User:Egm6321.f12.team4/Report5

Problem R5.1 – Proof of Exponentiation of a matrix
sec20-2b

Statement
Show that $$ exp [A^T] = exp[A]^T $$

Solution
We begin by assuming $$ A^T $$ as a 2x2 matrix ,


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$$ A^T= \begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix} $$     (5.1.1)
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Since,
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$$ exp A^T = \sum_{k=0}^{\infty} \frac{(A^T)^k}{k!} $$     (5.1.2)
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$$ A^T*A^T= \begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix}*\begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix} = \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix} $$     (5.1.3)
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$$ A^T*A^T*A^T= \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix}*\begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix}= \begin{bmatrix} 0 & t^3\\ -t^3 & 0 \end{bmatrix} $$     (5.1.4)
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Similarly,


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$$ (A^T)^k= \begin{bmatrix} -t^k & 0\\ 0 & -t^k \end{bmatrix} $$ if k is even (5.1.5)
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$$ =\begin{bmatrix} 0 & t^k\\ -t^k & 0 \end{bmatrix} $$ if k is odd (5.1.6)
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Putting the above in Eq. (5.1.2),
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$$ \begin{bmatrix} 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. & -\frac{t^1}{1!}+\frac{t^3}{3!}-\frac{t^5}{5!}+..\\ +\frac{t^1}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}+.. & 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. \end{bmatrix} $$ (5.1.7)
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Since,
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$$ sin t = t-\frac{t^3}{3!} + \frac{t^5}{5!} + \frac{t^7}{7!} + ... $$ (5.1.8)
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$$ cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} + \frac{t^6}{6!} + ... $$    (5.1.9)
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Eq. (5.1.7) can be rewritten in terms of (5.1.7) and (5.1.8) as ,


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$$ exp[A^T]=\begin{bmatrix} cos t & -sin t\\ sin t & cos t \end{bmatrix} $$ (5.1.8)
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Taking transpose of $$ A^T $$


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$$ A= \begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix} $$     (5.1.9)
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Since,
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$$ exp A = \sum_{k=0}^{\infty} \frac{(A)^k}{k!} $$     (5.1.10)
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$$ A*A= \begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix}*\begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix} = \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix} $$     (5.1.11)
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$$ A^T*A^T*A^T= \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix}*\begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix}= \begin{bmatrix} 0 & -t^3\\ t^3 & 0 \end{bmatrix} $$     (5.1.12)
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Similarly,


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$$ (A^T)^k= \begin{bmatrix} -t^k & 0\\ 0 & -t^k \end{bmatrix} $$ if k is even (5.1.13)
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$$ =\begin{bmatrix} 0 & -t^k\\ t^k & 0 \end{bmatrix} $$ if k is odd (5.1.14)
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Putting the above in Eq. (5.1.2),
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$$ \begin{bmatrix} 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. & \frac{t^1}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}+..\\ -\frac{t^1}{1!}+\frac{t^3}{3!}-\frac{t^5}{5!}+.. & 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. \end{bmatrix} $$ (5.1.15) which can be rewritten as ,
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$$ exp[A]=\begin{bmatrix} cos t & sin t\\ -sin t & cos t \end{bmatrix} $$ (5.1.16)
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From Eq. (5.1.8) & (5.1.16), it is obvious that


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$$ \begin{bmatrix} cos t & -sin t\\ sin t & cos t \end{bmatrix} = \begin{bmatrix} cos t & sin t\\ -sin t & cos t \end{bmatrix}^T $$ (5.1.17)
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Hence,
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$$ exp [A^T] = exp[A]^T $$ (5.1.18)
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Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Jason Alphonso

Problem R5.2 – Proof of Exponentiation of a diagonal matrix
sec20-2b

Statement
Consider a diagonal matrix $$ [D]=\begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} $$ where $$ d_1, d_2 , ... , d_n $$ are complex numbers.

