align:right"> (R2.11)

Solution
$$ y''-10y'+25y=r(x) \!$$<p style="text-align:right"> (R2.12)

The characteristic equation is:

$$ \lambda^2-10\lambda+25=0\!$$<p style="text-align:right"> (R2.13)

The first homogeneous root is:

$$ \lambda=5 \!$$<p style="text-align:right"> (R2.14)

Hence the homogeneous solution is:

$$ y_h(x)=c_1e^{5x}+c_2xe^{5x} \!$$<p style="text-align:right"> (R2.15)

Apply the initial conditions y(0)=1 and y'(0)=0:

$$ y_h(0)= c_1 =1 \!$$<p style="text-align:right"> (R2.16)

$$ y_h'(x)= 5c_1e^{5x}+5c_2xe^{5x}+c_2 \!$$<p style="text-align:right"> (R2.17)

$$ y_h'(0)=5+c_2 \!$$<p style="text-align:right"> (R2.18)

Hence:

$$c_1=1 \!$$<p style="text-align:right"> (R2.19)

$$ c_2=-5 \!$$<p style="text-align:right"> (R2.20)

The homogeneous equation is:

$$ y_h(x)=e^{5x}-5xe^{5x} \!$$<p style="text-align:right"> (R2.21)

Since r(x)=0 the particular equation is zero. Hence the solution to the problem is:

$$ y(x)=e^{5x}-5xe^{5x} \!$$<p style="text-align:right"> (R2.22)

The plot of the function is:



References:

Section 5 Notes

Author: Kendall Ryser Reviewed : Frank Portillo

Problem Statement
3.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 6y' + 8.96y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is

$$ \lambda^2 + a\lambda + b = 0\,\!$$<p style="text-align:right">(3.1a)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 6\lambda + 8.96 = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$<p style="text-align:right">(3.2a)

$$ \Delta = 36 - 4(8.96) = 0.16\,\!$$

Since the discriminant is positive, there will be two distinct real solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$<p style="text-align:right">(3.3a)

$$ \lambda_{1,2} = \frac{-6 \pm \sqrt{0.16}}{2}\,\!$$ $$ = \frac{-6 \pm 0.4}{2}\,\!$$ $$ = -3 \pm 0.2\,\!$$

Therefore,

$$ \lambda_{1} = -2.8 \,\!$$<p style="text-align:right">(3.4a)

$$ \lambda_{2} = -3.2 \,\!$$<p style="text-align:right">(3.5a)

The two distinct, linearly independent, homogeneous solutions for the case with two real roots are

$$ y_{h,1}(x) = e^{\lambda_1 x}\,\!$$<p style="text-align:right">(3.6a)

$$ y_{h,2}(x) = e^{\lambda_2 x}\,\!$$<p style="text-align:right">(3.7a)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$<p style="text-align:right">(3.8a)

$$ y_{h}(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}\,\!$$

The final general solution is therefore

$$ y_{h}(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$<p style="text-align:right">(3.9a)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}\,\!$$<p style="text-align:right">(3.10a)

$$ y''(x) = 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x}\,\!$$<p style="text-align:right">(3.11a)

Substituting back into the original ODE gives

$$ y'' + 6y' + 8.96y = 0\,\!$$<p style="text-align:right">(3.12a)

$$ 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} + 6(-2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}) + 8.96(c_1 e^{-2.8 x} + c_2 e^{-3.2 x}) = 0\,\!$$

$$ \cancelto{0}{(7.84 - 16.8 + 8.96)c_1 e^{-2.8x}} + \cancelto{0}{(10.24 - 19.2 + 8.96)c_2 e^{-3.2x}} = 0\,\!$$

Since all terms cancel to 0, $$ y(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$ is a general solution to the original ODE.

