University of Florida/Egm4313/s12.team5.R2

Question
Given the two roots and the initial conditions:


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$$\displaystyle \lambda_1=-2, \lambda_2 = +5$$ (1.0)
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$$\displaystyle y(0)=1, y'(0)=0$$ (1.1)
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Part 1.

Find the non-homogeneous L2-ODE-CC in standard form. Find the solution in terms of the initial conditions and the general excitation $$r(x).\!$$ Now with no excitation and plot the solution:


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$$\displaystyle r(x) = 0$$ (1.2) Part 2.
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Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $$\lambda_1=-2, \lambda_2 = +5\!$$ as the two roots of the corresponding characteristic equation.

Solution
Part 1.

Characteristic Equation:
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$$\displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) = 0$$ (1.3)
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$$\displaystyle (\lambda-(-2))(\lambda-5) = 0$$ (1.4)
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$$\displaystyle \lambda^2-3\lambda-10=0$$ (1.5) Non-Homogeneous L2-ODE-CC:
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$$ \displaystyle y''-3y'-10y=r(x)\rightarrow standard form$$ (1.6)
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Homogenous Solution:
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$$\displaystyle y_h(x)=c_1e^{-2x} + c_2e^{5x}$$ (1.7) Overall Solution:
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$$\displaystyle y(x)=c_1e^{-2x} + c_2e^{5x}+y_p(x)$$ (1.8)
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$$\displaystyle y'(x)=-2c_1e^{-2x} + 5c_2e^{5x}+y'_p(x)$$ (1.9) Satisfy Initial Conditions:
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$$\displaystyle y(0)=1=c_1 + c_2 +y_p(0)$$ (1.10)
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$$\displaystyle y'(0)=0=-2c_1 + 5c_2 +y'_p(0)$$ (1.11) No excitation:
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$$\displaystyle r(x) = 0 \rightarrow y_p(x)=0 \rightarrow y'_p(x)=0$$ (1.12) From (1.10):
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$$\displaystyle c_1+c_2 = 1 \rightarrow c_1=1-c_2$$ (1.13) From (1.11):
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$$\displaystyle -2c_1+5c_2 = 0$$ (1.14)
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$$\displaystyle 5c_2=2c_1$$ (1.15)
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$$\displaystyle \frac{5}{2}c_2=c_1$$ (1.16) Plug (1.13) into (1.16):
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$$\displaystyle \frac{5}{2}c_2=1-c_2$$ (1.17) Solve for $$c_2\!$$:
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$$\displaystyle c_2=\frac{2}{7}$$ (1.18) Plug (1.18) into (1.13) and solve for $$c_1\!$$:
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$$\displaystyle c_1=\frac{5}{7}$$ (1.19) Therefore, the final solution in terms of the initial conditions and the general excitation $$r(x)\!$$ is:
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$$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}$$ (1.20)
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Plot of Solution:

Figure 1

Part 2.

Three non-standard and non-homogeneous solutions using the same roots given above:

1.
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$$\displaystyle 2(\lambda-(-2))(\lambda-5) = 0$$ (1.21)
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$$\displaystyle 2\lambda^2-6\lambda-20=0$$ (1.22)
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2.
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$$\displaystyle 3(\lambda-(-2))(\lambda-5) = 0$$ (1.23)
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$$\displaystyle 3\lambda^2-9\lambda-30=0$$ (1.24)
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3.
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$$\displaystyle 4(\lambda-(-2))(\lambda-5) = 0$$ (1.25)
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$$\displaystyle 4\lambda^2-12\lambda-40=0$$ (1.26)
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Author
This problem was solved and uploaded by [Derik Bell]

This problem was proofread by: David Herrick

Question
From Section 5 in the notes, pg. 5-6

Solve the L2-ODE-CC from pg. 5-5, equation (4)

$$ \displaystyle \ y'' - 10y' +25y = r(x)\ $$

For initial conditions:

$$ \displaystyle \ y(0) = 1, y'(0) = 0\ $$

Where there is no excitation, i.e. $$ \displaystyle \ r(x) = 0\ $$

Solution
$$ \displaystyle \ \lambda^2 - 10\lambda + 25 = 0\ $$

Factoring:

$$ \displaystyle \ (\lambda - 5)^2 = 0\ $$

Thus, there is a double root, where $$ \displaystyle \ \lambda = 5\ $$

Because there is no excitation, the homogenous solution will be the same as the final solution. Thus, the solution will be:

$$ \displaystyle \ y_h(x) = y(x) = C_1e^{5x} + C_2xe^{5x}\ $$

Plugging in the initial condition,

$$ \displaystyle \ y(0) = 1 = C_1e^{0} + C_2 (0) e^{0}\ $$

$$ \displaystyle \ C_1 = 1\ $$

In order to solve for the second constant, we need to take a derivative.

