University of Florida/Egm4313/s12.teamboss/R3

Problem R3.1: Using the Method of Undetermined Coefficients and the Modification Rule to find a solution

Statement

Given the double root $\lambda =5$ and the excitation $r(x)=7e^{5x}-2x^2$, with the initial conditions $y(0)=4,y^{\prime }(0)=5$ find the solution $y(x)$.
Plot this solution and the solution to the same problem except with the excitation $r(x)=7e^{5x}$.

Solution

The solution $y(x)$ is composed of a general solution and a particular solution so that $y(x)=y_g(x)+y_p(x)$.
Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:

$(\lambda -5{)}^2={\lambda }^2-10\lambda +25=0$

(1.0)

Homogeneous solution:

$y^{″}-10y^{\prime }+25y=r(x)$

(1.1)

$y^{″}-10y^{\prime }+25y=7e^{5x}-2x^2$

(1.2)

First, by using the Modification Rule, we find that the general equation associated with the given double root is:

$y_g(x)=C_1{e}^{5x}+C_{2}x{e}^{5x}$

(1.3)

We need to find the particular solution to the excitation $r(x)=7e^{5x}-2x^2$. In analyzing the excitation, it is found that the particular solution looks like this:

$y_p(x)=C{x}^2{e}^{5x}-[K_2{x}^2+K_{1}x+K_0]$

(1.4)

Now, we need to find the values for the constants $C,K_2,K_1,K_0$, by taking the first and second derivatives of the particular solution:

$y_p(x)=C{x}^2{e}^{5x}-[K_2{x}^2+K_{1}x+K_0]$

(1.4)

$y_p(x{)}^{\prime }=2Cx{e}^{5x}+5C{x}^2{e}^{5x}-[2K_{2}x+K_1]$

(1.5)

$y_p(x{)}^{″}=2C{e}^{5x}+20Cx{e}^{5x}+25C{x}^2{e}^{5x}-[2K_2]$

(1.6)

Now, plug these into the homogeneous solution (1.2) and simplify:

$2C{e}^{5x}-2K_2+20K_{2}x+10K_1-25K_2{x}^2-25K_{1}x-25K_0=7e^{5x}-2x^2$

(1.7)

Comparing the coefficients for $e^{5x},x^2$ and $x$ allows us to solve for the unknown coefficients:

$e^{5x}:2C=7\to C=7/2$

(1.8)

$x^2:-25K_2=-2\to K_2=0.08$

(1.9)

$x:20K_2-25K_1=0\to K_1=0.064$

(1.10)

$-2K_2+10K_1-25K_0=0\to K_0=0.0192$

(1.11)

Plug into $y_p(x)$ (1.4):

$y_p(x)=\frac{7}{2}{x}^2{e}^{5x}-[0.08x^2+0.064x+0.0192]$

(1.12)

Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, $y_g(x)$, and then we will be able to solve for the final solution $y(x)$:

$y(x)=C_1{e}^{5x}+C_{2}x{e}^{5x}+\frac{7}{2}{x}^2{e}^{5x}-[0.08x^2+0.064x+0.0192]$

(1.13)

Take derivative of (1.13):

$y(x{)}^{\prime }=5C_1{e}^{5x}+C_2{e}^{5x}+5C_{2}x{e}^{5x}+7x{e}^{5x}+\frac{35}{2}{x}^2{e}^{5x}-[0.16x+0.064]$

(1.14)

Using the given initial conditions and equations (1.13) and (1.14):

$y(0)=4\to 4=C_1-0.0192\to C_1=4.0192$

(1.15)

$y^{\prime }(0)=5\to 5=5C_1+C_2-0.064\to C_2=-15.032$

(1.16)

Plug these constants back into the solution (1.13) to obtain the final solution:

$y(x)=4.0192e^{5x}-15.032x{e}^{5x}+\frac{7}{2}{x}^2{e}^{5x}-[0.08x^2+0.064x+0.0192]$

(1.17)

The solution to the same problem but with excitation $r(x)=7e^{5x}$ is:

$y(x)=4e^{5x}-25x{e}^{5x}+\frac{7}{2}{x}^2{e}^{5x}$

(1.18)

This is the plot for (1.17) and (1.18):

File:R3.1PLOT.jpg

Author

Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)

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Problem R3.2 Perturbation Method for Double Real Root

Statement

Perturbation method for double real root:
Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see Sec7 p. 7-5 ). Consider two distinct real roots of the form:

${\lambda }_1=\lambda ,{\lambda }_2=\lambda +ϵ$

1) Find the homogeneous L2-ODE-CC having the above distinct roots.
2) Show that the following is a homogeneous solution:
$\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}$

The fraction in (3) p.7-5, for small $ϵ$, is a finite difference formula that approximates the derivative

$\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}\approx \frac{d}{d\lambda }{e}^{\lambda x}$

In fact,
$\frac{d}{d\lambda }{e}^{\lambda x}=\underset{ϵ\to 0}{\lim }\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}$

3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)

4)Take the derivative of $e^{\lambda x}$ with respect to $\lambda$
5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters.
6)Numerical experiment: Compute (3) p.7-5 using at $\lambda =5$ and with $ϵ=0.001$, and compare to get the value obtained from the exact 2nd homogeneous solution.

