# User:Egm4313.s12.team1.stewart

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## 6.6

From Lecture 10 Pg. 9

 ${\displaystyle \displaystyle y''_{p}+4y'_{p}+13y_{p}=2e^{-2x}cos(3x)}$ (6.1)

To find the particular solution of an ODE we can use the method of undetermined coefficients. In report 3 we used the table 2.1 below to find the standard form of the particular solution given the form of the excitation ${\displaystyle r(x)\!}$.

Table 2.1

So the general form of the particular solution is:

 ${\displaystyle \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x]}$ (6.2)

Task 1 asks to simplify they first term of the given ODE which is ${\displaystyle y_{p}''(x)\!}$. This is done by taking the second derivative of ${\displaystyle y_{p}(x)\!}$ and plugging it into equation 6.1

 ${\displaystyle \displaystyle y''_{p}(x)={\frac {\partial }{\partial x}}{\frac {\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}}}$ (6.3)

${\displaystyle y_{p}''(x)\!}$ simplifies to:

 ${\displaystyle \displaystyle y''_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])}$ (6.4)

Now we must do the same for the second term ${\displaystyle 4y_{p}'(x)\!}$.

 ${\displaystyle \displaystyle 4y'_{p}(x)={\frac {\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}}}$ (6.5)
 ${\displaystyle \displaystyle 4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]}$ (6.6)

Add the second term to the first to continue building the original ODE (equation 6.1).

 ${\displaystyle \displaystyle y''_{p}(x)+4y'_{p}(x)={\frac {\partial }{\partial x}}{\frac {\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}}+{\frac {\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}}}$ (6.7)
 ${\displaystyle \displaystyle y''_{p}(x)+4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])}$ (6.8)

Finally add the third term ${\displaystyle 13y_{p}(x)\!}$ to the other two to have the lhs of the ODE 6.1

 ${\displaystyle \displaystyle 13y_{p}(x)=13xe^{-2x}[Mcos3x+Nsin3x]}$ (6.9)
 ${\displaystyle \displaystyle y''_{p}(x)+4y'_{p}(x)+13y_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])+4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+13xe^{-2x}[Mcos3x+Nsin3x]}$ (6.10)

This simplifies to the particular solution below:

 ${\displaystyle \displaystyle y_{p}(x)=6e^{-2x}[Ncos(3x)-Msin(3x)]}$ (6.11)

### R5

 ${\displaystyle \mathbf {v} =4\mathbf {e} _{1}+2\mathbf {e} _{2}=c_{1}\mathbf {b} _{1}+c_{2}\mathbf {b} _{2}}$

The oblique basis vectors b1,b2 are:

 ${\displaystyle \mathbf {b} _{1}=2\mathbf {e} _{1}+7\mathbf {e} _{2}}$
 ${\displaystyle \mathbf {b} _{2}=1.5\mathbf {e} _{1}+3\mathbf {e} _{2}}$

1. Find the components c1, c2 using the Gramian matrix.
2. Verify the result found above.

### R4.3

 ${\displaystyle \displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0&0&0&0\\0&2&-6&6&0&0&0&0&0&0&0\\0&0&2&-9&12&0&0&0&0&0&0\\0&0&0&2&-12&20&0&0&0&0&0\\0&0&0&0&2&-15&30&0&0&0&0\\0&0&0&0&0&2&-18&42&0&0&0\\0&0&0&0&0&0&2&-21&56&0&0\\0&0&0&0&0&0&0&2&-24&72&0\\0&0&0&0&0&0&0&0&2&-27&90\\0&0&0&0&0&0&0&0&0&2&-30\\0&0&0&0&0&0&0&0&0&0&2\\\end{bmatrix}}\cdot {\begin{pmatrix}{A}_{0}\\{A}_{1}\\{A}_{2}\\{A}_{3}\\{A}_{4}\\{A}_{5}\\{A}_{6}\\{A}_{7}\\{A}_{8}\\{A}_{9}\\{A}_{10}\\\end{pmatrix}}={\begin{Bmatrix}0\\1\\-1/2\\1/6\\-1/24\\1/120\\-1/720\\1/5040\\1/40320\\1/362880\\1/3628800\\\end{Bmatrix}}}$ (3.30)

### Statement

4.1 from lecture notes R4.1 Lect. 7c pgs. 19-22

Given the general form of polynomial excitation.