$$ \displaystyle \ y'_h(x) = y'(x) = 5C_1e^{5x} + C_2e^{5x} + 5C_2xe^{5x}\ $$

Plugging in C1 and the other initial condition:

$$ \displaystyle \ y'_h(0) = y'(0) = 0 = 5(1)e^{0} + C_2e^{0} + 5C_2(0)e^{0} \ $$

$$ \displaystyle \ 5 + C_2 = 0 \Rightarrow C_2 = -5\ $$

So, the final solution is:

$$ \displaystyle \ y(x) = e^{5x} - 5xe^{5x}\ $$

Matlab Code:

x = 0:0.001:1;

f = exp(5*x) - 5*x.*exp(5*x);

plot(x,f)

xlabel ('x')

ylabel ('y(x)')



Author
This problem was solved and uploaded by [David Herrick] This problem was checked by [William Knapper]

Question
Complete problems 3 and 4 from p.59 of K 2011.

Solution
The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

Problem 3
$$ \displaystyle 3. \; {y}''+6{y}' +8.96y = 0 $$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

$$ \displaystyle \lambda ^{2} + a\lambda + b = 0$$

where a = 6 and b = 8.96. Taking the discriminant, we find that

$$ \displaystyle 6^2 - 4(8.96) = 0.16 > 0 $$

A positive discriminant implies that there exists two real roots, with a general solution of the form:

$$ \displaystyle y = c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x} $$

To find the unknowns, we find the roots of the above equation:

$$ \displaystyle \lambda ^{2} + 6\lambda + 8.96 = 0$$

$$ \displaystyle \lambda _{1} = -3.2, \lambda _{2} = -2.8 $$

Therefore, the correct general solution is:

$$ \displaystyle y=c_{1}e^{-3.2x} + c_{2}e^{-2.8x} $$

To check that this solution is correct, we take the first and second derivatives:

$$ \displaystyle {y}' = -3.2c_{1}e^{-3.2x} - 2.8c_{2}e^{-2.8x} $$

$$ \displaystyle {y}'' = 10.24c_{1}e^{-3.2x} + 7.84c_{2}e^{-2.8x} $$

Now, we plug these values into the original equation:

$$ \displaystyle 10.24c_{1}e^{-3.2x} + 7.84c_{2}e^{-2.8x} - 19.2c_{1}e^{-3.2x} - 16.8c_{2}e^{-2.8x} + 8.96c_{1}e^{-3.2x} + 8.96c_{2}e^{-2.8x} = 0 $$

Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.

Problem 4
$$ \displaystyle 4. \; {y}'' + 4{y}' + (\pi ^{2}+4)y = 0$$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

$$ \displaystyle \lambda ^{2} + a\lambda + b = 0$$

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

$$ \displaystyle 4^{2} - 4(\pi ^{2} + 4) = -4\pi ^{2} < 0 $$

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

$$ \displaystyle y = e^{-2x}(Acos(\omega x)+Bsin(\omega x)) $$

Where

$$ \displaystyle \omega = \sqrt{b-\frac{1}{4}a^{2}} = \sqrt{(\pi ^{2}+4)-(\frac{1}{4})4^{2}} = \pi  $$

This yields

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$

To verify that this is the correct solution, we first find the first and second derivatives of the solution:

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}' = e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)] $$

We then plug these values into the original problem to get this somewhat lengthy equation:

$$ \displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x) +2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+$$ $$ \displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-$$ $$ \displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x) = 0 $$

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

Author
This problem was solved and uploaded by: (Will Knapper)

This problem was proofread by: David Herrick

Question
K 2011 p.59 pbs.5,6. Find a general solution. Check your answer by substitution.