Solution

In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:

$({\lambda }^{\prime }-\lambda )({\lambda }^{\prime }-\lambda -ϵ)=0$

(2.0)

$({\lambda }^{\prime }{)}^2-2{\lambda }^{\prime }\lambda +{\lambda }^2-{\lambda }^{\prime }ϵ+\lambda ϵ=({\lambda }^{\prime }{)}^2-(2\lambda +ϵ){\lambda }^{\prime }+({\lambda }^2+\lambda ϵ)=0$

(2.1)

Therefore the corresponding L2-ODE-CC is:

$y^{″}-(2\lambda +ϵ)y^{\prime }+({\lambda }^2+\lambda ϵ)y=0$

(2.2)

Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.

To show that the following is a homogeneous solution, the first and second derivatives must be taken:

$y(x)=\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}$

(2.3)

$y^{\prime }(x)=\frac{(\lambda +ϵ)e^{(\lambda +ϵ)x}-\lambda e^{\lambda x}}{ϵ}$

(2.4)

$y^{″}(x)=\frac{(\lambda +ϵ{)}^2{e}^{(\lambda +ϵ)x}-{\lambda }^2{e}^{\lambda x}}{ϵ}$

(2.5)

Using these values in 2.2 yields:

$\frac{(\lambda +ϵ{)}^2{e}^{(\lambda +ϵ)x}-{\lambda }^2{e}^{\lambda x}}{ϵ}-\frac{(2\lambda +ϵ)[(\lambda +ϵ)e^{(\lambda +ϵ)x}-\lambda e^{\lambda x}]}{ϵ}+\frac{({\lambda }^2+\lambda ϵ)[e^{(\lambda +ϵ)x}-e^{\lambda x}]}{ϵ}=0$

(2.6)

Rearranging and simplifying yields:

$\frac{[{\lambda }^2+2\lambda ϵ+{ϵ}^2-2{\lambda }^2-3\lambda ϵ-{ϵ}^2+{\lambda }^2+\lambda ϵ]e^{(\lambda +ϵ)x}+[-{\lambda }^2+2{\lambda }^2+\lambda ϵ-{\lambda }^2-\lambda ϵ]e^{\lambda x}}{ϵ}=0$

(2.7)

All of the terms in the above equation cancel to yield:

$\frac{0e^{(\lambda +ϵ)x}+0e^{\lambda x}}{ϵ}=0$

(2.8)

Using l'Hopital's rule:

$\underset{ϵ\to 0}{\lim }\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}=\underset{ϵ\to 0}{\lim }\frac{\frac{d}{dϵ}(e^{(\lambda +ϵ)x}-e^{\lambda x})}{\frac{d}{dϵ}(ϵ)}=\underset{ϵ\to 0}{\lim }\frac{\frac{d}{dϵ}(e^{(\lambda +ϵ)x})-0}{1}$

(2.9)

Simplifying further:

$\underset{ϵ\to 0}{\lim }\frac{d}{dϵ}(e^{(\lambda +ϵ)x})=\underset{ϵ\to 0}{\lim }x{e}^{(\lambda +ϵ)x}=x\underset{ϵ\to 0}{\lim }{e}^{(\lambda +ϵ)x}$

(2.10)

This ultimately yields:

$\underset{ϵ\to 0}{\lim }\frac{e^{(\lambda +ϵ)x}-e^{\lambda x}}{ϵ}=x{e}^{\lambda x}$

(2.11)

The following should also be taken into consideration:

$\frac{d}{d\lambda }(e^{\lambda x})=x{e}^{\lambda x}$

(2.12)

Clearly, the results of 2.11 and 2.12 are equivalent. This shows that $y(x)=x{e}^{\lambda x}$ is an appropriate solution to a homogeneous L2-ODE-CC having one double root.

To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test, $\lambda =5$ and $ϵ=0.001$

$y_{approx}(x)=\frac{e^{(5+0.001)x}-e^{5x}}{0.001}=1000(e^{0.001x}-1)e^{5x}$

(2.13)

$y_{exact}(x)=x{e}^{5x}$

(2.14)

If the above derivations are true, then the following approximation must also be true:

$1000(e^{0.001x}-1)\approx x$

(2.15)

Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x.

Figure 3.2-1

File:Fig3.2.1.png

Author

Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)

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Problem R3.3 Finding Solution of ODE with Polynomial Excitation

Statement

Find the complete solution for $y^{″}-3y^{\prime }+2y=4x^2$, with the initial conditions

$y(0)=1,y^{\prime }(0)=0$

Plot the solution $y(x).$

Solution

First we create the characteristic equation in standard form:

${\lambda }^2-3\lambda +2=0$

(3.0)

Then, by setting it equal to zero, we can find what $\lambda$ equals:

$(\lambda -2)(\lambda -1)=0$

(3.1)

$\lambda =2,\lambda =1$

(3.2)

Given two, distinct, real roots, the general solution looks like this:

$y_g(x)=C_1{e}^{2x}+C_2{e}^x$

(3.3)