We have to show that $$ exp([D])=\begin{bmatrix} e^d_1 & 0 & 0 & .. &0 \\ 0& e^d_2 &  &  & \\ .& & e^d_3 &  & \\ .& &  & . & \\  0&  &  &  &e^d_n \end{bmatrix} $$

Solution
We start by considering


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$$ exp D = \sum_{k=0}^{\infty} \frac{(D)^k}{k!} $$     (5.2.1)
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Evaluating individual terms ,


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$$ D*D = \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} * \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} = \begin{bmatrix} d_1^2 & 0 & 0 & .. &0 \\ 0& d_2^2 &  &  & \\ .& & d_3^2 &  & \\ .& &  & . & \\  0&  &  &  &d_n^2 \end{bmatrix} $$     (5.2.2)
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$$ D*D*D = \begin{bmatrix} d_1^2 & 0 & 0 & .. &0 \\ 0& d_2^2 &  &  & \\ .& & d_3^2 &  & \\ .& &  & . & \\  0&  &  &  &d_n^2 \end{bmatrix}* \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} = \begin{bmatrix} d_1^3 & 0 & 0 & .. &0 \\ 0& d_2^3 &  &  & \\ .& & d_3^3 &  & \\ .& &  & . & \\  0&  &  &  &d_n^3 \end{bmatrix} $$     (5.2.3)
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Similarly ,


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$$ D^k = \begin{bmatrix} d_1^k & 0 & 0 & .. &0 \\ 0& d_2^k &  &  & \\ .& & d_3^k &  & \\ .& &  & . & \\  0&  &  &  &d_n^k \end{bmatrix} $$    (5.2.4)
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Putting Eq (5.2.2), Eq (5.2.3), Eq (5.2.4) in Eq. (5.2.1) ,


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$$ exp (D) = \begin{bmatrix} 1+d_1+\frac{d_1^2}{2!}+\frac{d_1^3}{3!}+...+\frac{d_1^k}{k!} & 0 & 0 & .. &0 \\ 0& 1+d_2+\frac{d_2^2}{2!}+\frac{d_2^3}{3!}+...+\frac{d_2^k}{k!} &  &  & \\ .& & 1+d_3+\frac{d_3^2}{2!}+\frac{d_3^3}{3!}+...+\frac{d_3^k}{k!} &  & \\ .& &  & . & \\  0&  &  &  &1+d_n+\frac{d_n^2}{2!} + \frac{d_n^3}{3!}+...+\frac{d_n^k}{k!} \end{bmatrix} $$    (5.2.5)
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Since we know,


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$$ exp (x) = 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + .. + \frac{x^n }{n!} $$    (5.2.6)
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Using the above in Eq. (5.2.5)


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$$ exp (D) = \begin{bmatrix} e^{d_1} & 0 & 0 & .. &0 \\ 0& e^{d_2} &  &  & \\ .& & e^{d_3} &  & \\ .& &  & . & \\  0&  &  &  &e^{d_n} \end{bmatrix} $$    (5.2.7)
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Hence proved.

Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Jason Alphonso

Problem R5.3 – Exponentiation of Diagonalizable Matrix
sec20-4

Statement
Show that
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$$ exp[\mathbf A] = \Phi Diag[e^{\lambda_1}, e^{\lambda_2},...,e^{\lambda_n}] \Phi ^{-1} $$     (5.3.1)
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when


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$$ [ \mathbf \Lambda] = Diag[\lambda_1, \lambda_2, ... \lambda_n] \in \mathbb C^{n \times n} $$ (5.3.2)
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Given
From the solution of R5.2 above, it can be shown that
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$$ exp[D] = Diag [e^{d_1}, e^{d_2}, ..., e^{d_n}] \in \mathbb C^{n \times n} $$ (5.3.3)
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when


