University of Florida/Egm4313/s12.teamboss/R5

Statement
Find the radius of convergence for:


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

$$\displaystyle r(x)=\sum_{k=0}^{\infty } (k + 1) k x^k $$     (1.0)
 * 1)
 * 1)
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle r(x)=\sum_{k=0}^{\infty }\frac{-1^{k}}{\gamma ^{k}} x^{2k} $$     (1.1)
 * 2)
 * 2)
 * 
 * }

And find the radius of convergence for the Taylor series of 3) Sin(x) about x=0, 4) log(1+x) about x=0, and 5)log(1+x) about x=1.

Solution
1)--- First, we establish what $$ d_{k}$$ is. As found in the notes (section 7-c), we consider an infinite power series of the form:
 * {| style="width:100%" border="0"

r(x)=\sum_{k=0}^{\infty }d_kx^k $$     (1.2) The radius is then calculated using the formula below:
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{d_{k+1}}{d_{k}} \right | \right ] ^{-1} $$     (1.3)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

In the case of equation (1.0), $$ d_{k}$$ is equals to:
 * {| style="width:100%" border="0"

d_{k}= (k+1)(k) $$     (1.4)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This means, from equation (1.3):


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(k+2)(k+1)}{(k+1)(k)} \right | \right ] ^{-1} $$     (1.5)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This simplifies to:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(k+2)}{(k)} \right | \right ] ^{-1} $$     (1.6)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{1}{1} \right | \right ] ^{-1} $$     (1.7)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This limit just becomes 1 and the radius of convergence is then established as 1.

2)--- In the case of equation (1.1), after making a bounds change where j=2k and then making k=j,$$ d_{k}$$ is equal to:
 * {| style="width:100%" border="0"

d_{k}= \frac {-1^{k/2}}{\gamma ^{k/2}} $$     (1.8) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{k/2+1/2})(\gamma^{k/2}) }{(\gamma ^{k/2+1/2})(-1^{k/2})} \right | \right ] ^{-1} $$     (1.9)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

$$ \gamma $$ is a constant in this case, and the radius of convergence simplifies to this form:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{1/2})}{(\gamma ^{1/2} )} \right | \right ] ^{-1} $$     (1.10)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

When taking the limit at infinity, the radius of convergence becomes $$ - \gamma^{1/2} $$.

3)--- The Taylor series that represents sin(x) about x=0 is:
 * {| style="width:100%" border="0"

\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} $$     (1.11)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

We put this in the form found in equation (1.2) first. This will allow us to get a $$ d_{k}$$. To put (1.11) in the form found in (1.2), we let k=2n+1:
 * {| style="width:100%" border="0"

r(x)=\sum_{k=1}^{\infty }\frac{-1^{ \frac{k+1}{2}}}{k!} x^k $$     (1.12)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This means $$ d_{k} $$ would equal:
 * {| style="width:100%" border="0"

\frac{-1^{ \frac{k+1}{2}}}{k!} $$     (1.13)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This means, from equation (1.3):


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{ (-1^{ \frac{k+2}{2}}) (k)! } {(k+1)! (-1^{ \frac{k+1}{2}})} \right | \right ] ^{-1} $$     (1.14)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This simplifies to:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{.5})}{(k+1)} \right | \right ] ^{-1} $$     (1.15)
 * $$\displaystyle
 * $$\displaystyle
 * 
 * }

This inverted limit makes $$ R_{c}= \infty $$. This is the best possible case, as this means the series converges for all values of x. 4)--- The Taylor series that represents log(1+x) about x=0 is:
 * {| style="width:100%" border="0"

\sin x=\sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k $$     (1.16)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

So $$d_k $$ would be defined as shown below:
 * {| style="width:100%" border="0"

d_{k}= \frac{(-1)^k}{k} $$     (1.17) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{k+1})(k) }{(k+1)(-1^k)} \right | \right ] ^{-1} $$     (1.18)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This simplifies to:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1)(k) }{(k+1)} \right | \right ] ^{-1} $$     (1.19)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{-1}{1} \right | \right ] ^{-1} $$     (1.20)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This absolute, inverted limit just becomes 1 and the radius of convergence is then established as 1.

5)--- The Taylor series that represents log(1+x) about x=1 is:
 * {| style="width:100%" border="0"

\sin x= - \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k $$     (1.21)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

So $$d_k $$ would be defined as shown below:
 * {| style="width:100%" border="0"

d_{k}= - \frac{(-1)^k}{k} $$     (1.22) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | -\frac{(-1^{k+1})(k) }{(k+1)(-1^k)} \right | \right ] ^{-1} $$     (1.23)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This simplifies to:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | -\frac{(-1)(k) }{(k+1)} \right | \right ] ^{-1} $$     (1.24)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:


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

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{1}{1} \right | \right ] ^{-1} $$     (1.25)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This absolute, inverted limit just becomes 1 and the radius of convergence is then established as 1.

Author
Solved and Typed By -Egm4313.s12.team1.silvestri (talk) 17:26, 24 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.durrance (talk) 17:33, 30 March 2012 (UTC)

Statement
Determine whether the following functions are linear independent using the Wronskian.