By using the method of undetermined coefficients, the excitation $4x^2$ is analyzed to yield a particular solution:
In assessing a polynomial with a second power, the form of the particular solution will look like this:

$y_p(x)=A_2{x}^2+A_{1}x+A_0$

(3.5)

It's derivative would look like this:

${y_p}^{\prime }(x)=2A_{2}x+A_1$

(3.6)

And the second derivative to follow would then become:

${y_p}^{″}(x)=2A_2$

(3.7)

Based on the coefficients, the following system of equations exists:

$2A_2=4$

(3.8)

$-6A_2+A_1=0$

(3.9)

$2A_2-3A_1+A_0=0$

(3.10)

The results of this set of equations make the coefficients of A's:

$A_2=2$

$A_1=12$

$A_0=32$

The resulting particular equation looks like this:

$y_p(x)=2x^2+12x+32$

(3.11)

By adding the particular and general solutions, we get the complete solution:

$2x^2+12x+32+C_1{e}^{2x}+C_2{e}^x=y$

(3.12)

We consider the initial conditions by taking the first derivative of the complete solution:

$4x+12+2C_1{e}^{2x}+C_2{e}^x=y^{\prime }$

(3.13)

By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants $C_1,C_2$:

$y(0)=1=2*0^2+12*0+32+C_1{e}^{2*0}+C_2{e}^0=32+C_1+C_2=1$

(3.14)

$y^{\prime }(0)=0=4*0+12+2C_1{e}^{2*0}+C_2{e}^0=12+2C_1+C_2=0$

(3.15)

Solving the equations proves that $C_1=19,C_2=-50$:
The resulting complete solution with consideration for initial conditions then becomes:

$2x^2+12x+32+19e^{2x}-50e^x=y$

(3.12)

y plotted looks like this:

File:R33graph.jpg

Author

Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC)
Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)

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Problem R3.4 Solving for Particular Solution in Summation Form

Statement

From R3.4 in Sec 3 p. 7-11; Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for

$y^{″}-3y^{\prime }+2y=4x^2-6x^5$

(4.1)

is of the form

$y_p(x)=\sum _{j=0}^n{c}_j{x}_j$

(4.2)

i.e.,

$y_p(x)=\sum _{j=0}^5{c}_j{x}_j$

(4.3)

Basic Rule: Select $y_p(x)$ from the table and determine the coefficients by substituting $y_p(x)$ in

$y^{″}+a{y}^{\prime }+by=r(x)$

(4.4)

Sum Rule: If $r(x)$ is the sum of the terms in the 1st column of table 2.1 then $y_p(x)$ is the sum of the corresponding terms in the 2nd column of this table.

Table 2.1

File:Table2.1.JPG

Solution

According to the table the Homogeneous equation has two $r(x)$ values of the form $k{x}^n(n=0,1,2...)$. Using the Basic Rule this means that there is a particular solution of the form $K_n{x}^n+K_{n-1}{x}^{n-1}+...+K_{1}x+K_0$ for each $r(x)$.

$r_1(x)=4x^2\to r_1(x)=K{x}^n(n=2)$

(4.5)

So the particular solution for this $r_1(x)$ should be:

$y_{p1}=K_2{x}^2+K_1{x}^1+K_0$

(4.6)

Where $K=4$. Which simplifies to:

$y_{p1}=\sum _{j=0}^2{K}_j{x}^j$

(4.7)

For the second $r_2(x)$:

$r_2(x)=-6x^5\to r_2(x)=C{x}^n(n=5)$

(4.8)

So the particular solution for this $r_2(x)$ should be:

$y_{p2}=C_5{x}^5+C_4{x}^4+C_3{x}^3+C_2{x}^2+C_{1}x+C_0$

(4.9)

Where $C=4$. Which simplifies to:

$y_{p1}=\sum _{j=0}^5{C}_j{x}^j$

(4.10)

Using the sum rule:

$y_p=\sum _{j=0}^2{K}_j{x}^j+\sum _{j=0}^{5}C{x}^j$

(4.11)

Now using the Sum Rule which just states if there are two $r(x)$ values in any form on the left side of the table, then the particular solution is the sum of the solutions for $r(x)$ on the right side of the table.

So the particular solution for $y^{″}-3y^{\prime }+2y=4x^2-6x^5$ where $c=K+C$

$y_p=\sum _{j=0}^5{c}_j{x}^j$

(4.12)

Author

Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)

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Problem R3.5: Finding Particular Solutions by Combining Linear Coefficient Series Expansions

Statement

Given:

$y^{″}-3y^{\prime }+2y=4x^2-6x^5$

(5.0)

$y_p(x)=\sum _{j=0}^5{c}_j{x}^j=4x^2-6x^5$

(5.1)

Find:
-Coefficients of particular solution using series expansion and matrix back-substitution
-Solution $y(x)$ using initial conditions $y(0)=1,y^{\prime }(0)=0$
Plot solution.

