User:Egm4313.s12.team1.stewart

Chris Stewart's Wiki Page

Team Page

Course Wiki

Reports
Report 1

Report 2

6.6
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec10-1.djvu Lecture 10 Pg. 9]


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

$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}=2e^{-2x}cos(3x) $$     (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 $$r(x)\!$$. Table 2.1
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

So the general form of the particular solution is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x] $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Task 1 asks to simplify they first term of the given ODE which is $$y_{p}''(x)\!$$. This is done by taking the second derivative of $$y_{p}(x)\!$$ and plugging it into equation 6.1
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

$$y_{p}''(x)\!$$ simplifies to:
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Now we must do the same for the second term $$4y_{p}'(x)\!$$.
 * {| style="width:100%" border="0"

$$  \displaystyle 4y'_{p}(x)=\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle 4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)] $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Add the second term to the first to continue building the original ODE (equation 6.1).
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}+\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Finally add the third term $$13y_{p}(x)\!$$ to the other two to have the lhs of the ODE 6.1
 * {| style="width:100%" border="0"

$$  \displaystyle 13y_{p}(x)=13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle y''_{p}(x)+4y'_{p}(x)+13y_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])+4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

This simplifies to the particular solution below:
 * {| style="width:100%" border="1"

$$  \displaystyle y_{p}(x)=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.11)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

R5
'''[https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec8b.djvu See R5.7 Lect. 8b pg. 11]:'''


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

$$\mathbf{v}=4\mathbf e_1+2\mathbf e_2=c_1\mathbf b_1+c_2\mathbf b_2$$
 * style="width:95%" |
 * style="width:95%" |
 * }

The oblique basis vectors b1,b2 are:
 * {| style="width:100%" border="0"

$$\mathbf b_1 =2\mathbf e_1 + 7\mathbf e_2$$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$\mathbf b_2 =1.5\mathbf e_1 + 3\mathbf e_2$$ 1. Find the components c1, c2 using the Gramian matrix. 2. Verify the result found above.
 * style="width:95%" |
 * style="width:95%" |
 * }

R4.3



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

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & -12 & 20 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & -15 & 30 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 2 & -18 & 42 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 &-21 & 56 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -24 & 72 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -27 & 90\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -30\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}\\ {A}_{7}\\ {A}_{8}\\ {A}_{9}\\ {A}_{10}\\ \end{pmatrix} = \begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720 \\ 1/5040\\ 1/40320\\ 1/362880\\ 1/3628800\\ \end{Bmatrix} $$     (3.30)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Statement
4.1 from lecture notes [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c-1.djvu R4.1 Lect. 7c pgs. 19-22]

Given the general form of polynomial excitation.


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

$$  \displaystyle y''+ay'+by=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The particular solution that satisfies:


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

$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x^{j} $$     (1.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

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


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

$$  \displaystyle y_{p}'(x)=\sum_{j=0}^{n-1}c_{j+1}(j+1) x^{j} $$     (1.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle y_{p}''(x)=\sum_{j=0}^{n-2}c_{j+2}(j+2)(j+1) x^{j} $$     (1.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

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


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

$$  \displaystyle \sum_{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

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

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


 * {| style="width:100%" border="1"

$$  \displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0} $$     (1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking $$j=1\!$$
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}} $$     (1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x $$     (1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The equation associated with $$d_{1}\!$$, coefficients of $$x\!$$
 * {| style="width:100%" border="1"

$$  \displaystyle 6c_{3}+2ac_{2}+bc_{1}=d_{1} $$     (1.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking $$j=2\!$$
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}} $$     (1.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2} $$     (1.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The equation associated with $$d_{2}\!$$, coefficients of $$x^{2}\!$$
 * {| style="width:100%" border="1"

$$  \displaystyle 12c_{4}+3ac_{3}+bc_{2}=d_{2} $$     (1.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking $$j=n-2\!$$
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The equation associated with $$d_{n-2}\!$$, coefficients of $$x^{n-2}\!$$
 * {| style="width:100%" border="1"

$$  \displaystyle c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}=d_{n-2} $$     (1.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking $$j=n-1\!$$
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{(n-1)+2}(n-1+2)(n-1+1)+ac_{(n-1)+1}(n-1+1)+bc_{(n-1)}]x^{n-1}+ac_{n}nx^{n-1}+bc_{n-1}x^{n-1}=d_{n-1}x^{n-1} $$     (1.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle [c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1} $$     (1.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The equation associated with $$d_{n-1}\!$$, coefficients of $$x^{n-1}\!$$
 * {| style="width:100%" border="1"

$$  \displaystyle c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}=d_{n-1} $$     (1.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking $$j=n\!$$.
 * {| style="width:100%" border="0"