 ${\displaystyle \displaystyle y''+ay'+by=\sum _{j=0}^{n}d_{j}x^{j}}$ (1.1)

The particular solution that satisfies:

 ${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x^{j}}$ (1.2)

The first and second derivative of the particular solution that solves the original polynomial excitation equation.

 ${\displaystyle \displaystyle y_{p}'(x)=\sum _{j=0}^{n-1}c_{j+1}(j+1)x^{j}}$ (1.3)
 ${\displaystyle \displaystyle y_{p}''(x)=\sum _{j=0}^{n-2}c_{j+2}(j+2)(j+1)x^{j}}$ (1.4)

The particular solutions are put into the polynomial excitation equation to give the general summation form:

 ${\displaystyle \displaystyle \sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum _{j=0}^{n}d_{j}x^{j}}$ (1.5)

Obtain the equations associated with ${\displaystyle d_{1}\!}$, coefficients of ${\displaystyle x\!}$; ${\displaystyle d_{2}\!}$, coefficients of ${\displaystyle x^{2}\!}$; ${\displaystyle d_{n-2}\!}$, coefficients of ${\displaystyle x^{n-2}\!}$; ${\displaystyle d_{n-1}\!}$, coefficients of ${\displaystyle x^{n-1}\!}$; ${\displaystyle d_{n}\!}$, coefficients of ${\displaystyle x^{n}\!}$. Five total equations for coefficients.

Also set up the matrix ${\displaystyle \mathbf {A} \!}$ that satisfies ${\displaystyle \mathbf {A} \mathbf {c} =\mathbf {d} \!}$.

### Solution

The given equation associated with ${\displaystyle d_{0}\!}$ taking ${\displaystyle j=0\!}$

 ${\displaystyle \displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0}}$ (1.6)

Taking ${\displaystyle j=1\!}$

 ${\displaystyle \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}}}$ (1.7)

 ${\displaystyle \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x}$ (1.8)

The equation associated with ${\displaystyle d_{1}\!}$ , coefficients of ${\displaystyle x\!}$

 ${\displaystyle \displaystyle 6c_{3}+2ac_{2}+bc_{1}=d_{1}}$ (1.9)

Taking ${\displaystyle j=2\!}$

 ${\displaystyle \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}}}$ (1.10)
 ${\displaystyle \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2}}$ (1.11)

The equation associated with ${\displaystyle d_{2}\!}$ , coefficients of ${\displaystyle x^{2}\!}$

 ${\displaystyle \displaystyle 12c_{4}+3ac_{3}+bc_{2}=d_{2}}$ (1.12)

Taking ${\displaystyle j=n-2\!}$

 ${\displaystyle \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}$ (1.13)
 ${\displaystyle \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2}}$ (1.14)

The equation associated with ${\displaystyle d_{n-2}\!}$ , coefficients of ${\displaystyle x^{n-2}\!}$

 ${\displaystyle \displaystyle c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}=d_{n-2}}$ (1.15)

Taking ${\displaystyle j=n-1\!}$

 ${\displaystyle \displaystyle [c_{(n-1)+2}(n-1+2)(n-1+1)+ac_{(n-1)+1}(n-1+1)+bc_{(n-1)}]x^{n-1}+ac_{n}nx^{n-1}+bc_{n-1}x^{n-1}=d_{n-1}x^{n-1}}$ (1.16)
 ${\displaystyle \displaystyle [c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1}}$ (1.17)

The equation associated with ${\displaystyle d_{n-1}\!}$ , coefficients of ${\displaystyle x^{n-1}\!}$

 ${\displaystyle \displaystyle c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}=d_{n-1}}$ (1.18)

Taking ${\displaystyle j=n\!}$.