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

f(x)=x^2, g(x)=x^4 $$     (2.0)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

f(x)=\cos(x), g(x)=\sin(3x) $$     (2.1) Using the Gramian over interval [a,b] = [1,1] to come to the same conclusion.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Wronskian Solution (1)
The Wronskian is defined as:
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=fg'-gf' $$     (2.2) If the Wronskian does not equal zero, then the equations are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

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

f'=2x $$     (2.3)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

g'=4x^3 $$     (2.4) Substituting all of the conditions into Eq. (2.2):
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=x^2(4x^3)-2x(x^4)=2x^{5}\neq 0 $$     (2.5) Because the Wronskian does not equal zero, Eqs. (2.0) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Wronskian Solution (2)
The Wronskian is defined as:
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=fg'-gf' $$     (2.2) If the Wronskian does not equal zero, then the equations are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

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

f'=-\sin(x) $$     (2.6)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

g'=3\cos(3x) $$     (2.7) Substituting all of the conditions into Eq. (5.2):
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

W(f,g)=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=3\cos(x)(\cos(3x))-(-\sin(x))(\sin(3x))\neq 0 $$     (2.8) Because the Wronskian does not equal zero, Eqs. (2.1) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Gramian Solution (1)
The Gramian is defined as:
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (2.9) Where the scalar product is defined as:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

\langle f, g \rangle := \int_a^b f(x) g(x) \,dx $$     (2.10) When the Gramian does not equal zero, the functions are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

So calculating the scalar products:
 * {| style="width:100%" border="0"

\langle f, f \rangle = \int_{-1}^1 (x^2)(x^2)\,dx = 2/5$$ $$\displaystyle \langle f, g \rangle = \int_{-1}^1 (x^2)(x^4)\,dx = 2/7$$ $$\displaystyle \langle g, f \rangle = \int_{-1}^1 (x^4)(x^2)\,dx = 2/7$$ $$\displaystyle \langle g, g \rangle = \int_{-1}^1 (x^4)(x^4)\,dx = 2/9 $$     (2.11) Substituting those values into Eq. (2.9) and calculating the determinant:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix}= (2/5)(2/9)-(2/7)(2/7)=16/2205\neq 0 $$     (2.12) Because the determinant does not equal zero, Eqs. (2.0) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Gramian Solution (2)
The Gramian is defined as:
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (2.9) Where the scalar product is defined as:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

\langle f, g \rangle := \int_a^b f(x) g(x) \,dx $$     (2.10) When the Gramian does not equal zero, the functions are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

So calculating the scalar products:
 * {| style="width:100%" border="0"

\langle f, f \rangle = \int_{-1}^1 \cos(x)\cos(x)\,dx = 1.9998$$ $$\displaystyle \langle f, g \rangle = \int_{-1}^1 \cos(x)\sin(3x)\,dx = 0$$ $$\displaystyle \langle g, f \rangle = \int_{-1}^1 \sin(3x)\cos(x)\,dx = 0$$ $$\displaystyle \langle g, g \rangle = \int_{-1}^1 \sin(3x)\sin(3x)\,dx = 0.0018 $$     (2.13) Substituting those values into Eq. (2.9) and calculating the determinant:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix}= 1.998*0.0018-0=0.0036\neq 0 $$     (2.14) Because the determinant does not equal zero, Eqs. (2.1) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Author
Solved and Typed By - --Egm4313.s12.team1.durrance (talk) 18:35, 26 March 2012 (UTC)--128.227.12.77 18:17, 26 March 2012 (UTC) Reviewed By ---Egm4313.s12.team1.stewart (talk) 19:03, 26 March 2012 (UTC)

Statement
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c.djvu Lect. 7c Pg. 38] Verify that $$\mathbf{b_{1}},\mathbf{b_{2}}\!$$ are linearly independent using the Gramian.


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

$$  \displaystyle \mathbf{b_{1}}=2\mathbf{e}_{1}+7\mathbf{e}_{2} $$     (3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \mathbf{b_{2}}=1.5\mathbf{e}_{1}+3\mathbf{e}_{2} $$     (3.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (3.3)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

But for the vectors given the Gramian looks like this:


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

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}}):=\det \begin{bmatrix} \langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle \\ \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle \end{bmatrix} $$     (3.4)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

And:


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

$$     (3.5)
 * $$\langle \mathbf{b_{i}}, \mathbf{b_{j}} \rangle \equiv \mathbf{b_{i}} \cdot \mathbf{b_{j}}
 * $$\langle \mathbf{b_{i}}, \mathbf{b_{j}} \rangle \equiv \mathbf{b_{i}} \cdot \mathbf{b_{j}}
 * <p style="text-align:right">
 * }

And for the scalar dot product:
 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf e_1, \mathbf e_2 \rangle = 0 $$     (3.6) Therefore:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

\langle \mathbf{b_{1}}, \mathbf{b_{1}} \rangle = (2\mathbf{e}_{1}+7\mathbf{e}_{2}) \cdot (2\mathbf{e}_{1}+7\mathbf{e}_{2})=4\mathbf{e}_{1}+49\mathbf{e}_{2}=4+49=53 $$     (3.7)
 * <p style="text-align:right">
 * }


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

\langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle = (2\mathbf{e}_{1}+7\mathbf{e}_{2}) \cdot (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) = 3\mathbf{e}_{1}+21\mathbf{e}_{2}=3+21=24 $$     (3.8)
 * <p style="text-align:right">
 * }


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

\langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle = (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) \cdot (2\mathbf{e}_{1}+7\mathbf{e}_{2})= 3\mathbf{e}_{1}+21\mathbf{e}_{2}=3+21=24 $$     (3.9)
 * <p style="text-align:right">
 * }


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

\langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle = (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) \cdot (1.5\mathbf{e}_{1}+3\mathbf{e}_{2})= 2.25\mathbf{e}_{1}+9\mathbf{e}_{2}=2.25+9=11.25 $$     (3.10)
 * <p style="text-align:right">
 * }

The determinant to solve for the Gramian is:


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

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}}):=\det \begin{bmatrix} \langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle \\ \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle \end{bmatrix}=\langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle\cdot \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle -\langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle\cdot \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle $$     (3.11)
 * <p style="text-align:right">
 * }

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

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=(53)\cdot (11.25)-(24)\cdot (24) $$     (3.12)
 * <p style="text-align:right">
 * }


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

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=596.25-576 $$     (3.13)
 * <p style="text-align:right">
 * }


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

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=20.25 $$     (3.14) Qualifications for a linearly independent system of vectors:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma } \neq 0 \Rightarrow \mathbf{\Gamma^{-1} } exists \Rightarrow \mathbf{c}=\mathbf{\Gamma } ^{-1}\mathbf{d} $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