Section 3 Lecture Notes

Section 5 Lecture Notes

Author: Gabrielle Steinberg Reviewed : Frank Portillo

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

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

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is

$$ \lambda^2 + a\lambda + b = 0\,\!$$<p style="text-align:right">(3.1b)

For the ODE in this problem, the characteristic equation becomes

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

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$<p style="text-align:right">(3.2b)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$<p style="text-align:right">(3.3b)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$<p style="text-align:right">(3.4b)

$$ \lambda_{2} = -2 - \pi i \,\!$$<p style="text-align:right">(3.5b)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$<p style="text-align:right">(3.6b)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$<p style="text-align:right">(3.7b)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$<p style="text-align:right">(3.8b)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$<p style="text-align:right">(3.9b)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$<p style="text-align:right">(3.10b)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$<p style="text-align:right">(3.11b)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$<p style="text-align:right">(3.12b)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 5 Lecture Notes

Section 6 Lecture Notes

Section 5 Notes

Author: Gabrielle Steinberg Reviewed : Frank Portillo

Problem Statement
5. Find a general solution. Check by substitution.

$$  \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dx}\ + \pi ^{2}y=0   $$ <p style="text-align:right">(4.0a)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ <p style="text-align:right">(4.1a)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ <p style="text-align:right">(4.2a)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2xe ^ { \lambda _2 x}  $$ <p style="text-align:right">(4.3a)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ <p style="text-align:right">(4.4a)

Check by Substitution:

To check by substitution, you need to find $$  \; y, y', y'' $$ and plug them into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$<p style="text-align:right">(4.5a)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ <p style="text-align:right">(4.6a)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 5 Lecture Notes

Section 5 Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem Statement
6. Find the general solution. Check by substitution.

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

Solution
The original problem can be re-written as:
 * {| style="width:100%" border="0"

$$  \displaystyle 10\frac{d^2y}{dx^2}-32\frac{dy}{dx}+25.6y=0 $$     (4.0b)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \frac{d}{dx}=\lambda $$     (4.1b) By substituting equation 4.1b into 4.0b we get:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle 10\lambda^2-32\lambda+25.6=0 $$     (4.2b) Solve for $$ \; \lambda $$ using the quadratic equation.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (4.3b) By using equation 4.3b it is shown that $$ \lambda=\frac{8}{5}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since $$ \; \lambda$$ is a solution to the ODE then
 * {| style="width:100%" border="0"

$$  \displaystyle y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x} $$ (4.4b) is the solution of the ODE with real and distinct roots.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

By plugging the value of $$ \; \lambda $$ into equation 4.4b we come to the general solution.

General Solution $$ y=e^{\frac{8}{5}x}({c_1+c_2 x}) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.


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

$$  \displaystyle y'=\frac{8}{5}e^{\frac{8}{5}x}(c_1+c_2 x)+c_2 e^{\frac{8}{5}x} $$     (4.5b)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y''=\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+\frac{16}{5}c_2 e^{\frac{8}{5}x} $$     (4.6b) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 4.5b, and 4.6b back into the original equation.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

After substitution we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)-\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+32c_2e^{\frac{8}{5}x}-32c_2e^{\frac{8}{5}x}=?? $$ This equation clearly equals  zero  which proves our general solution is  correct 
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Section 5 Lecture Notes

Section 5 Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem R2.5a
Problem 16 on pg 59-Find an ODE $$ \displaystyle y'' + ay' +by=0$$ for the given basis of  $$ \displaystyle e^{2.6x}, e^{-4.3x}$$ This is a case with two distinct Real-Roots. Therefore the general equation for the ODE will be


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

$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$     (5.0a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Where $$\displaystyle {\lambda_1x}=2.6x $$  and   $$ \displaystyle {\lambda_2x} = -4.3x $$ so that


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

$$  \displaystyle y=c_1e^{2.6}+c_2e^{-4.3} $$     (5.1a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The auxiliary equation is therefore


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

$$  \displaystyle ({\lambda}-2.6)({\lambda}+4.6)=0 $$     (5.2a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle {\lambda^2} + 1.7{\lambda}-11.18=0 $$     (5.3a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

which results in


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

$$  \displaystyle y'' + 1.7y'-11.18y=0 $$     (5.4a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution

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

$$\displaystyle y'' + 1.7y'-11.18y=0 $$
 * style="width:95%" |
 * style="width:95%" |

Section 5 Notes

Author: Amy Mclarty Reviewed : Frank Portillo

Problem Statement
17. Find an ODE for the given solutions.

$$ e ^{- \sqrt{5} x} $$, $$ e ^{- \sqrt{5} x} $$

using $$ \; {y}''+a {y}'+b y=0 $$ for the given basis.