Solution

Using Sec7b-1 p.7-13 Eq. (1):

$\sum _{j=0}^3{c}_{j+2}(j+2)(j+1)x^j-3\sum _{j=0}^4{c}_{j+1}(j+1)x^j+2\sum _{j=0}^5{c}_j{x}^j=4x^2-6x^5$

(5.2)

Which is a combined series expansion for the general coefficient series expansion of Sec7b-1 p.7-12 Eq. (4):

$\sum _{j=2}^5{c}_{j}j(j-1)x^{j-2}-3\sum _{j=1}^5{c}_{j}j{x}^{j-1}+2\sum _{j=0}^5{c}_j{x}^j=4x^2-6x^5$

(5.3)

We can find a simultaneous linear system of equations to solve for the coefficients of $y_p(x)$.

Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:

$\sum _{j=0}^3[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_j]x^j+(2c_4-15c_5)x^4+2c_5{x}^5=4x^2-6x^5$

(5.4)

Therefore providing a system of equations by equating coefficients from the summations to the $y_p(x)$:

$j=0:[c_2(2)(1)-3c_1(1)+2c_0]x^0=0x^0=[2c_0-3c_1+2c_2]x^0$

$j=1:[c_3(3)(2)-3c_2(2)+2c_1]x^1=0x^1=[2c_1-6c_2+6c_3]x^1$

$j=2:[c_4(4)(3)-3c_3(3)+2c_2]x^2=4x^2=[2c_2-9c_3+12c_4]x^2$

$j=3:[c_5(5)(4)-3c_4(4)+2c_3]x^3=0x^3=[2c_3-12c_4+20c_5]x^3$

$[2c_4-15c_5]x^4=0x^4$

$2c_5{x}^5=-6x^5$

(5.5)

Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:

$j=0:2c_0{x}^0=2c_0$

$j=1:2c_1{x}^1-3c_1(1)x^0=2c_{1}x-3c_1$

$j=2:2c_2{x}^2-3c_2(2)x^1+c_2(2)(1)x^0=2c_2{x}^2-6c_{2}x+3c_2$

$j=3:2c_3{x}^3-3c_3(3)x^2+c_3(3)(2)x^1=2c_3{x}^3-9c_3{x}^2+6c_{3}x$

$j=4:2c_4{x}^4-3c_4(4)x^3+c_4(4)(3)x^2=2c_4{x}^4-12c_4{x}^3+12c_4{x}^2$

$j=5:2c_5{x}^5-3c_5(5)x^4+c_5(5)(4)x^3=2c_5{x}^5-15c_5{x}^4+20c_5{x}^3$

(5.6)

By summing the terms in (5.6) and grouping like terms, then equating it to the $y_p(x)$, we find the coefficient equations are the same as Eqs. (5.5):

$[2c_0-3c_1+2c_2]x^0+[2c_1-6c_2+6c_3]x^1+[2c_2-9c_3+12c_4]x^2+[2c_3-12c_4+20c_5]x^3+[2c_4-15c_5]x^4+2c_5{x}^5=0x^0+0x^1+4x^2+0x^3+0x^4-6x^5$

(5.7)

The linear system of equations in Eqs. (5.5) can be placed in matrix form:

$\left[\begin{array}{cccccc}2& -3& 2& 0& 0& 0\\ 0& 2& -6& 6& 0& 0\\ 0& 0& 2& -9& 12& 0\\ 0& 0& 0& 2& -12& 20\\ 0& 0& 0& 0& 2& -15\\ 0& 0& 0& 0& 0& 2\end{array}\right]$ $\left(\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5\end{array}\right)=\left[\begin{array}{c}0\\ 0\\ 4\\ 0\\ 0\\ -6\end{array}\right]$

(5.8)

Solving this system of equations by back-substitution yields the values of the coefficients for the $y_p(x)$:

$c=\{-701.75,-691.5,-335.5,-105,-22.5,-3\}$

(5.9)

Substituting these coefficients into Eq. (5.1):

$y_p(x)=-701.75x^0-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5$

(5.10)

The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:

$y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5+C_1{e}^{2x}+C_2{e}^x$

(5.11)

Considering the initial value condition $y(0)=1$ in the general solution:

$y(0)=1=-701.75+C_1{e}^0+C_2{e}^0=-701.75+C_1+C_2$

(5.12)

Taking the first derivative of Eq. (5.11) to consider the initial value condition $y^{\prime }(0)=0$:

$y^{\prime }(x)=-691.5-671x-315x^2-90x^3-15x^4+2C_1{e}^{2x}+C_2{e}^x$

(5.13)

$y^{\prime }(0)=0=-691.5+2C_1+C_2$

(5.14)

Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

$C_1=-11.25,C_2=714$

(5.15)

This yields our final solution by substituting the coefficients into Eq. (5.11): Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

$y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5-11.25e^{2x}+714e^x$

(5.16)

A plot of the solution:

Figure 3.5-1

Author

Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)--
Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)

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Problem R3.6: Superimposing Particular Solutions

Statement

Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODEs-CC (see p.7-2b):

$y_{p^{″},1}-3y_{p^{\prime },1}+2y_{p,1}=r_1(x):=4x^2$

(6.0)

$y_{p^{″},2}-3y_{p^{\prime },2}+2y_{p,2}=r_2(x):=-6x^5$

(6.1)

The particular solution $y_{p,1}$ had been found in R3.3 p.7-11. Find the particular solution $y_{p,2}$, and then obtain the solution $y$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

Compare the result with that obtained in R3.5.