So the vectors $$\mathbf{b_{1}},\mathbf{b_{2}}\!$$ are linearly independent because:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma }= 20.25 \neq 0 $$     (3.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Author
Solved and Typed By - Egm4313.s12.team1.stewart (talk) 18:20, 25 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.rosenberg (talk) 05:20, 28 March 2012 (UTC)

Statement
Show that $$ y_p(x) = \sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''_{p,i} + p(x)y'_{p,i} + q(x)y_{p,i} = r_i(x)\!$$ with the excitation $$ r(x) = r_1(x) + r_2(x) + ... + r_n(x) = \sum_{i=0}^{n}r_i(x)\!$$. Discuss the choice of $$ y_p(x)\!$$, in example for $$ r(x) = Kcos(wx)\!$$. Why would you need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$?

Solution
The following represent particular solutions and their derivatives, equated into a summation:
 * {| style="width:100%" border="0"

y_p(x) = y_{p,1}(x) + y_{p,2}(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}(x) $$     (4.0)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


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

y_p'(x) = y_{p,1}'(x) + y_{p,2}'(x) + ... + y_{p,n}'(x) \rightarrow y_p'(x) = \sum_{i=0}^{n}y_{p,i}'(x) $$     (4.1)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


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

y_p(x) = y_{p,1}(x) + y_{p,2}''(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}''(x) $$     (4.2)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Because the given ODE is in the form of L2-ODE-VC, it is linearly independent. Each $$ y_{p,i}\! $$ is a solution for each $$ r_i\! $$, and when there are multiple excitations, the solution to a sum of these excitations is the sum of the particular solutions. Using these particular solutions, we can show that they are the solutions to the following L2-ODE-VC (4.3) with the given excitation:


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

y_{p,i}'' + p(x)y_{p,i}' + q(x)y_{p,i} = r_i(x) $$     (4.3)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


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

\sum_{i=0}^{n}y_{p,i}''(x) + p(x)\sum_{i=0}^{n}y_{p,i}'(x) + q(x)\sum_{i=0}^{n}y_{p,i}(x) = \sum_{i=0}^{n}r_i(x) $$     (4.4)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

For $$ r(x) = Kcos(wx)\!$$. You need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$ because when solving for the particular solution, it is necessary to take derivatives of the particular solution, and in the case of $$ r(x) = Kcos(wx)\!$$ the derivatives will produce extra terms since the derivative of $$ cos(wx)\!$$ will produce both $$ sin(wx)\!$$ and $$ cos(wx)\! $$ terms. Having both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ is necessary to eliminate the extra terms.

Author
Solved and Typed By - --Egm4313.s12.team1.wyattling (talk) 19:17, 26 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.armanious (talk) 05:25, 30 March 2012 (UTC)

Statement
1. Show that $$\cos 7x$$ and $$\sin 7x$$ are linearly independent using the Wronskian and the Gramian(1 period). 2. Find 2 equations for the two unknowns M, N, and solve for M, N. 3. Find the overall solution $$y(x)$$ that corresponds to the initial condition (3b) p3-7. $$y(0)=1, y'(0)=0$$ Plot the solution over 3 periods.