Problem 5

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

Solution
Determine the characteristic equation and the nature of the determinate:
 * $$ y = e^{-\lambda x} \ $$
 * $$ y' = -\lambda e^{-\lambda x} \ $$
 * $$ y'' = \lambda^2 e^{-\lambda x} \ $$

Characteristic Equation:
 * $$ \lambda^2 - 2\pi \lambda + \pi^2\lambda = 0 \ $$
 * $$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0 \ $$

0 indicates a real double root, therefore:
 * $$ y_g = e^{-\frac{a}{2}x} (C_1 + C_2 x) \ $$

$$ a = 2\pi \ $$ from the original equation. Therefore: $$ y_g = e^{-\pi x} (C_1 + C_2 x) \ $$
 * $$ y'_g = e^{-\pi x} (C_2 - \pi C_1 - \pi C_2 x) \ $$
 * $$ y''_g = e^{-\pi x} (C_1 \pi^2 + C_2 \pi^2 x - 2C_2 \pi) \ $$

Substituting into the original equation gives:
 * $$ e^{-\pi x} [(C_1 \pi^2 + C_2 \pi^2 x - 2C_2 \pi) + 2\pi (C_2 - \pi C_1 - \pi C_2 x) + \pi^2 (C_1 + C_2 x)] = 0 \ $$
 * $$ = e^{-\pi x} [C_1\pi^2 + C_2\pi^2 x - 2C_2\pi + 2C_2\pi -2C_1\pi^2 - 2C_2\pi^2 x +C_1\pi^2 + C_2\pi^2 x] \ $$

Canceling gives 0 = 0, confirming the general solution found above.

Author
This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

Problem 5

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

Solution
Reduce original equation so that the first coefficient is 1:
 * $$ y'' - 3.2y' + 2.56y = 0 \ $$

Determine the characteristic equation and the nature of the determinate:
 * $$ y = e^{-\lambda x} \ $$
 * $$ y' = -\lambda e^{-\lambda x} \ $$
 * $$ y'' = \lambda^2 e^{-\lambda x} \ $$

Characteristic Equation:
 * $$ \lambda^2 - 3.2\lambda + 2.56 = 0 \ $$
 * $$ a^2 - 4b = 3.2^2 - 4(2.56) = 10.24 - 10.24 = 0 \ $$

0 indicates a real double root, therefore:
 * $$ y_g = e^{-\frac{a}{2}x} (C_1 + C_2 x) \ $$

$$ a = \frac{-(-3.2)}{2} = 1.6 $$ from the original equation. Therefore: $$ y_g = e^{1.6x} (C_1 + C_2 x) \ $$
 * $$ y'_g = e^{1.6x} (1.6C_1 + 1.6C_2 x + C_2) \ $$
 * $$ y''_g = e^{1.6x} (2.56C_1 + 2.56C_2 x + 3.2C_2) \ $$

Substituting into the original equation gives:
 * $$ e^{1.6x} [(2.56C_1 + 2.56C_2 x + 3.2C_2) -3.2(1.6C_1 + 1.6C_2 x + C_2) + 2.56(C_1 + C_2 x)] = 0 \ $$
 * $$ = e^{1.6x} [(2.56C_1 + 2.56C_2 x + 3.2C_2 - 5.12C_1 - 5.12C_2 x - 3.2C_2 + 2.56C_1 + 2.56C_2 x] = 0 \ $$

Canceling gives 0 = 0, confirming that general solution found above.

Author
This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

Question
Problems 16 and 17 from pg. 59 of the textbook. The two questions have the same instructions: Find an ODE of the form $$ \displaystyle {y}'' + a{y}' + by = 0 $$ for the given basis.

Problem 16
$$ \displaystyle 16. \; e^{2.6x}, e^{-4.3x} $$ This basis implies that there are two distinct, real roots: 2.6 and -4.3. We put these roots into the characteristic equation to get: $$ \displaystyle (\lambda - 2.6)(\lambda + 4.3)=0 \rightarrow \lambda ^{2} + 1.7\lambda -11.18=0 $$ Then, you convert the characteristic equation to a regular ODE: $$ \displaystyle {y}'' + 1.7{y}' -11.18y = 0 $$

Problem 17
$$ \displaystyle 17. \; e^{-\sqrt{5}x} \;, xe^{-\sqrt{5}x} $$ A basis of this form implies that the solution is of the form: $$ \displaystyle y = (c_{1} + c_{2}x)e^{-ax/2} $$ The real double root can be found by this equation: $$ \displaystyle \lambda = -a/2 = -\sqrt{5} $$ That means that the characteristic equation can be found by solving: $$ \displaystyle (\lambda + \sqrt{5} )^{2} = \lambda ^{2} + 2 \sqrt{5}\lambda + 5 $$ Converting the characteristic equation to the ODE, we get: $$ \displaystyle {y}'' + 2 \sqrt{5} {y}' + 5y = 0 $$

Author
This problem was solved and uploaded by: (Will Knapper) This problem was proofread by (Josh House)

Question
Realize spring-dashpot-mass system in series as shown in Fig. p.1-4 with the similar characteristic as in (3)p.5-5, but with the double real root $$ \displaystyle \lambda=-3\ $$, i.e., find values for parameters k,c,m.