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$$ [D]= Diag [d_1, d_2, ... d_n] $$     (5.3.4)
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Solution
To solve this problem, one should be aware of the matrix theory
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$$ (TAT^{-1})^n=TA^nT^{-1} $$     (5.3.5)
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This theory can be shown as follows:
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$$ (TAT^{-1})^n=(TAT^{-1})(TAT^{-1})...(TAT^{-1})=TA(T^{-1}T)A(T^{-1}T)...A(T^{-1}T)AT^{-1} $$
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Since $$ TT^{-1} = 1 $$
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$$ TA(T^{-1}T)A(T^{-1}T)...A(T^{-1}T)AT^{-1}=TAAA...AT^{-1}=TA^nT^{-1} $$
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The exponentiation of $$ \mathbf A $$ can be written as
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$$ exp[\mathbf A]=\sum_{n=0}^{\infty} \frac{1}{k!} A^k $$     (5.3.6)
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Taking $$ \mathbf A $$ to be a diagonalizable matrix, $$ \mathbf A $$ can be decomposed as $$ [\mathbf A] = [\Phi \mathbf \Lambda \Phi^{-1}] $$. Therefore
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$$ exp[\mathbf A]=exp[\Phi \mathbf \Lambda \Phi^{-1}]= \sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda \Phi^{-1}]^k $$     (5.3.7)
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From the matrix theory described above in equation 5.3.5, one can see that
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$$ \sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda \Phi^{-1}]^k=\sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda ^k \Phi^{-1}] $$     (5.3.8) Since $$ \Phi $$ and $$ \Phi ^{-1} $$ do not depend on k, they can be pulled outside of the summation as follows
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$$ \Phi \sum_{n=0}^{\infty} \frac{1}{k!} [\mathbf \Lambda ^k ] \Phi^{-1} $$     (5.3.9)
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The summation in equation 5.3.9 can be written in the form shown in equation 5.3.6:
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$$ exp[\mathbf A] = \Phi exp[\mathbf \Lambda] \Phi ^{-1} $$     (5.3.10)
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From equations 5.3.2-5.3.4, one can see that
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$$ exp[\mathbf A] = \Phi Diag[e^{\lambda_1}, e^{\lambda_2}, ..., e^{\lambda_n}] \Phi ^{-1} $$
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Thus equation 5.3.1 is valid.

Author and References

 * Solved and Typed by -- Kaitlin Harris
 * Reviewed by -- Seong Hyeon Hong

Problem R5.4 – An Example of a Matrix Exponential
sec20-5

Statement
Show the process of equation (1) and (2)

Solution
We know that matrix A with real value components can always be in the following form,

Comparing this with equation (1), we can say that,

 (a) Finding Eigenvalues 

Let's say that we only know matrix A and we need to find eigenvalues and eigenvectors. Then we have,

From the above matrix we can have the following relations,

In order to hold equation 5-4-6 to be true, either $$ \Phi $$ has to be a zero matrix or $$A - \Lambda$$ has to be linearly independent. Since we want $$ \Phi $$ to be non-trivial we can come up with the following equation.

Now we can solve for eigenvalues using the above equation.

(b)Finding Eigenvectors Let $$ \phi_1 = \begin{bmatrix} x_1 \\ y_1\end{bmatrix} $$ and $$ \phi_2 = \begin{bmatrix} x_2 \\ y_2\end{bmatrix} $$

If we let $$ x_1 = 1 $$ then $$ y_1 = -i $$

In the same way, we can find $$\phi_2$$

If we let $$ x_2 = 1 $$ then $$ y_2 = i $$

Finally we have the following eigenvalue matrix and its inverse.

Finally, we came up with equation (1) which is,

(c) Taking the exponential of A

From R5.2 and R5.3 we already know that,

Therefore we can write the equation as below and carry out the matrix multiplication.

Recall: Euler's Formula

If we apply Euler's Formula we now have the following.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Kaitlin Harris

Problem R*5.5 – Generate a class of exact L2-ODE-VC
sec21-5

Statement
Derive the following equation,

from (5.5.1) and (5.5.2) below,

Solution
Comparing equations (5.5.1) and (5.5.2) the coefficients of y can be equated. This is not true for the coefficient of the y and y' terms hence,

Integrating the wrt to p,

where h(x,y) is a function only of x and y only.