Solution
Part 1a When using the Wronskian, if the solution does not equal to 0, then the two are linearly independent of each other. The Wronskian can be defined as:
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\displaystyle W(f,g):=det\begin{bmatrix} f &g \\ f'&g' \end{bmatrix}=fg'-gf' $$     (5.1) Lets set $$f = \cos (7x)$$ and $$g = \sin (7x)$$ so that we can find $$f', g'$$
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\displaystyle f' = -7\sin (7x), g' = 7\cos (7x) $$     (5.2) Plugging these values into the Wronskian equation yields:
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\displaystyle W(f,g):=det\begin{bmatrix} \cos (7x) &\sin (7x) \\ -7\sin (7x)&7\cos (7x) \end{bmatrix}=7\cos ^2 (7x) - 7\sin ^2 (7x) $$     (5.3)
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\displaystyle 7\cos ^2 (7x) - 7\sin ^2 (7x) \neq 0 $$     (5.4)
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Thus f and g are in fact linearly independent of each other. Part 1b Now we need to solve it using the Gramian Gramian can be defined as:
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\displaystyle \Gamma (f,g):=det\begin{bmatrix} <f,f> &<f,g> \\ <g,f> &<g,g> \end{bmatrix} $$     (5.5) Where like with the Wronskian, f and g are linearly independent of each other when $$\Gamma (f,g)\neq 0 $$ Integrating over one period implies that our boundaries will be $$(0, \frac{2\pi }{7})$$
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\displaystyle <f,f>=\int_{0}^{\frac{2\pi }{7}}\cos ^{2}(7x)dx $$     (5.6) Setting $$u=7x$$ gives us $$du=7dx$$ which also changes our integration factors to be $$(0,2\pi )$$ giving us:
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\displaystyle <f,f>=\frac{1}{7}\int_{0}^{2\pi }\cos ^{2}(u)du $$     (5.7) Integrating gives us:
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\displaystyle \frac{1}{7}[\frac{u}{2} +\frac{1}{4} \sin 2u\mathbf{\mid} _{0}^{2\pi }] $$     (5.8) Which equals:
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\displaystyle \frac{\pi }{7} $$     (5.9)
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\displaystyle <g,g>=\int_{0}^{\frac{2\pi }{7}}\sin ^{2}(7x)dx $$     (5.10) Setting $$u=7x$$ gives us $$du=7dx$$ which also changes our integration factors to be $$(0,2\pi )$$ giving us:
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\displaystyle <g,g>=\frac{1}{7}\int_{0}^{2\pi }\sin ^{2}(u)du $$     (5.11) Integrating gives us:
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\displaystyle \frac{1}{7}[\frac{u}{2} -\frac{1}{4} \sin 2u\mathbf{\mid} _{0}^{2\pi }] $$     (5.12) Which equals:
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\displaystyle \frac{\pi }{7} $$     (5.13)
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\displaystyle <f,g>=<g,f>=\int_{0}^{\frac{2\pi }{7}}\cos (7x)\sin (7x)dx $$     (5.14) Because cos and sin are orthogonal of each other, without going through with the integration, we can say that $$<f,g>=<g,f>=0$$ Plugging these values into our Gramian equation gives us:
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\displaystyle \Gamma (f,g):=det\begin{bmatrix} \frac{\pi }{7}&0 \\ 0 &\frac{\pi }{7} \end{bmatrix} =\frac {\pi ^2}{49} \neq 0 $$     (5.15) Thus, we again see that f and g are linearly independent of each other. Part 2 From the notes, we are given the following information:
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\displaystyle y''-3y'-10y=3\cos (7x) $$     (5.16)
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\displaystyle y_{p}(x)=M\cos (7x)+N\sin (7x) $$     (5.17)
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\displaystyle y'_{p}(x)=-M7\sin (7x)+N7\cos (7x) $$     (5.18)
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\displaystyle y''_{p}(x)=-M49\cos (7x)-N49\sin (7x) $$     (5.19) Plugging 5.17-19 back into our original equation in 5.16 yields:
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\displaystyle -M49\cos (7x)-N49\sin (7x)+M21\sin (7x)-N21\cos (7x)-M10\cos (7x)-N10\sin (7x)=3\cos (7x) $$     (5.20) Collecting like terms leaves us with:
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\displaystyle -59M\cos (7x)-59N\sin (7x)+21M\sin (7x)-21N\cos (7x)=3\cos (7x) $$     (5.21) Now we can equate coefficients to solve of M and N:
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\displaystyle -59M = 3,M=\frac{-3}{59} $$     (5.22)
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\displaystyle 21N = 3,N=\frac{1}{7} $$     (5.23) Part 3 The overall solution to the equation is expressed as $$ y(x)=y_{p}(x)+y_{h}(x)$$. We must now find the homogeneous equation. Writing our given equation in homogeneous form gives us:
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\displaystyle y''-3y'-10=0 $$     (5.24) rewriting in characteristic form:
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\displaystyle \lambda ^2-3\lambda -10 = 0 $$     (5.25) We can solve for our roots using simple factoring:
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\displaystyle (\lambda -5)(\lambda +2) = 0 $$     (5.26) Thus:
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\displaystyle \lambda_{1,2}=(5,-2) $$     (5.27) Yielding:
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\displaystyle y_{h}(x)=c_{1} e^{5x}+c_{2} e^{-2x} $$     (5.28) Now using the given initial conditions, we can solve for $$c_{1},c_{2}$$
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\displaystyle y'_{h}(x)=5c_{1}e^{5x}-2c_{2}e^{-2x} $$     (5.29) Plugging in initial conditions we have:
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\displaystyle y(0)=c_{1}+c_{2}=1, y'(0)=5c_{1}-2c_{2}=1 $$     (5.30) Solving for each gives us:
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\displaystyle c_{1}=\frac{2}{7}, c_{2}=\frac{5}{7} $$     (5.31) Therfore:
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\displaystyle y_{h}(x)=\frac{2}{7} e^{5x}+\frac{5}{7} e^{-2x} $$     (5.32) Now plugging in our answers from Step 2 into the general particular equation we get that:
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\displaystyle y_{p}(x)=\frac{-3}{59} \cos (7x) + \frac{1}{7} \sin (7x) $$     (5.33) so our final equation to is:
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\displaystyle y(x)=\frac{2}{7} e^{5x}+\frac{5}{7} e^{-2x} +\frac{-3}{59} \cos (7x) + \frac{1}{7} \sin (7x) $$     (5.34) To plot over 3 periods, we will plot from $$(0,\frac{6\pi }{7})$$
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Author
Solved and Typed By - Egm4313.s12.team1.rosenberg (talk) 18:45, 30 March 2012 (UTC) Reviewed By -Egm4313.s12.team1.silvestri (talk) 18:44, 30 March 2012 (UTC)