Solution
First we put the double root into the characteristic equation: $$ \displaystyle (\lambda+3)^2 =0 $$ $$ \displaystyle \lambda^2 + 6\lambda + 9 = 0 $$ The corresponding L2-ODE is: $$ \displaystyle {y}'' + 6{y}' + 9y = 0 $$ The equation of motion of the spring-dashpot-mass system is: $$ \displaystyle m \left(y_k'' + \frac{k}{c}y_k' \right) + ky_k = 0 $$ or: $$ \displaystyle my_k'' + m \frac{k}{c}y_k' +ky_k = 0 $$ Solving for the parameters k,c,m: $$ \displaystyle m = 1 $$ $$ \displaystyle m \frac{k}{c} = 6 $$ $$ \displaystyle k = 9 $$ We get: $$ \displaystyle m = 1 $$ $$ \displaystyle k = 9 $$ $$ \displaystyle c = \frac{3}{2} $$

Author
This problem was solved and uploaded by: (Radina Dikova)

This problem was proofread by: Mike Wallace

Question
Develop the McLauren series (Taylor series at t=0) for e^t, cos(t), and sin(t).

Solution
The general form of the Taylor series expansion at point a is
 * $$ f(a) + \frac{f'(a)} {1!} (x-a) + \frac{f(a)} {2!} (x-a)^2 + \frac{f'(a)} {3!} (x-a)^3 + ... $$

for $$ e^t $$ at t=0:
 * $$ e^0 + \frac{e^0} {1!} (x) + \frac{e^0} {2!} (x)^2 + \frac{e^0} {3!} (x)^3 + ... $$

$$ = 1 + x + \frac{x^2} {2!} + \frac{x^3} {3!} + ... $$ for $$ \cos(t) $$ at t=0:
 * $$ \cos(0) + \frac{-\sin(0)} {1!} (x) + \frac{-\cos(0)} {2!} (x)^2 + \frac{sin(0)} {3!} (x)^3 + \frac{cos(0)} {4!} (x)^4 + ... $$

$$ = 1 - \frac{x^2} {2!} + \frac{x^4} {4!} + ... $$ for $$ \sin(t) $$ at t=0:
 * $$ \sin(0) + \frac{cos(0)} {1!} (x) + \frac{-\sin(0)} {2!} (x)^2 + \frac{-cos(0)} {3!} (x)^3 + \frac{\sin(x)} {4!} (x)^4 + \frac{\cos(0)} {5!} (x)^5 + ... $$

$$ = x - \frac{x^3} {3!} + \frac{x^5} {5!} + ... $$

Author
This problem was solved and uploaded by: (John North) This problem was checked by: William Knapper

Question
Find a general solution. Check your answer by substitution.

8. $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

15. $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
8. $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Writing the characteristic equation:
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

Which is now in the form:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Solving for the two roots:
 * $$ \displaystyle \lambda _{1} = \frac{1}{2}(-a+\sqrt{a^{2}-4b}), \lambda _{2}=\frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

We will find the discriminant to be less than 0, leading to complex conjugate roots:
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Which leads to a solution of the form:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Where $$ \displaystyle \omega^{2} = b-\frac{1}{4}a^{2} $$, therefore:
 * $$ \displaystyle \omega^{2} = 3.25-\frac{1}{4}(1)^{2} =3 \rightarrow \omega=\sqrt{3} $$

Giving us a solution of: $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking our solution:
 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Plugging all of this into the original differential equation and combining like terms gives us:
 * $$\displaystyle [\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Acos(\sqrt{3}x) $$
 * $$\displaystyle [\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}-\sqrt{3}]e^{-x/2}Asin(\sqrt{3}x) $$
 * $$\displaystyle [-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\sqrt{3}]e^{-x/2}Bcos(\sqrt{3}x) $$
 * $$\displaystyle [\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Bsin(\sqrt{3}x) $$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