Similarly $$ \phi_{y}$$,

Rearranging the terms in (5.5.5)and(5.5.6) in (5.5.1)

Further rearranging the terms within the equation, (also realizing p=y')

Comparing now with (5.5.2), the terms in $$ R(x)y = \frac{\partial h(x,y)}{\partial x} $$ and $$ Q(x)= P'(x)+ \frac{\partial h(x,y)}{\partial y}$$

To find $$\phi$$, integrate all terms with respect to x

Therefore substituting value of h(x,y) from (5.5.9) into equation (5.5.4)

which is equivalent as the problem statement,

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Rui Che

Problem R*5.6 - Solve a L2-ODE
sec18-2 sec21-6

Statement
L2-ODE-VC

Find
1.Show equation 5.6.1 is exact

2.Find Φ

3.Slove for $$y(x)$$

1.Show equation 5.6.1 is exact
Let

We obtain

It satisfies the 1st exactness condition

Recall the 2nd exactness condition for L2-ODE-VC:

We substitute all equations above into (4.4.3) and (4.4.4), we obtain

It satisfies the 2nd exactness condition

It is proved

End

2.Find Φ
We know (5.6.1) is exact,so we obtain

We obtain

Finally, we obtain

End

3.Solve for $$y(x)$$
Recall that

From (5.6.5) and (5.6.6), we obtain

Rearrange, we have

Use integration factor

Finally, we obtain

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Seong Hyeon Hong

Problem R*5.7 – Symmetry of mixed partial derivatives
sec22-4

Statement
Using the general form of the 2nd exactness condition for Nn-ODE, show the equivalence to symmetry of mixed 2nd partial derivatives of the first integral $$ \phi $$

Solution
We know that $$ \phi $$ is a function of x,y,y'.

Taking the total derivative of $$ \phi $$ w.r.t x.


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$$ \frac{d}{dx}\phi= \phi_x + \phi_y y' + \phi_{y'} y'' $$ (5.7.1)
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Since,


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$$ g_i = \frac{\partial }{\partial y^{i}}(\frac{d}{dx}\phi) $$ (5.7.2)
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$$ g_0= \frac{\partial}{\partial y} (\frac{d\phi}{dx}) = \phi_{xy}+\phi_{yy}y'+\phi_{yy'}y'' $$ (5.7.3)
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$$ g_1= \frac{\partial}{\partial y'} (\frac{d\phi}{dx}) = \phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y'' $$ (5.7.4)
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$$ g_2= \frac{\partial}{\partial y''} (\frac{d\phi}{dx}) = \phi_y' $$ (5.7.5)
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For the exactness condition,


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$$ g_0 - \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$ (5.7.6)
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Differentiating $$ g_2$$ with respect to x once,


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$$ \frac{d}{dx} g_2 = \phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y'' $$ (5.7.7)
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Putting in the values of $$ g_0, g_1, g_2 $$ from Eqns. (5.7.2 - 5.7.4) and 5.7.7,


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$$ g_0 - \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$
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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} (\phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y) + \frac{d}{dx} (\phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y) =0 $$ (5.7.8)
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The above can be rearranged to be written as


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (5.7.9)
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Considering the part outside the derivative,


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y = (\phi_{xy}+\phi_{yy}y'+\phi_{yy'}y) - (\phi_{yx}+\phi_{yy}y'+\phi_{yy'}y)$$ (5.7.10)
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Putting (5.7.10) in (5.7.9),


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$$ \phi_{xy}-\phi_{yx} - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (5.7.11)
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For the terms to cancel out we should have the following relations,


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$$ \phi_{xy}=\phi_{yx}, \phi_{xy'}=\phi_{y'x}, \phi_{yy'}= \phi_{y'y} $$
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Hence Proved.

Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Kaitlin Harris

Problem R*5.8-Show equivalence of two forms of 2nd exactness condition
sec16-5 sec21-7 sec22-5 sec22-6

Statement
Case n=2:N2-ODE

Form2

Form1

or equivalently 2nd exactness condition for N2-ODEs

Show this equivalence

Solution
From sec22-5(1)

Define

From sec21-7(3)

We obtain

Derivation

From Eq(5.8.6)

From Eq(5.8.7)

Put Eq(5.8.5), Eq(5.8.8) and Eq(5.8.9) into Eq(5.8.1), yields

Define

So, Eq(5.8.10) can be transferred to

Since 1 and q are linearly independent, we must have

Finally we have

It is proved. The same to sec16-5(2)

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Rui Che

Problem R5.9-Use Taylor series to derive
sec64-4 sec64-5 sec64-7

Statement
Use Taylor series at $$x=0$$ (aka Maclaurin series) to derive

and

Solution
First, recall Taylor series

At $$x=0$$, it means that $$a=0$$ in Eq(5.9.1), yields

i) $$f(x)=(1-x)^{-a}$$
Now consider the derivation

We obtain

Substitute Eq(5.9.3) into Eq(5.9.2), we obtain

Recall Pochhammer's symbol

So, let $$n=k$$ Eq(5.9.4) can be transferred to

Recall hybergeometric function

Finally, we obtain

End

ii) $$f(x)=\frac{1}{x}\arctan(1+x)$$
Before I start to solve this problem, Rui Che and I recognize that the equation in the lecture(sec64-7) is totally wrong. So we modify it in our assumption.

We are solving $$f(x)=\frac{1}{x}\arctan(x)$$ in stead of $$f(x)=\frac{1}{x}\arctan(1+x)$$

Recall

We obtain

Now consider the derivation

We obtain

Put Eq(5.9.6) and Eq(5.9.7) into Eq(5.9.2), yields

Let $$ n=2k $$

Rearrange Eq(5.9.9), we obtain

Finally, we obtain

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Rui Che

Problem R5.10 – Plotting Hypergeometric Function
sec64-9b

Statement
Use matlab to plot F(5, 10; 1; x) near x = 0 to display the local maximum (or maxima) in this region.

Show that

Solution
(a) Matlab Code 

(b) Matlab Result 

(c) Graph 

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Kaitlin Harris

Problem R5.11 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral in


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

$$ \int_{z(0)=0}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (5.11.1)
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 * <p style="text-align:right">
 * }


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$$ \int \frac{dz}{az^n+b}= \frac{1}{b}z\ _2F_1 \left(1,\frac{1}{n};1+ \frac{1}{n};-a\frac{z^n}{b}\right)+k $$     (5.11.2)
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 * <p style="text-align:right">
 * }

Let value of n=3, a=2 and b=10

Find,

(1) The altitude z(t) for different values of time (t) (2) Plot z(t) vs t (3) Find time where projectile returns to ground. i.e. t when z(t)= 0

Solution
Part 1 - To find the altitude z(t) for different values of time (t)

Substituting the values of n=3, a=2 and b=10 in equation (5.11.2),


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

$$ \int \frac{dz}{2z^3+10}= \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{3};1+ \frac{1}{3};-2\frac{z^3}{10}\right)+k $$     (5.11.3)
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 * }


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$$ \int \frac{dz}{2z^3+10}= \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right)+k $$     (5.11.4)
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 * <p style="text-align:right">
 * }


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

$$ \int_{z(0)=0}^{z(t)=z}\frac{dz}{2z^3+10}= \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.5)
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 * <p style="text-align:right">
 * }

From equation (5.11.1) this equates to -t,


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

$$ -t = \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.6)
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 * <p style="text-align:right">
 * }


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

$$ t = - \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.7) which is the solution for Part 1 of the problem.
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 * }

Part 2 - To draw the graphs of z(t) vs time.

The graph as represented by equation (5.11.7) shown above,

NOTE: As seen from the graph, the solution includes real and imaginary solution which is not representative of projectile motion. We try a different approach as detailed below.