Statement
Consider the following L2-ODE-CC; see p.6-6:
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$$\displaystyle{y}''-4{y}'+13{y}=2e^{-2x}\cos(3x)$$ (6.0) Homogeneous solution:
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$$\displaystyle{y}_{h}(x)=e^{-2x}[A\cos3x+B\sin3x]$$ (6.1) Particular solution:
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$$\displaystyle{y}_{p}(x)=xe^{-2x}[M\cos3x+N\sin3x]$$ (6.2) Complete the solution for this problem.
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Find the overall solution $$y(x)$$ that corresponds to the initial condition (3b) p.3-7
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$$\displaystyle y(0)=1,y'(0)=0$$ (6.3)
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Solution
Start by finding $$y_{p}'$$ and $$y_{p}''$$.
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$$\displaystyle {y}_{p}'={{e}^{-2x}}[(-3Mx-2Nx+N)\sin 3x+(-2Mx+3Nx+M)\cos 3x]$$ (6.4)
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$$\displaystyle {y}_{p}''={{e}^{-2x}}[(12Mx-6M-5Nx-4N)\sin 3x+(-15Mx-12M-12Nx+6N)\cos 3x]$$ (6.5) Substitute $$y_p$$ and its derivatives into (6.0) to find $$M$$ and $$N$$.
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$$\displaystyle {y}_{p}''+4{y}_{p}'+13{y}_{p}={{e}^{-2x}}[(-6M)\sin 3x+(-10Mx-8M+6N)\cos 3x]$$ (6.6) Separating terms and setting equal to the excitation from (6.0):
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$$\displaystyle -6M{{e}^{-2x}}\sin 3x+(-8M+6N){{e}^{-2x}}\cos 3x-10Mx{{e}^{-2x}}\cos 3x=2{{e}^{-2x}}\cos 3x$$
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(6.7) From (6.7), we solve coefficients to get
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$$\displaystyle -6M=0$$ $$\displaystyle -8M+6N=2$$ $$\displaystyle -10M=0$$ (6.8) Solving for $$M$$ and $$N$$:
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$$\displaystyle M=0$$ $$\displaystyle N=\frac{1}{3}$$ (6.9) Which gives us the particular solution:
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$$\displaystyle {y}_{p}=\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.10) For the general solution,
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$$\displaystyle y={{y}_{h}}+{{y}_{p}}$$ (6.11)
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$$\displaystyle y=e^{-2x}[A\cos3x+B\sin3x]+\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.12)
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$$\displaystyle y=e^{-2x}[A\cos3x+(B+\frac{1}{3}x)\sin3x]$$ (6.13) To solve for $$A$$ and $$B$$, we use initial conditions from (6.3):
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$$\displaystyle y(0)=e^{-2(0)}[A\cos3(0)+(B+\frac{1}{3}(0))\sin3(0)]=1$$ (6.14) Which simplifies to:
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$$\displaystyle A=1$$ (6.15) For the second initial condition from (6.3):
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$$\displaystyle y'={{e}^{-2x}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}x)\sin 3x+(3B+x-2)\cos 3x]$$ (6.16)
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$$\displaystyle y'(0)={{e}^{-2(0)}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}(0))\sin 3(0)+(3B+(0)-2)\cos 3(0)]=0$$ (6.17)
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$$\displaystyle 3B-2=0$$ (6.18)
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$$\displaystyle B=\frac{2}{3}$$ (6.19) We can now write $$y$$ with all coefficients known.
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$$\displaystyle y={{e}^{-2x}}[(1)\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.20) Final Equation
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$$\displaystyle y={{e}^{-2x}}[\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.21)
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Author
Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 00:28, 20 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 18:29, 25 March 2012 (UTC)

Statement
'''[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]:'''


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$$\mathbf{v}=4\mathbf e_1+2\mathbf e_2=c_1\mathbf b_1+c_2\mathbf b_2$$
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The oblique basis vectors b1,b2 are:
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$$\mathbf b_1 =2\mathbf e_1 + 7\mathbf e_2$$
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$$\mathbf b_2 =1.5\mathbf e_1 + 3\mathbf e_2$$ 1. Find the components c1, c2 using the Gramian matrix. 2. Verify the result found above.
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Solution
To find the components of the oblique basis vectors, the Gram matrix must be used. The Gram matrix is defined as such:
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$$  \displaystyle \boldsymbol \Gamma (\mathbf b_1, \mathbf b_2) := \begin{bmatrix} \langle \mathbf b_1, \mathbf b_1 \rangle & \langle \mathbf b_1, \mathbf b_2 \rangle \\ \langle \mathbf b_2, \mathbf b_1 \rangle & \langle \mathbf b_2, \mathbf b_2 \rangle \end{bmatrix} $$     (7.0)
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This matrix requires several scalar products to be found. one important feature of scalar products will be presented here and implicitly used throughout these calculations:
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$$  \displaystyle \langle \mathbf e_1, \mathbf e_2 \rangle = 0 $$     (7.1)
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Equation 7.1 states that the scalar product of two perpendicular vectors is zero. To find the Gram matrix, three scalar products must be found:
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$$  \displaystyle \langle \mathbf b_1, \mathbf b_1 \rangle = (2)(2)+(7)(7)=53 $$     (7.2)
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$$  \displaystyle \langle \mathbf b_1, \mathbf b_2 \rangle =\langle \mathbf b_2, \mathbf b_1 \rangle =(2)(1.5)+(7)(3)=24 $$     (7.3)
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$$  \displaystyle \langle \mathbf b_2, \mathbf b_2 \rangle = (1.5)(1.5)+(3)(3)=11.25 $$     (7.4) Using the above values the Gram matrix is found to be:
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$$  \displaystyle \boldsymbol \Gamma (\mathbf b_1, \mathbf b_2) = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix} $$     (7.5) To find the components, the Gram matrix must be used to solve the following equation:
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$$  \displaystyle \begin{bmatrix} \langle \mathbf b_1, \mathbf b_1 \rangle & \langle \mathbf b_1, \mathbf b_2 \rangle \\ \langle \mathbf b_2, \mathbf b_1 \rangle & \langle \mathbf b_2, \mathbf b_2 \rangle \end{bmatrix}\begin{Bmatrix}c_1\\c_2 \end{Bmatrix}=\begin{Bmatrix}\langle \mathbf b_1, \mathbf v \rangle\\\langle \mathbf b_2, \mathbf v \rangle \end{Bmatrix} $$     (7.6)
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The right hand side of the equation can be found using:
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$$  \displaystyle \langle \mathbf b_1, \mathbf v \rangle =(2)(4)+(7)(2)=22 $$     (7.7)
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$$  \displaystyle \langle \mathbf b_2, \mathbf v \rangle =(1.5)(4)+(3)(2)=12 $$     (7.8) Using known values, 7.6 becomes:
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$$  \displaystyle \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}\begin{Bmatrix}c_1\\c_2 \end{Bmatrix}=\begin{Bmatrix}22\\12 \end{Bmatrix} $$     (7.9) To find the components, the Gramian (the determinant of the Gram matrix) and the inverse of the Gram must be found.
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$$  \displaystyle \Gamma = \begin{vmatrix} 53 & 24 \\ 24 & 11.25 \end{vmatrix}=(53)(11.25)-(24)(24)=20.25 $$     (7.10)
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$$  \displaystyle \boldsymbol \Gamma ^{-1} = \frac{1}{20.25}\begin{bmatrix} 11.25 & -24 \\ -24 & 53 \end{bmatrix}=\begin{bmatrix} \frac{5}{9} & -\frac{32}{27} \\ -\frac{32}{27} & \frac{212}{81} \end{bmatrix} $$     (7.11) This can be used in the matrix equation to solve for the components:
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$$  \displaystyle \begin{Bmatrix}c_1\\c_2 \end{Bmatrix}= \begin{bmatrix} \frac{5}{9} & -\frac{32}{27} \\ -\frac{32}{27} & \frac{212}{81} \end{bmatrix}\begin{Bmatrix}22\\12 \end{Bmatrix}=\begin{Bmatrix}-2\\ \frac{16}{3} \end{Bmatrix} $$     (7.12) Therefore the matrix v with respect to the oblique vectors is:
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$$  \displaystyle \mathbf v = -2 \mathbf b_1 + \frac{16}{3} \mathbf b_2 $$     (7.13) As a check, the definitions of each vectors with respect to the basis e1 and e2:
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$$  \displaystyle 4\mathbf e_1 + 2\mathbf e_2 = -2 (2\mathbf e_1 + 7\mathbf e_2) + \frac{16}{3} (1.5\mathbf e_1 + 3\mathbf e_2)=-4\mathbf e_1 -14\mathbf e_2 + 8\mathbf e_1 + 16\mathbf e_2 $$     (7.14) Which reduces to:
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$$  \displaystyle 4\mathbf e_1 + 2\mathbf e_2 = 4\mathbf e_1 + 2\mathbf e_2 $$     (7.15)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 01:40, 25 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 18:31, 25 March 2012 (UTC)