Solution
Case: Real double root ODE

where $$ \; \lambda = -a/2 $$

<p style="text-align:right">(2.5b.0) and the general solution is $$ \; y= e ^ {-ax} (c_1+ c_2x)  $$

<p style="text-align:right">(2.5b.1) So $$   \; e ^ { \lambda x} =  e ^ {- \sqrt {5} x}  $$

Plugging in $$ \; \lambda = -a/2 $$

<p style="text-align:right">(2.5b.2) $$  \; e ^ {-ax/2} =  e ^ {- \sqrt {5} x}  $$

Solving for a,

<p style="text-align:right">(2.5b.3) $$  \;  -ax/2= - \sqrt {5} x  $$

$$  \; a= 2 \sqrt {5}  $$

<p style="text-align:right">(2.5b.4) By plugging a into the determinant $$ \; a^2-4b=0 $$, Solve for b.

<p style="text-align:right">(2.5b.5) $$  \; {({2 \sqrt {5}})}^2 -4b=0  $$

<p style="text-align:right">(2.5b.6) $$  \; 4(5)=4b $$

$$  \; b=5  $$

By plugging in a and b, the ODE is:

<p style="text-align:right">(2.5b.7) $$  \; {y}''+ 2 \sqrt {5}{y}'+ 5y=0 $$

Section 5 Lecture Notes

Section 5 Notes

Author: Neisha Ramnarine Reviewed : Frank Portillo

Problem Statement
Using the spring-dashpot-mass system shown below (from Fig p1.4) and following similar characteristics of (3)p5.5 but using the double real root $$\displaystyle \lambda = -3 $$, find values for the parameters k,c,m.




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

$$  \displaystyle {(\lambda -5)}^{2} = {\lambda }^{2}-10\lambda +25 = 0 $$     ((3)p5.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewriting ((3)p5.5) with the double real root $$\displaystyle \lambda = -3 $$ gives the new characteristic equation


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

$$  \displaystyle {(\lambda -3)}^{2} = {\lambda }^{2}-6\lambda +9 = 0 $$     (6.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From kinematics:


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

$$  \displaystyle y = {y}_{k}+{y}_{c} $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From kinetics:


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

$$  \displaystyle my''+{f}_{1} = f(t) $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle {f}_{1} = {f}_{k} = {f}_{c} $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Relationships:


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

$$  \displaystyle {f}_{k} = k{y}_{k} $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle {f}_{c} = c{y'}_{c} $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From equation (6.1) take the second derivative


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

$$  \displaystyle y = {y}_{k} = {y''}_{c} $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From equations (6.3), (6.4) & (6.5)


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

$$  \displaystyle {f}_{k} = {f}_{c} \Rightarrow k{y}_{k} = c{y'}_{c} \Rightarrow {y'}_{c} = \frac{k}{c}{y}_{k} $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From equations (6.6) & (6.7)


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

$$  \displaystyle y = {y}_{k}+({y'}_{c})' \Rightarrow {y''}_{k}+(\frac{k}{c}{y}_{k})' $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From equations (6.2), (6.3) & (6.8)


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

$$  \displaystyle m({y''}_{k}+\frac{k}{c}{y'}_{k})+k{y}_{k} = f(t) $$     (6.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From the characteristic equation (6.0), the corresponding homogeneous L2-ODE-CC is


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

$$  \displaystyle y''+6y'+9y = 0 $$     (6.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now take equation (6.9) and divide through by m