Solution

Beginning with equation (6.1), we find that a particular solution has the form

$y_{p,2}(x)=\sum _{j=0}^n{K}_j{x}^j$

(6.2)

where $n=5$. That is,

$y_{p,2}(x)=K_0+K_{1}x+K_2{x}^2+K_3{x}^3+K_4{x}^4+K_5{x}^5$

(6.3)

Differentiating twice:

$y_{p^{\prime },2}(x)=K_1+2K_{2}x+3K_3{x}^2+4K_4{x}^3+5K_5{x}^4$

(6.4)

$y_{p^{″},2}(x)=2K_2+6K_{3}x+12K_4{x}^2+20K_5{x}^3$

(6.5)

Substitute (6.3-5) into (6.1) to obtain

$(2K_2+6K_{3}x+12K_4{x}^2+20K_5{x}^3)-3(K_1+2K_{2}x+3K_3{x}^2+4K_4{x}^3+5K_5{x}^4)$ $+2(K_0+K_{1}x+K_2{x}^2+K_3{x}^3+K_4{x}^4+K_5{x}^5)=-6x^5$

(6.6)

Rearranging terms with respect to $x$ power:

$(2K_2-3K_1+2K_0)+x(6K_3-6K_2+2K_1)+x^2(12K_4-9K_3+2K_2)$ $+x^3(20K_5-12K_4+2K_3)+x^4(-15K_5+2K_4)+x^5(2K_5)=-6x^5$

(6.7)

In matrix form:

$\left[\begin{array}{cccccc}2& -3& 2& 0& 0& 0\\ 0& 2& -6& 6& 0& 0\\ 0& 0& 2& -9& 12& 0\\ 0& 0& 0& 2& -12& 20\\ 0& 0& 0& 0& 2& -15\\ 0& 0& 0& 0& 0& 2\end{array}\right]$ $\left(\begin{array}{c}{K}_0\\ K_1\\ K_2\\ K_3\\ K_4\\ K_5\end{array}\right)=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ -6\end{array}\right]$

(6.8)

Solving for $K$, using MATLAB, yields

$K=\{-708.75,-697.5,-337.5,-105,-22.5,-3\}$

(6.9)

which means that

$y_{p,2}=-708.75-697.5x-337.5x^2-105x^3-22.5x^4-3x^5$

(6.10)

From R3.3:

$y_{p,1}=32+12x+2x^2$

(6.11)

Summing for the final solution:

$y=C_1{y}_{p,1}+C_2{y}_{p,2}$

(6.12)

$y=C_1(32+12x+2x^2)$ $+C_2(-708.75-697.5x-337.5x^2-105x^3-22.5x^4-3x^5)$

(6.13)

For initial condition $y(0)=-5$

$y(0)=-5=C_1(32)+C_2(-708.75)$

(6.14)

Similarly, for initial condition $y^{\prime }(0)=2$

$y^{\prime }(0)=2=C_1(12)+C_2(-697.5)$

(6.15)

Equations (6.14) and (6.15) yield the following matrix equation:

$\left[\begin{array}{cc}32& -708.75\\ 12& -697.5\end{array}\right]$ $\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)=\left[\begin{array}{c}-5\\ 2\end{array}\right]$

(6.16)

Solving (6.16) in MATLAB yields

$C=\{-0.355,-0.009\}$

(6.17)

Therefore the combined solution of $y$ is

$y(x)=-0.355(32+12x+2x^2)$ $-0.009(-708.75-697.5x-337.5x^2-105x^3-22.5x^4-3x^5)$

(6.18)

Which can be simplified to Final Equation

$y(x)=-4.98125+2.0175x+2.3275x^2+0.945x^3+0.2025x^4+0.027x^5$

(6.19)

Author

Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)

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Problem R3.7 : Verifying series representation for method of undetermined coefficients

Statement

Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.
These equalities are:

$\sum _{j=2}^5{C}_{j}j(j-1)x^{{}^{j-2}}=\sum _{j=0}^3{C}_{j+2}(j+2)(j+1)x^{{}^j}$

(7.0)

$\sum _{j=1}^5{C}_{j}j{x}^{{}^{j-1}}=\sum _{j=0}^4{C}_{j+1}(j+1)x^{{}^j}$

(7.1)

Solution

This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:

$\sum _{k=0}^3{C}_{k+2}(k+2)(k+1)x^{{}^k}$

(7.3)

We can then represent the variable k with a j. This new summation looks like this:

$\sum _{j=0}^3{C}_{j+2}(j+2)(j+1)x^{{}^j}$

(7.4)

Expanding both sides, (7.0 and 7.4) yields the same result:

$C_5(5)(4)x^{{}^3}+C_4(4)(3)x^{{}^2}+C_3(3)(2)x^{{}^1}+C_2(2)(1)x^{{}^0}$

(7.5)

We follow the same process for equality (7.1):
This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:

$\sum _{k=0}^4{C}_{k+1}(k+1)x^{{}^k}$

(7.6)

We can then represent the variable k with a j. This new summation looks like this:

$\sum _{j=0}^4{C}_{j+1}(j+1)x^{{}^j}$

(7.7)

Expanding both sides, (7.1 and 7.7)yields the same result:

$C_5(5)x^{{}^4}+C_4(4)x^{{}^3}+C_3(3)x^{{}^2}+C_2(2)x^{{}^1}+C_1$

(7.8)

Author

Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC)
Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)

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Problem R3.8: Finding general solutions to Non-homogeneous Linear ODEs using the Method of Undetermined Coefficients and the Basic Rule

Statement

Find a general solution for the following two problems:

Kreyszig 2011 p.84 Problem 5

Given

Homogeneous solution: $y^{″}+4y^{\prime }+4y=e^{-x}cosx$

Solution

The solution $y(x)$ is composed of a general solution and a particular solution so that $y(x)=y_g(x)+y_p(x)$.
First, we will find the general solution, $y_g(x)$, by finding the roots of the characteristic equation:

${\lambda }^2+4\lambda +4\lambda =0\to (\lambda +2{)}^2$

(8.0)

Which means that the characteristic equation has a double root of $\lambda =-2$.
Based on the double root, then, the general solution is:

$y_g(x)=(C_1+C_{2}x)e^{-\frac{ax}{2}}\to y_g(x)=(C_1+C_{2}x)e^{-2x}$

(8.1)

Next, we need to find the particular solution, and based on analysis of the excitation, $r(x)=e^{-x}cosx$, we find that the particular solution is the following:

$y_p(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\to y_p(x)=e^{-x}Kcosx+e^{-x}Msinx$

(8.2)

Now, we need to find the values for the constants $K,M$, by taking the first and second derivatives of the particular solution (8.2):

$y_p(x)=e^{-x}Kcosx+e^{-x}Msinx$

(8.2)

$y_p(x{)}^{\prime }=-e^{-x}Ksinx-e^{-x}Kcosx+e^{-x}Mcosx-e^{-x}Msinx$

(8.3)

$y_p(x{)}^{″}=-e^{-x}Kcosx+e^{-x}Ksinx+e^{-x}Ksinx+e^{-x}Kcosx-e^{-x}Msinx-e^{-x}Mcosx-e^{-x}Mcosx+e^{-x}Msinx$

(8.4)

Now, plug these into the homogeneous solution and simplify:

$-e^{-x}(Kcosx-Ksinx-Ksinx-Kcosx+Msinx+Mcosx+Mcosx-Msinx)+4[-e^{-x}(Ksinx+Kcosx-Mcosx+Msinx)]+4[e^{-x}(Kcosx+Msinx)]=e^{-x}cosx$

(8.5)

$-e^{-x}[-2Ksinx+2Mcosx]-e^{-x}[4Ksinx+4Kcosx-4Mcosx+4Msinx]+e^{-x}[4Kcosx+4Msinx]=e^{-x}cosx$

(8.6)

Pull $e^{-x}$ out of equation (8.6) and simplify:

$e^{-x}[4Kcosx+4Msinx-4Ksinx-4Kcosx+4Mcosx-4Msinx+2Ksinx-2Mcosx]=e^{-x}cosx$

(8.7)

$e^{-x}[-2Ksinx+2Mcosx]=e^{-x}cosx$

(8.8)

Now, solve for the unknown coefficients:

$2M=1\to M=\frac{1}{2}$

(8.9)

$-2K=0\to K=0$

(8.10)

Plug into $y_p(x)$ (8.2):

$y_p(x)=\frac{1}{2}{e}^{-x}sinx$

(8.11)

Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution, $y(x)$

$y(x)=(C_1+C_{2}x)e^{-2x}+\frac{1}{2}{e}^{-x}sinx$

(8.12)

Kreyszig 2011 p.84 Problem 6

Given

Homogeneous solution: $y^{″}+y^{\prime }({\pi }^2+\frac{1}{4})=e^{-\frac{x}{2}}sin\pi x$

Solution

The solution $y(x)$ is composed of a general solution and a particular solution so that $y(x)=y_g(x)+y_p(x)$.
First, we will find the general solution, $y_g(x)$, by finding the roots of the characteristic equation, using the quadratic equation:

${\lambda }^2+\lambda +({\pi }^2+\frac{1}{4})=0\to \lambda =\frac{-1\pm \sqrt{1-4(1)({\pi }^2+\frac{1}{4})}}{2}$

(8.13)

$\lambda =-\frac{1}{2}+i\pi$

(8.14)

The roots of the characteristic equation are complex conjugates, meaning that the general solution, $y_g(x)$, is the following:

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

(8.15)

Next, we need to find the particular solution, and based on analysis of the excitation, $r(x)=e^{-\frac{x}{2}}sin\pi x$, we find that the particular solution is the following:

$y_p(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\to y_p(x)=e^{-\frac{x}{2}}Kcos\pi x+e^{-\frac{x}{2}}Msin\pi x$

(8.16)