15. $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Writing the characteristic equation:
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

Which is now in the form:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Solving for the two roots:
 * $$ \displaystyle \lambda _{1} = \frac{1}{2}(-a+\sqrt{a^{2}-4b}), \lambda _{2}=\frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

We will find the discriminant to be less than 0, leading to complex conjugate roots:
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Which leads to a solution of the form:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Where $$ \displaystyle \omega^{2} = b-\frac{1}{4}a^{2} $$, therefore:
 * $$ \displaystyle \omega^{2} = (0.0729-\pi)-\frac{1}{4}(0.54)^{2} =\pi \rightarrow \omega=\sqrt{\pi} $$

Giving us a solution of: $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking our solution:
 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$
 * $$\displaystyle y^{'}=-0.27e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$
 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Plugging all of this into the original differential equation and combining like terms gives us:
 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$
 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$
 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$
 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

Author
This problem was solved and uploaded by: (Josh House) This problem was proofread by: (Radina Dikova)

Question
For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to

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

In another figure superpose this figure, Fig. from R2.6 p.5-6, and the Fig. from R2-1 p.3-7.

This corresponds to the L2-ODE-CC in standard form

$$ \displaystyle \ y'' + 4y' + 13y = r(x) \ $$

Given
Initial conditions

$$ \displaystyle \ y(0) = 1, y'(0) = 0 \ $$

Where there is no excitation

$$ \displaystyle \ r(x) = 0 \ $$

Solution
Solving the quadratic equation for $$ \displaystyle \ \lambda \ $$ determines the roots of the ODE

$$ \displaystyle \ \lambda_1 = \frac{-b + \sqrt{b^2-(4ac)}}{2a} \! = \frac{-4 + \sqrt{(4)^2-4(1)(13)}}{2(1)} \!$$

$$ \displaystyle \ \lambda_2 = \frac{-b - \sqrt{b^2-(4ac)}}{2a} \! = \frac{-4 - \sqrt{(4)^2-4(1)(13)}}{2(1)} \!$$

The roots of the quadratic equation come out to be:

$$ \displaystyle \ \lambda = (-2 \pm 3i) \ $$

The general homogeneous solution of the ODE will resemble the following form.

$$ \displaystyle \ y = C_1e^{a_1x}cosb_1x + C_2e^{a_2x}sinb_2x \ $$

Plugging in the roots into the equation we get

$$ \displaystyle \ y = C_1e^{-2x}cos3x + C_2e^{-2x}sin3x \ $$

Implementing initial conditions where: $$ \displaystyle \ y(0) = 1,   y'(0) = 0 \ $$

$$ \displaystyle \ y(0) = C_1e^{-2(0)}cos3(0) + C_2e^{-2(0)}sin3(0)= 1 \ $$

$$ \displaystyle \ y'(0)= -2C_1e^{-2x}cos3x - 3C_1e^{-2x}sin3x - 2C_2e^{-2x}sin3x + 3C_2e^{-2x}cos3x = 0 \ $$

$$ \displaystyle \ y'(0)= -2(1)e^{-2(0)}cos3(0) + 3C_2e^{-2(0)}cos3(0) = -2 + 3C_2 = 0; \ $$

So from our initial conditions the constants are determined to be

$$ \displaystyle \ C_1 = 1, C_2 = 2/3 \ $$

The final solution becomes

$$ \displaystyle \ y(x) = e^{-2x}cos3x + \frac{2}{3}e^{-2x}sin3x \ $$

Author
This problem was solved and uploaded by: Mike Wallace This problem was proofread by John North

Contribution Summary
Problem 2 was solved and Problems 1, 3, and 4 were proofread by David Herrick 15:56, 6 February 2012 (UTC)

Problem 8 was solved and Problem 5 was proofread by Josh House 23:41, 7 February 2012 (UTC)

Problem 6 was solved and Problem 8 was proofread by Radina Dikova 16:26, 7 February 2012 (UTC)

Problem 9 was solved and Problem 6 was proofread by Mike Wallace 23:35, 7 February 2012 (UTC)

Problems 3 and 5 were solved and Problems 2 and 7 were proofread by William Knapper 05:04, 8 February 2012 (UTC)

Problem 1 was solved by Derik Bell 15:44, 8 February 2012 (UTC)

Problems 4 and 7 were solved and Problem 9 was proofread by John North 15:47, 8 February 2012 (UTC)