Statement
'''[https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec8b.djvu see R5.8 Lect. 8b pg. 16]:''' Find the integral: $$\int x^n \log(1+x) dx\!$$

Solution

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$$  \displaystyle \int x^n \log(1+x) dx $$ (8.0) To find this integral, integration by parts must be used. The formula for integration by parts is:
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$$  \displaystyle \int udv=uv-\int vdu + C $$ (8.1)
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$$  \displaystyle u = \log(1+x) $$     (8.2)
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$$  \displaystyle du=\frac{dx}{1+x} $$     (8.3)
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$$  \displaystyle dv=x^ndx $$     (8.4)
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$$  \displaystyle v=\frac{x^{n+1}}{n+1} $$     (8.5) Using these expressions in the formula yields:
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$$  \displaystyle \int x^n \log(1+x)dx=\frac{x^{n+1}}{n+1}\log(1+x)-\frac{1}{n+1}\int \frac{x^{n+1}}{1+x}dx + C $$ (8.6) Now, the integral $$\int \frac{x^{n+1}}{1+x} dx\!$$ must be found. To start, the fraction must be expanded using long division:
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$$  \displaystyle \frac{x^{n+1}}{1+x}=x^n-x^{n-1}+x^{n-2}-x^{n-3}+...+\frac{(-1)^{n+1}}{x+1}=\frac{(-1)^{n+1}}{x+1}+\sum_{k=0}^{n}(-1)^kx^{n-k} $$     (8.7) This expression can now be easily integrated to yield the following:
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$$  \displaystyle \int \frac{x^{n+1}}{1+x}dx=\int\frac{(-1)^{n+1}}{x+1}dx+\int\sum_{k=0}^{n}(-1)^kx^{n-k}dx $$     (8.8)
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$$  \displaystyle \int\frac{(-1)^{n+1}}{x+1}dx=(-1)^{n+1}\log(1+x) $$     (8.9)
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$$  \displaystyle \int\sum_{k=0}^{n}(-1)^kx^{n-k}dx=\sum_{k=0}^{n}\int(-1)^kx^{n-k}dx= \sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1} $$     (8.10) Therefore:
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$$  \displaystyle \int \frac{x^{n+1}}{1+x} dx= (-1)^{n+1}\log(1+x)+\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1} $$     (8.11) Substituting into the original equation:
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$$  \displaystyle \int x^n \log(1+x)dx=\frac{x^{n+1}}{n+1}\log(1+x)-\frac{1}{n+1}((-1)^{n+1}\log(1+x)+\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1}) + C $$ (8.12) Simplifying this yields:
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 * {| style="width:100%" border="1"