$$  \displaystyle [c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+bc_{n}]x^{n}+bc_{n}x^{n}=d_{n}x^{^{n}} $$     (1.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The equation associated with $$d_{n}\!$$, coefficients of $$x^{n}\!$$
 * {| style="width:100%" border="1"

$$  \displaystyle c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+2bc_{n}=d_{n} $$     (1.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now set up the equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{A}\mathbf{c}=\mathbf{d} $$     (1.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \begin{bmatrix} Equation for d_{0}\\ Equation for d_{1}\\ Equation for d_{2}\\ ...\\ Equation for d_{n-2}\\ Equation for d_{n-1}\\ Equation for d_{n} \end{bmatrix}\begin{bmatrix} c_{0}\\ c_{1}\\ c_{2}\\ ...\\ c_{n-2}\\ c_{n-1}\\ c_{n} \end{bmatrix}=\begin{bmatrix} d_{0}\\ d_{1}\\ d_{2}\\ ...\\ d_{n-2}\\ d_{n-1}\\ d_{n} \end{bmatrix} $$     (1.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore the matrix $$\mathbf{A}\!$$ that satisfies the matrix equation is:
 * {| style="width:100%" border="1"

$$  \displaystyle \mathbf{A}=\begin{bmatrix} b& a& 2&  &  &  & \\ & b&  2a&  6&  &  & \\ & &  b&  3a&  12&  & \\ ...& ...&  ...&  ...&  ...&  ...& ...\\  &  &  &  &  b&  a(n-1)& n(n-1)\\ & &  &  &  &  2b& 2an\\ & &  &  &  &  & 2b \end{bmatrix} $$     (1.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle y''-3y'+2y=4x^{2}-6x^{5} $$     (4.1) is of the form
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x_{j} $$     (4.2) i.e.,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x_{j} $$     (4.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y''+ay'+by=r(x) $$     (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
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
According to the table the Homogeneous equation has two $$r(x)\!$$ values of the form $$kx^{n}(n=0,1,2...)$$. 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 $$r(x)\!$$.


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

$$  \displaystyle r_{1}(x)=4x^{2} $$     (4.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y_{p1}= 4x^{2}+4x^{1}+4 $$     (4.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which simplifies to:


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

$$  \displaystyle y_{p1}= \sum_{j=0}^{2}4x^{j} $$     (4.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle r_{2}(x)=-6x^{5} $$     (4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y_{p2}= -6x^{5}-6x^{4}-6x^{3}-6x^{2}-6x-6 $$     (4.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which simplifies to:


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

$$  \displaystyle y_{p1}= \sum_{j=0}^{5}-6x^{j} $$     (4.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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 $$y''-3y'+2y=4x^{2}-6x^{5}\!$$ is:


 * {| style="width:100%" border="1"

$$  \displaystyle y_{p}=\sum_{j=0}^{2}4x^{j}+ \sum_{j=0}^{5}-6x^{j} $$     (4.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Statement
From K 2011 pg. 59 #3,4

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


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

$$  \displaystyle y''+6y'+8.96y=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }


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

$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }

Solution
Given:

The Homogeneous equation is given:


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

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & -12 & 20 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & -15 & 30 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 2 & -18 & 42 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 &-21 & 56 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -24 & 72 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -27 & 90\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -30\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}\\ {A}_{7}\\ {A}_{8}\\ {A}_{9}\\ {A}_{10}\\ \end{pmatrix} = \begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720 \\ 1/5040\\ 1/40320\\ 1/362880\\ 1/3628800\\ \end{Bmatrix} $$     (3.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle \lambda ^2+6\lambda +8.96=0 $$     (3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle (\lambda ^2+6\lambda +9)+8.96=0+9 $$     (3.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle (\lambda +3)^2=.04 $$     (3.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The roots of the characteristic equation:


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

$$  \displaystyle \lambda =-3\pm .2 $$     (3.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y=C_{1}e^{\lambda x}+C_{2}e^{\lambda x} $$ (3.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


 * {| style="width:100%" border="1"

$$  \displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$     (3.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The Homogeneous equation is given:


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

$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$     (3.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle \lambda ^2+4\lambda +(\pi ^2+4)=0 $$     (3.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle (\lambda ^2+4\lambda +4)+(\pi ^2+4)=0+4 $$     (3.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle (\lambda +2)^2=4-(\pi ^2+4) $$     (3.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle (\lambda +2)^2=-\pi ^2 $$     (3.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle \lambda =-2\pm \pi i $$ (3.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle \lambda =a\pm bi $$ (3.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle y=C_{1}e^{ax}cos(bx)+C_{2}e^{ax}sin(bx) $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle y=C_{1}e^{-2x}cos(\pi x)+C_{2}e^{-2x}sin(\pi x) $$ (3.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

Edited By -