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

$$  \displaystyle {y''}_{k}+\frac{k}{cm}{y'}_{k}+\frac{k}{m}{y}_{k} = \frac{f(t)}{m} = 0 $$     (6.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Comparing to equation (6.10) gives


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

$$  \displaystyle \frac{k}{cm} = 6 $$     (6.12) &
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{k}{m} = 9 $$     (6.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

There are an infinite number of solutions to these two equations, one of these is

$$   \displaystyle k = 9\;\;m = 1\;\;c = \frac{3}{2}$$

Section 1 Lecture Notes

Section 5 Lecture Notes

Section 5 Notes

Author: Steven Rossmeissl Reviewed : Frank Portillo

Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for

$$e^{t} \qquad \sin t \qquad \ cos t\ $$

To develop any MacLaurin series, it is important to follow the following formula:

<p style="text-align:right">(7.0) $$ f(x) = f(a) + f'(a)\frac{(x-a)}{1!}^{1}\ + f(a)\frac{(x-a)}{2!}^{2} +f'(a)\frac{(x-a)}{3!}^{3}\ + \cdots f^{n}\frac{(x-a)}{n!}^{n}$$

This equation can be simplified by

<p style="text-align:right">(7.1) $$ \sum_{n=0}^{\infty} f^{n}(a)\frac{(x-a)}{n!}^{n}$$

When writing the Maclaurin series for $$ e^{t}$$, using equation 7.0, it looks like

<p style="text-align:right">(7.2) $$ 1 + (1)\frac{(x-0)}{1!}^{1} + (1)\frac{(x-0)}{2!}^{2} + (1)\frac{(x-0)}{3!}^{3}$$

which can be simplified to

Solution R2.7a
<p style="text-align:right">(7.3) $$ e^{t} = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\cdots\frac{x^n}{n!} $$

Similarly, starting with the function

$$ \sin t\ $$

the MacLaurin series looks like:

<p style="text-align:right">(7.4) $$ 1\frac{(t-0)}{1!}^{1}+ 0 \frac{(t-0)}{2!}^{2} +1\frac{(t-0)}{3!}^{3} + 0\frac{(t-0)}{4!}^{4} $$

which can be simplified into

Solution R2.7b
<p style="text-align:right">(7.5) $$ \sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} +\cdots - \frac{t^{n-1}}{(t-1)!} + \frac{t^n}{n!} $$

Finally, starting with

<p style="text-align:right">(7.6) $$ \cos t\ $$, the MacLaurin series looks like

<p style="text-align:right">(7.7) $$ 1 - (0)\frac{(x-0)}{1!}^{1} - (1)\frac{(x-0)}{2!}^{2} + (0)\frac{(x-0)}{3!}^{3} + (1)\frac{(x-0)}{4!}^{4}$$

which can be simplified into

Solution R2.7c
<p style="text-align:right">(7.8) $$ \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} +\cdots - \frac{t^{n-1}}{(t-1)!} + \frac{t^n}{n!} $$

Problem Statement
8. Find the general solution. Check by substitution.

$$ y''+y'+3.25y=0 $$

Solution
The original problem can be re-written as:
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}+3.25y=0 $$     (8.0a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \frac{d}{dx}=\lambda $$     (8.1a) By substituting equation 8.1a into 8.0a we get:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+\lambda+3.25=0 $$     (8.2a) Solve for $$ \; \lambda $$ using the quadratic equation.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (8.3a) By using equation 8.3a it is shown that $$ \lambda=\frac{-1\pm i \sqrt{12}}{2}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since $$ \; \lambda$$ is a solution to the ODE then
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.4a) is the solution of the ODE with real and distinct roots.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

General Solution $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.


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

$$  \displaystyle y'=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x) $$     (8.5a)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y''=\frac{1}{4}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)$$
 * style="width:95%" |
 * style="width:95%" |

$$+\frac{1}{2}e^{-\frac{1}{2}x}(-12c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.6a) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 8.5a, and 8.6a back into the original equation.
 * <p style="text-align:right">
 * }

After substitution we get:
 * {| style="width:100%" border="0"

$$  \displaystyle y''+y'+3.25y=0 $$ So the solution to the ODE is: $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Section 6 Lecture Notes

Author: Ryan Hagberg Reviewed : Frank Portillo

Problem Statement
15. Find a general solution. Check by substitution.