$$  \displaystyle \int x^n \log(1+x)dx=\frac{1}{n+1}[(x^{n+1}-(-1)^{n+1})\log(1+x)-\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1}]+C $$     (8.13) To illustrate this, two test cases with n=0 and n=1 will be used. For n=0
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$$  \displaystyle \int x^0 \log(1+x)dx=\frac{1}{0+1}[(x^{0+1}-(-1)^{0+1})\log(1+x)-\frac{(-1)^0}{0-0+1}x^{0-0+1}]+C $$     (8.14) This simplifies to:
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$$  \displaystyle \int \log(1+x)dx=(x+1)\log(1+x)-x+C $$     (8.15) For n=1
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$$  \displaystyle \int x^1 \log(1+x)dx=\frac{1}{1+1}[(x^{1+1}-(-1)^{1+1})\log(1+x)-\frac{(-1)^0}{1-0+1}x^{1-0+1}-\frac{(-1)^1}{1-1+1}x^{1-1+1}]+C $$     (8.16) This simplifies to:
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 * }
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$$  \displaystyle \int x \log(1+x)dx=\frac{1}{2}[(x^{2}-1)\log(1+x)-\frac{1}{2}x^{2}+x]+C $$     (8.17) In fact, this formula can be further generalized for any $$\int x^n \log(r+x)dx \!$$ where r is any real number. The most notable step that changes is the long division expansion. Each term in the expansion increases by a factor of $$r^{k} \!$$. The result is:
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$$  \displaystyle \frac{x^{n+1}}{r+x}=x^n-rx^{n-1}+r^2x^{n-2}-r^3x^{n-3}+...+\frac{(-r)^{n+1}}{x+r}=\frac{(-r)^{n+1}}{x+r}+\sum_{k=0}^{n}(-r)^kx^{n-k} $$     (8.18) It is important to note that this particular step will fail when r=0 because $$0^{0} \!$$ is undefined. A special case with r=0 will also be shown for full generality. The rest of the process is identical to that shown above, with every $$\log (1+x) \!$$ term replaced with $$\log (r+x) \!$$. The final result of the integration yields:
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 * }
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$$  \displaystyle \int x^n \log(r+x)dx=\frac{1}{n+1}[(x^{n+1}-(-r)^{n+1})\log(r+x)-\sum_{k=0}^{n}\frac{(-r)^k}{n-k+1}x^{n-k+1}]+C; r\neq 0 $$     (8.19) When r=0, the integral is of the form $$\int x^n \log(x)dx \!$$, which can be integrated using integration by parts.
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$$  \displaystyle u = \log(x) $$     (8.20)
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$$  \displaystyle du=\frac{dx}{x} $$     (8.21)
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$$  \displaystyle dv=x^ndx $$     (8.22)
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$$  \displaystyle v=\frac{x^{n+1}}{n+1} $$     (8.23) Using these in (8.1) yields:
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$$  \displaystyle \int x^n \log(x)dx=\frac{x^{n+1}}{n+1}\log(x)-\frac{1}{n+1}\int \frac{x^{n+1}}{x}dx + C $$ (8.24) Now, the integral $$\int \frac{x^{n+1}}{x} dx\!$$ must be found. This is simply:
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$$  \displaystyle \int \frac{x^{n+1}}{x}dx=\int x^ndx=\frac{x^{n+1}}{n+1}+C $$     (8.25) Substituting (8.25) into (8.24) yields:
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$$  \displaystyle \int x^n \log(x)dx=\frac{x^{n+1}}{n+1}\log(x)-\frac{x^{n+1}}{(n+1)^2} + C $$ (8.26) Thus, the overall solution for any real value of r is
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$$  \displaystyle \int x^n \log(r+x)dx=\begin{cases} \frac{1}{n+1}[(x^{n+1}-(-r)^{n+1})\log(r+x)-\sum_{k=0}^{n}\frac{(-r)^k}{n-k+1}x^{n-k+1}]+C & \text{ if } r\neq 0\\ \frac{x^{n+1}}{n+1}\log(x)-\frac{x^{n+1}}{(n+1)^2} + C& \text{ if } r=0 \end{cases} $$     (8.27)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 02:51, 25 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri (talk) 18:29, 30 March 2012 (UTC)

Statement
Consider the following L2-ODE-CC with log(1+x) as the excitation:
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r(x)=y''-3y'+2y $$     (9.0)
 * $$\displaystyle
 * $$\displaystyle
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 * }


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r(x)=log(1+x) $$     (9.1)
 * $$\displaystyle
 * $$\displaystyle
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 * }

Also, consider the initial conditions:
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y(- \frac{3}{4})=1, y'(- \frac{3}{4})=0 $$     (9.2) 1)Project the excitation r(x) on the polynomial basis
 * $$\displaystyle
 * $$\displaystyle
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 * }


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\left \{ b_{j}(x) = x^j, j=0,1,...n \right \} $$     (9.3) i.e., find $$ d_{j} $$ such that:
 * $$\displaystyle
 * $$\displaystyle
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r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j $$     (9.4) for x in [-.75, 3], and for n= 0,1. Plot $$ r(x) $$ and $$ r_n(x) $$ to show uniform approximation and convergence. Note that $$ <x^i, r>=\int_{a}^{b} x^i\log(1+x) dx $$ In a separate series of plots, compare the approximation of the function log(1+x) by 2 methods:
 * $$\displaystyle
 * $$\displaystyle
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B. Taylor series expansion about $$ \hat x =0 $$
 * A. Projection on Polynomial basis (1) p8-17
 * A. Projection on Polynomial basis (1) p8-17


 * }

Observe and discuss the pros and cons of each method.

2) Find $$ y_n (x) $$ such that:
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y''_n +ay'_n + by_n = r_n(x) $$     (9.5)
 * $$\displaystyle
 * $$\displaystyle
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 * }