 ${\displaystyle \displaystyle [c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+bc_{n}]x^{n}+bc_{n}x^{n}=d_{n}x^{^{n}}}$ (1.19)

The equation associated with ${\displaystyle d_{n}\!}$ , coefficients of ${\displaystyle x^{n}\!}$

 ${\displaystyle \displaystyle c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+2bc_{n}=d_{n}}$ (1.20)

Now set up the equation:

 ${\displaystyle \displaystyle \mathbf {A} \mathbf {c} =\mathbf {d} }$ (1.21)
 ${\displaystyle \displaystyle {\begin{bmatrix}Equationford_{0}\\Equationford_{1}\\Equationford_{2}\\...\\Equationford_{n-2}\\Equationford_{n-1}\\Equationford_{n}\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\...\\c_{n-2}\\c_{n-1}\\c_{n}\end{bmatrix}}={\begin{bmatrix}d_{0}\\d_{1}\\d_{2}\\...\\d_{n-2}\\d_{n-1}\\d_{n}\end{bmatrix}}}$ (1.22)

Therefore the matrix ${\displaystyle \mathbf {A} \!}$ that satisfies the matrix equation is:

 ${\displaystyle \displaystyle \mathbf {A} ={\begin{bmatrix}b&a&2&&&&\\&b&2a&6&&&\\&&b&3a&12&&\\...&...&...&...&...&...&...\\&&&&b&a(n-1)&n(n-1)\\&&&&&2b&2an\\&&&&&&2b\end{bmatrix}}}$ (1.23)

### Author

Solved and Typed By - Chris Stewart User:Egm4313.s12.team1.stewart -- 21:12, 11 March 2012 (UTC)
Reviewed By -

### R3

Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for

 ${\displaystyle \displaystyle y''-3y'+2y=4x^{2}-6x^{5}}$ (4.1)

is of the form

 ${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x_{j}}$ (4.2)

i.e.,

 ${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x_{j}}$ (4.3)

Basic Rule: Select ${\displaystyle y_{p}(x)\!}$ from the table and determine the coefficients by substituting ${\displaystyle y_{p}(x)\!}$ in

 ${\displaystyle \displaystyle y''+ay'+by=r(x)}$ (4.4)

Sum Rule: If ${\displaystyle r(x)\!}$ is the sum of the terms in the 1st column of table 2.1 then ${\displaystyle y_{p}(x)\!}$ is the sum of the corresponding terms in the 2nd column of this table.

Table 2.1

### Solution

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

 ${\displaystyle \displaystyle r_{1}(x)=4x^{2}}$ (4.5)

So the particular solution for this ${\displaystyle r_{1}(x)\!}$ should be:

 ${\displaystyle \displaystyle y_{p1}=4x^{2}+4x^{1}+4}$ (4.6)

Which simplifies to:

 ${\displaystyle \displaystyle y_{p1}=\sum _{j=0}^{2}4x^{j}}$ (4.7)

For the second ${\displaystyle r_{2}(x)\!}$:

 ${\displaystyle \displaystyle r_{2}(x)=-6x^{5}}$ (4.8)

So the particular solution for this ${\displaystyle r_{2}(x)\!}$ should be:

 ${\displaystyle \displaystyle y_{p2}=-6x^{5}-6x^{4}-6x^{3}-6x^{2}-6x-6}$ (4.9)

Which simplifies to:

 ${\displaystyle \displaystyle y_{p1}=\sum _{j=0}^{5}-6x^{j}}$ (4.10)

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

So the particular solution for ${\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}$ is:

 ${\displaystyle \displaystyle y_{p}=\sum _{j=0}^{2}4x^{j}+\sum _{j=0}^{5}-6x^{j}}$ (4.11)

## Report 2 Progress

### Statement

From K 2011 pg. 59 #3,4

Find the General Solution to the two ODEs below. (See Sec.5 Pg.5-6 notes)

 ${\displaystyle \displaystyle y''+6y'+8.96y=0}$
 ${\displaystyle \displaystyle y''+4y'+(\pi ^{2}+4)y=0}$

### Solution

Given:

The Homogeneous equation is given:

 ${\displaystyle \displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0&0&0&0\\0&2&-6&6&0&0&0&0&0&0&0\\0&0&2&-9&12&0&0&0&0&0&0\\0&0&0&2&-12&20&0&0&0&0&0\\0&0&0&0&2&-15&30&0&0&0&0\\0&0&0&0&0&2&-18&42&0&0&0\\0&0&0&0&0&0&2&-21&56&0&0\\0&0&0&0&0&0&0&2&-24&72&0\\0&0&0&0&0&0&0&0&2&-27&90\\0&0&0&0&0&0&0&0&0&2&-30\\0&0&0&0&0&0&0&0&0&0&2\\\end{bmatrix}}\cdot {\begin{pmatrix}{A}_{0}\\{A}_{1}\\{A}_{2}\\{A}_{3}\\{A}_{4}\\{A}_{5}\\{A}_{6}\\{A}_{7}\\{A}_{8}\\{A}_{9}\\{A}_{10}\\\end{pmatrix}}={\begin{Bmatrix}0\\1\\-1/2\\1/6\\-1/24\\1/120\\-1/720\\1/5040\\1/40320\\1/362880\\1/3628800\\\end{Bmatrix}}}$ (3.0)

The first step is to change the equation into the characteristic equation.

 ${\displaystyle \displaystyle \lambda ^{2}+6\lambda +8.96=0}$ (3.1)

Next use either completing the square or the quadratic formula to solve for ${\displaystyle \lambda \!}$ (the roots of the characteristic equation). In this case completing the square is used.

 ${\displaystyle \displaystyle (\lambda ^{2}+6\lambda +9)+8.96=0+9}$ (3.3)
 ${\displaystyle \displaystyle (\lambda +3)^{2}=.04}$ (3.4)

The roots of the characteristic equation:

 ${\displaystyle \displaystyle \lambda =-3\pm .2}$ (3.5)

Now that the roots are known we can use the Homogeneous solutions in this equation:

 ${\displaystyle \displaystyle y=C_{1}e^{\lambda x}+C_{2}e^{\lambda x}}$ (3.6)

Finally plug in the known roots and this is the final answer and general solution:

 ${\displaystyle \displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x}}$ (3.7)

The Homogeneous equation is given:

 ${\displaystyle \displaystyle y''+4y'+(\pi ^{2}+4)y=0}$ (3.8)

Again, the first step is to convert to the characteristic equation.

 ${\displaystyle \displaystyle \lambda ^{2}+4\lambda +(\pi ^{2}+4)=0}$ (3.9)

In order to solve for the roots (${\displaystyle \lambda \!}$) either use the quadratic formula or completing the square. Once again in this case completing the was used to solve for ${\displaystyle \lambda \!}$.

 ${\displaystyle \displaystyle (\lambda ^{2}+4\lambda +4)+(\pi ^{2}+4)=0+4}$ (3.10)

 ${\displaystyle \displaystyle (\lambda +2)^{2}=4-(\pi ^{2}+4)}$ (3.11)

 ${\displaystyle \displaystyle (\lambda +2)^{2}=-\pi ^{2}}$ (3.12)

The root takes the square root of a negative one, therefore the complex number ${\displaystyle i\!}$ is in the solution. The roots of the characteristic equation is shown:

 ${\displaystyle \displaystyle \lambda =-2\pm \pi i}$ (3.13)

For complex roots the Homogeneous solution is different and the root must broken into components and plugged into the complex Homogeneous solution equation below.

 ${\displaystyle \displaystyle \lambda =a\pm bi}$ (3.14)

 ${\displaystyle \displaystyle y=C_{1}e^{ax}cos(bx)+C_{2}e^{ax}sin(bx)}$ (3.15)

When plugging in ${\displaystyle \lambda \!}$ the final answer and general solution is:

 ${\displaystyle \displaystyle y=C_{1}e^{-2x}cos(\pi x)+C_{2}e^{-2x}sin(\pi x)}$ (3.16)

### Author

Solved and typed by - Egm4313.s12.team1.stewart 02:03, 4 February 2012 (UTC)
Reviewed By -
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