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$

Solution
The given ODE can be rewritten as:

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y= \frac{d^2y}{dx^2}\ + 0.54 \frac{dy}{dx}\ + (0.00729+ \pi )y=0   $$

If  $$ \; \frac{d}{dx}\ = \lambda $$,

The ODE becomes:

<p style="text-align:right">(2.8b.0) $$ \; \lambda ^{2}+0.54 \lambda +(0.0729+ \pi)=0 $$

Solving for $$ \; \lambda $$ by using the quadratic formula,

<p style="text-align:right">(2.8b.1) $$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\,\!$$

<p style="text-align:right">(2.8b.2) $$ \lambda_{1} = -0.27 -{ \sqrt{ \pi } i} $$

<p style="text-align:right">(2.8b.3) $$ \lambda_{2} = -0.27 +{ \sqrt{ \pi } i} $$

<p style="text-align:right">(2.8b.4) In the standard form: $$ \; \lambda=a + bi $$.

General Solution:

The general homogeneous solution is:

<p style="text-align:right">(2.8b.5) $$ y= c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\ \!$$

When plugging in the values for a and b, the general solution becomes:

<p style="text-align:right">(2.8b.6) $$ y= c_1 e^{-0.27 x} \cos \sqrt{ \pi }x + c_2 e^{-0.27 x} \sin \sqrt{ \pi }x\ \!$$

<p style="text-align:right">(2.8b.7) So $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

Check by Substitution:

Solving for derivatives:

<p style="text-align:right">(2.8b.8) $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

<p style="text-align:right">(2.8b.9) $$ \; y'=e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

<p style="text-align:right">(2.8b.10) $$ \; y''=e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x) \ \!$$

Plugging in  $$   \; y, y', y'' $$ into

<p style="text-align:right">(2.8b.11) $$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$:

<p style="text-align:right">(2.8b.11) $$ \; [e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)]$$

<p style="text-align:right">(2.8b.12) $$ \; +0.54 [e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x +  \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)]+(0.0729+ \pi )[e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 5 Notes

Author: Neisha Ramnarine Reviewed : Frank Portillo

Problem Statement
With the given initial conditions, find and plot the solution to

$$ \lambda^2 + 4\lambda + 13 = 0 \!$$

The initial conditions are:

$$ y(0) = 1 \qquad y'(0) = 0 \!$$

It important to remember, there is no excitation in this problem,

$$ i.e. \qquad r(x)=0\ $$

According to Kreyszig, this solution can be found by using the quadratic to find the roots. These roots come out to be

<p style="text-align:right">(9.0) $$ \lambda_1 = -2 + 3i \qquad \lambda_2= -2 - 3i$$

We can now write the general solution to the problem as

<p style="text-align:right">(9.1) $$ y_h= e^{-2x} [Acos(3x) + Bsin(3x)] \!$$

it is now possible to apply the initial conditions to this solution, since we have two, it is necessary to take the derivative of the general solution.

<p style="text-align:right">(9.2) $$ y(0) = e^0 [A\cos(0) + B\sin(0) ]= 1 \color{Red} \rightarrow A=1 $$

Using the second pair of initial conditions yields

<p style="text-align:right">(9.3) $$y'(0) = 0 = -2 \color{Red} (1) \color{Black} e^{0} \cos(0) + 3\color{Red} B \color{Black} e^{0} \ cos(0) \color{Red} \rightarrow B = 2/3$$

Solution
$$ y_h= e^{-2x} [cos(3x) + \frac{2}{3}sin(3x)] \!$$



Section 6 Lecture Notes

Author: Frank Portillo Reviewed : Frank Portillo

= Contributing Members =

Table code referenced from Team 1