Now, we need to find the values for the constants $K,M$, by taking the first and second derivatives of the particular solution (8.16):

$y_p(x)=e^{\alpha x}(Kcos\omega x+Msin\omega x)\to y_p(x)=e^{-\frac{x}{2}}Kcos\pi x+e^{-\frac{x}{2}}Msin\pi x$

(8.16)

$y_p(x{)}^{\prime }=-\pi e^{\frac{-x}{2}}Ksin\pi x-\frac{e^{\frac{-x}{2}}}{2}Kcos\pi x+\pi e^{\frac{-x}{2}}Mcos\pi x+\frac{e^{\frac{-x}{2}}}{2}Msin\pi x$

(8.17)

$y_p(x{)}^{″}=-{\pi }^2{e}^{\frac{-x}{2}}Kcos\pi x+\pi \frac{e^{\frac{-x}{2}}}{2}Ksin\pi x+\pi \frac{e^{\frac{-x}{2}}}{2}Ksin\pi x+\frac{e^{\frac{-x}{2}}}{4}Kcos\pi x-{\pi }^2{e}^{\frac{-x}{2}}Msin\pi x-\pi \frac{e^{\frac{-x}{2}}}{2}Mcos\pi x-\pi \frac{e^{\frac{-x}{2}}}{2}Mcos\pi x+\frac{e^{\frac{-x}{2}}}{4}Msin\pi x$

(8.18)

Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, $M,K$ are both equal to $0$. This means that the final solution is equal to just the general solution:

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

(8.19)

Author

Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)

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Problem R3.9 Finding general solutions to Non-homogeneous Linear ODEs

Statement

K 2011 page 85 problems 13 and 14
Problem 13: Find the complete solution for $8y^{″}-6y^{\prime }+y=6\cosh x$, with the initial conditions
$y(0)=0.2,y^{\prime }(0)=0.05$

Problem 14: Find the complete solution for $y^{″}+4y^{\prime }+4y=e^{-2x}\sin 2x$, with the initial conditions
$y(0)=1,y^{\prime }(0)=-1.5$

Solution

Problem 13:
In order to solve the equation we must use the definition $y=y_h+y_p$. We start with the given equation:

$8y^{″}-6y^{\prime }+y=6\cosh x$

(9.1)

From this we get the homogeneous characteristic equation:

$8{\lambda }^2-6\lambda +1=0x$

(9.2)

Factoring the characteristic equation gives us:

$(4\lambda -1)(2\lambda -1)=0$

(9.3)

This give us the roots:

$\lambda =\frac{1}{4},\frac{1}{2}$

(9.4)

Plugging in the roots gives us the general homogeneous solution:

$y_h=c_1{e}^{\frac{1}{4}x}+c_2{e}^{\frac{1}{2}x}$

(9.5)

Now we must solve for the particular solution. So far we know:

$\cosh x=\frac{e^x+e^{-x}}{2}$

(9.6)

Now using the sum rule, $y_p=y_{p1}+y_{p2}$, gives us:

$y_p=C{e}^x+K{e}^{-x}$

(9.7)

Now we must take the first and second derivatives of equation 9.7:

$y_{p^{\prime }}=C{e}^x-K{e}^{-x}$

(9.8)

$y_{p^{″}}=C{e}^x+K{e}^{-x}$

(9.9)

Substituting 9.8 and 9.9 back into our original equation gives us:

$8(C{e}^x+K{e}^{-x})-6(C{e}^x-K{e}^{-x})+C{e}^x+K{e}^{-x}=6(\frac{e^x+e^{-x}}{2})$

(9.10)

Simplifying equation 9.10 give us:

$3C{e}^x+15K{e}^{-x}=3(e^x+e^{-x})$

(9.11)

From this we can deduce that:

$3C=3,15K=3$

(9.12)

Thus we get that:

$C=1,K=\frac{1}{5}$

(9.13)

We now have our particular equation:

$y_p=e^x+\frac{1}{5}{e}^{-x}$

(9.14)

Thus:

$y=c_1{e}^{\frac{1}{4}x}+c_2{e}^{\frac{1}{2}x}+e^x+\frac{1}{5}{e}^{-x}$

(9.15)

Now we can solve for $c_1$ and $c_2$ using the given initial values.
We are given that at $y(0)=0.2$, we can now plug this into our equation to get:

$0.2=c_1{e}^{\frac{1}{4}(0)}+c_2{e}^{\frac{1}{2}(0)}+e^{(0)}+\frac{1}{5}{e}^{-(0)}$

(9.16)

This simplifies to:

$0.2=c_1+c_2+1+\frac{1}{5}$

(9.17)

Giving us that:

$c_1+c_2=-1$

(9.18)

For our second condition, $y^{\prime }(0)=0.05$, we must we must take the derivative of our general solution:

$y^{\prime }=\frac{1}{4}{c}_1{e}^{\frac{1}{4}x}+\frac{1}{2}{c}_2{e}^{\frac{1}{2}x}+e^x-\frac{1}{5}{e}^{-x}$

(9.19)

At our initial condition gives us:

$0.05=\frac{1}{4}{c}_1+\frac{1}{2}{c}_2+1-\frac{1}{5}$

(9.20)