With the same initial conditions (9.2). Plot $$ y_n(x)$$ for n=0,1 for x in [-.75,3] In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution
Part 1 First to project the excitation r(x)=log(1+x) onto the polynomial basis. We know that $$ <b_i,b_j> \cdot d_j=<b_i,r> $$. $$ <b_0, b_0>$$ is the only term of concern for the $$ \gamma $$ matrix when n=0.
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<b_0,b_0> = \int_{-3/4}^{3} x^0x^1 dx=(3+3/4)=3.75 $$     (9.6)
 * $$\displaystyle
 * $$\displaystyle
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$$<b_i,r>$$ when n=0 is calculated below:
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<b_0,r> = \int_{-3/4}^{3} x^0log(1+x)dx=2.141751035 $$     (9.7) This means that, as stated in the opening sentence:
 * $$\displaystyle
 * $$\displaystyle
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<b_i,b_j> \cdot d_0=<bi,r> $$ $$\displaystyle 3.75 \cdot d_0=2.141751035 $$ $$\displaystyle d_0=.571136093 $$     (9.8) With that in mind, $$ r_n(x)$$ is developed from equation (9.4) as shown below:
 * $$\displaystyle
 * $$\displaystyle
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r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j= \sum_{j=0}^{0} d_jx^0=.571136093 $$     (9.9) Now projecting the excitation onto the polynomial basis with n=1: We know that (in matrix form) $$ \gamma \cdot d_j=<b_i,r> $$. Beginning with the $$ \gamma $$ matrix:
 * $$\displaystyle
 * $$\displaystyle
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\gamma = \begin{bmatrix} <b_0,b_0> & <b_0,b_1>\\ <b_1,b_0>& <b_1,b_1> \end{bmatrix} $$     (9.9) This matrix becomes:
 * $$\displaystyle
 * $$\displaystyle
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\gamma = \begin{bmatrix} 3.75 & 4.21875\\ 4.21875& 9.140625 \end{bmatrix} $$     (9.10) The matrix containing $$<b_i,r>$$ is shown below, defined as matrix c:
 * $$\displaystyle
 * $$\displaystyle
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c =\begin{Bmatrix} <b_o,r>\\ <b_1,r> \end{Bmatrix} ==== \begin{Bmatrix} \int_{-3/4}^{3} x^0log(1+x))\\ \int_{-3/4}^{3} x^1log(1+x)) \end{Bmatrix} ==== \begin{Bmatrix} 2.141751035\\ 5.007550553 \end{Bmatrix} $$     (9.10) We then find the $$ d_j $$ matrix in the following way:
 * $$\displaystyle
 * $$\displaystyle
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\gamma ^{-1} \cdot c=d_j =\begin{Bmatrix} -.0939750342\\ .5912076831 \end{Bmatrix} $$     (9.11) With that in mind, $$ r_n(x)$$ is developed from equation (9.4) as shown below:
 * $$\displaystyle
 * $$\displaystyle
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 * }
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r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j= \sum_{j=0}^{1} d_jx^j=-.0939750342+.5912076831x $$     (9.12) Graphed out compared to the actual excitation of log(1+x) [in red], the projections with n=0[in blue],1[in green] are compared below:
 * $$\displaystyle
 * $$\displaystyle
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Part 2 First we create the characteristic equation in standard form:
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$$\displaystyle{\lambda^2-3 \lambda +2=0}$$ (9.13) Then, by setting it equal to zero, we can find what $$ \lambda \!$$ equals:
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$$\displaystyle{(\lambda - 2)(\lambda -1) = 0}$$ (9.14)
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$$\displaystyle{\lambda = 2,\lambda =1}$$ (9.15)
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Given two, distinct, real roots, the homogeneous solution looks like this:
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$$  \displaystyle y_h(x)=C_1e^{2x} + C_2e^{x} $$     (9.16) By using the method of undetermined coefficients, for n=0, the excitation $$ 5.571136093$$ is analyzed to yield a particular solution: In assessing a polynomial with a power of 0, the form of the particular solution will look like this:
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$$  \displaystyle y_p (x)= A_0 $$     (9.17)
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The first and second derivatives of $$ A_0 $$, being a constant, are 0. So when plugging in $$ y_p (x)$$ into (9.0), $$ A_0 $$ is determined to be:
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.571136093=2 A_0 $$ $$\displaystyle .2855680465= A_0 $$     (9.18)
 * $$\displaystyle
 * $$\displaystyle
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 * }

The general solution, after adding $$ y_h $$ and $$ y_p $$ then becomes:
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y_g(x)=C_1e^{2x} + C_2e^{x} + .2855680465 $$     (9.18)
 * $$\displaystyle
 * $$\displaystyle
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 * }

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


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$$  \displaystyle 2C_1e^{2x} + C_2e^{x} = y'_g(x) $$     (9.19)
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By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
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$$  \displaystyle y_g(-3/4)=1 =C_1e^{2(-3/4)} + C_2e^{-3/4} + .2855680465 $$     (9.20)
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$$  \displaystyle y'_g(-3/4)=0= 2C_1e^{2(-3/4)} + C_2e^{-3/4} $$     (9.21)
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 * }

Solving the equations proves that $$C_1 = -3.201861878, C_2 = 3.024904915 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
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$$  \displaystyle y_g(x)=(-3.024904915)e^{2x} + (3.024904915)e^{x} + .2855680465 $$     (9.22) By using the method of undetermined coefficients, for n=1, the excitation $$ 5.571136093$$ is analyzed to yield a particular solution: In assessing a polynomial with a power of 0, the form of the particular solution will look like this:
 * style="width:95%" |
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 * <p style="text-align:right">
 * }


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$$  \displaystyle y_p (x)= A_1 x+ A_0 $$     (9.23)
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 * }

The first derivative of $$ y_p $$ is simply $$ A_1 $$ and the second derivative is 0. So when plugging in $$ y_p (x)$$ into (9.0), $$ A_0, A_1 $$ are determined to be:
 * {| style="width:100%" border="0"

.5912076831=2 A_1 $$ $$\displaystyle .2956038416= A_1 $$ $$\displaystyle -.0939750342= -3A_1+2A_0 $$ $$\displaystyle -.0939750342= -3(.2956038416)+2A_0 $$ $$\displaystyle .3964182452=A_0 $$     (9.24)
 * $$\displaystyle
 * $$\displaystyle
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 * }

The general solution, after adding $$ y_h $$ and $$ y_p $$ then becomes:
 * {| style="width:100%" border="0"

y_g(x)=C_1e^{2x} + C_2e^{x} + .2956038416x + .3964182452 $$     (9.25)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

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


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$$  \displaystyle 2C_1e^{2x} + C_2e^{x} + .2956038416 = y'_g(x) $$     (9.26)
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 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
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$$  \displaystyle y_g(-3/4)=1 =C_1e^{2(-3/4)} + C_2e^{-3/4} + .2956038416(-3/4) + .3964182452 $$     (9.27)
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 * }


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$$  \displaystyle y'_g(-3/4)=0= 2C_1e^{2(-3/4)} + C_2e^{-3/4} + .2956038416 $$     (9.28)
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 * }

Solving the equations proves that $$C_1 = -13.92092674, C_2 = 12.54701279 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle y_g(x)=(-13.92092674)e^{2x} + (12.54701279)e^{x} + .2956038416x + .3964182452 $$     (9.29) Graphed below, the approximations are shown. In green, the approximation with n=1, in blue, the approximation with n=0, and in red, the truncated Taylor series as n=1 is shown over the interval of [-3/4 to 3].
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 * }

Author
Solved and Typed By - Egm4313.s12.team1.silvestri (talk) 21:26, 29 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.durrance (talk) 19:02, 30 March 2012 (UTC)