Thus:

$c_1+2c_2=-3$

(9.21)

Combining 9.18 and 9.21 gives us that:

$c_1=1,c_2=-2$

(9.22)

Plugging these coefficients in gives us our final solution:

$y=e^{\frac{1}{4}x}-2e^{\frac{1}{2}x}+e^x+\frac{1}{5}{e}^{-x}$

(9.23)

Problem 14:
In order to solve the equation we must use the definition $y=y_h+y_p$. We start with the given equation:

$y^{″}+4y^{\prime }+4y=e^{-2x}\sin 2x$

(9.24)

From this we get the homogeneous characteristic equation:

${\lambda }^2+4\lambda +4=0$

(9.25)

Factoring the characteristic equation gives us:

$(\lambda +2)(\lambda +2)=0$

(9.26)

This give us the double root:

$\lambda =-2$

(9.27)

Plugging in the roots gives us the general homogeneous solution:

$y_h=c_1{e}^{-2x}+c_{2}x{e}^{-2x}$

(9.28)

Now we must solve for the particular solution. So far we know:

$e^{-2x}\sin 2x=Cx{e}^{-2x}\cos 2x+Kx{e}^{-2x}\sin 2x$

(9.29)

Now we must take the first and second derivatives of equation 9.29:

$y_{p^{\prime }}=-2e^{-2x}(Cx\cos 2x+Kx\sin 2x)+e^{-2x}(C\cos 2x-2Cx\sin 2x+K\sin 2x+2Kx\cos 2x)$

(9.30)

$y_{p^{″}}=(-4C+4K)e^{-2x}\cos 2x+(-4k-4C)e^{-2x}\sin 2x$

(9.31)

Substituting 9.30 and 9.31 back into our original equation gives us:

$(-3C+4K)e^{-2x}\cos 2x+(-3k-4C)e^{-2x}\sin 2x=e^{-2x}\sin 2x$

(9.32)

From this we can deduce that:

$-3C+4K=0,-3K-4C=1$

(9.33)

Thus we get that:

$C=\frac{-4}{25},K=\frac{-3}{25}$

(9.34)

We now have our particular equation:

$y_p=e^{-2x}(\frac{-4}{25}x\cos 2x-\frac{3}{25}x\sin 2x)$

(9.35)

Thus:

$y=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+e^{-2x}(\frac{-4}{25}x\cos 2x-\frac{3}{25}x\sin 2x)$

(9.36)

Now we can solve for $c_1$ and $c_2$ using the given initial values.
We are given that at $y(0)=1$, we can now plug this into our equation to get:

$y(0)=(c_1+c_{2}x)e^{-2(0)}+0=1$

(9.37)

Giving us that:

$c_1=1$

(9.38)

For our second condition, $y^{\prime }(0)=-1.5$, we must we must take the derivative of our general solution:

$y^{\prime }=-2c_1{e}^{-2x}+c_2(-2x{e}^{-2x}+e^{-2x})+\frac{-4}{25}{e}^{-2x}\cos 2x-\frac{3}{25}{e}^{-2x}\sin 2x+\frac{2}{25}{e}^{-2x}x\cos 2x+\frac{14}{25}{e}^{-2x}x\sin 2x$

(9.39)

At our initial condition we have:

$y^{\prime }(0)=-2+c_2-\frac{4}{25}-0+0+0=0$

(9.40)

Thus:

$c_2=\frac{54}{25}$

(9.41)

Plugging these coefficients in gives us our final solution:

$y=1e^{-2x}+\frac{54}{25}x{e}^{-2x}+e^{-2x}(\frac{-4}{25}x\cos 2x-\frac{3}{25}x\sin 2x)$

(9.42)

Author

Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)

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Contributing Members

Team Contribution Table
Problem Number	Lecture	Assigned To	Solved By	Typed By	Proofread By
3.1	R3.1 in Sec 3 p. 7-5	Wyatt Ling	Wyatt Ling	Wyatt Ling	George Armanious
3.2	R3.2 in Sec 3 p. 7-5	George Armanious	George Armanious	George Armanious	Jesse Durrance
3.3	R3.3 in Sec 3 p. 7-11	Emotion Silvestri	Emotion Silvestri	Emotion Silvestri	George Armanious
3.4	R3.4 in Sec 3 p. 7-11	Chris Stewart	Chris Stewart	Chris Stewart	Emotion Silvestri
3.5	R3.5 in Sec 7 p. 7-17 (Sakai)	Jesse Durrance	Jesse Durrance	Jesse Durrance	Emotion Silvestri
3.6	R3.6 in Sec 7 p. 7-17 (Sakai)	Eric Essenwein	Eric Essenwein	Eric Essenwein	Emotion Silvestri
3.7	R3.7 in Sec 7 p. 7-17 (Sakai)	Emotion	Emotion	Emotion	Wyatt Ling
3.8		Wyatt Ling	Wyatt Ling	Wyatt Ling	Steven Rosenberg
3.9		Steven Rosenberg	Steven Rosenberg	Steven Rosenberg	Eric Essenwein