Using the definition of hypergeometric function,
 * {| style="width:100%" border="0"

$$ t = -\frac{z}{10}\sum_{k=0}^{\infty} \frac {(1)_k(1/3)_k}{(4/3)_k} \frac{z^{3k}}{(-5)^kk!} $$     (5.11.8)
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 * <p style="text-align:right">
 * }


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

$$ t = -\frac{z}{10}(1 + \frac {(1)(1/3)}{(4/3)} \frac{z^{3}}{(-5)} + \frac {(1)(2)(1/3)(4/3)}{(4/3)(4/3+1)} \frac{z^{6}}{(-5)^22!} + \frac {(1)(2)(3)(1/3)(4/3)(7/3)}{(4/3)(7/3)(7/3+1)} \frac{z^{9}}{(-5)^33!} + ...+\frac {n!(1/3)(4/3)(7/3)..(1/3+n-1)}{(4/3)(7/3)...(1/3+n-1)(1/3+n)} \frac{z^{3n}}{(-5)^nn!} $$      (5.11.9)
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 * }

Canceling terms we get,


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

$$ t = -\frac{z}{10}\sum_{k=0}^{\infty} \frac {1}{1+3n} \frac{z^{3k}}{(-5)^k} $$     (5.11.10)
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 * <p style="text-align:right">
 * }


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

$$ 10t + \sum_{k=0}^{\infty} \frac {1}{1+3n} \frac{z^{3k+1}}{(-5)^k} = 0 $$     (5.11.11)
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 * <p style="text-align:right">
 * }

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Pushkar Mishra

Statement
Given a hypergeometric differential equation:

(1) Verify whether it is exact;

(2) Verify whether it is in the power form:


 * where

(3)Verify that $$F(a,b;c;x)$$ in (1) p.64-4 as shown below is indeed a solution of (5.12.0.1):


 * where Pochhammer's symbol is denoted as:

Solution (1)
First exactness condition:

Equation (5.12.0.1) can be written in the form:

where

Then it is obvious that the first exactness condition holds.

For the second exactness condition:

The partial derivatives in these two equations can be derived as:

Since $$f$$ is only a function of x,

On the other hand,

Thus,

Substituting (5.12.1.6)~(5.12.1.12) into (5.12.1.3) and (5.12.1.4), these two relations of the second exactness condition can be written as:

Then the first relation does not hold. So the second exactness condition does not hold and equation (5.12.0.1) is not exact.

Solution (2)
By comparing (5.12.0.1) and (5.12.0.2), it is clear that the term $$f(x)=x(1-x)$$ can not be written in the form of $$\alpha x^r$$, because the former expression has two linearly independent terms (with different exponents 1 and 2) while there is only one exponent in the latter term.

Similarly, $$[c-(a+b+1)x]y'$$ cannot also be written in the form of $$\beta x^sy'$$.

Although the term $$-aby$$ can be written as $$\gamma x^ty$$ if we let $$\gamma=-ab$$ and $$t=0$$, the overall equation is not in power form.

Solution (3)
Assume

as is shown in (5.12.0.3).

Then we can substitute it into the left part of equation (5.12.0.1):

Here the derivative for k=0,1 in the first term equal to 0, since they are the derivative of a constant:

For k=0:

For k=1:

Similarly the second term in (5.12.3.2) for k=0 is also equal to 0.

Then we can further reorganize (5.12.3.2) as:

To verify that this expression equals to zero, each coefficient of the linearly independent polynomial terms($$1,x,x^2,x^3,\cdots$$) should equal to zero. In order to evaluate these coefficients, we shall first make a index transformation for the first and the third terms of (5.12.3.5):

Then (5.12.3.5) becomes:

Then it can be further written as:

Because of the derivative process, the first and the fourth terms exists for only k>0 (same for k'), while second term exists for only k>1. Thus we shall examine k=0, k=1 and k>1 separately.

For k=0, namely for the coefficient of polynomial term '1':

For k=1, namely for the coefficient of polynomial term 'x':

For any term k>1, a general coefficient form for each of the $$x^k$$ polynomial from (5.12.3.8):

It can be further reorganized as:

Thus, any coefficient for the polynomial terms is equal to 0. Then the overall expression for the left part of (5.12.0.1), namely the expression (5.12.3.2) equals to 0.

Then we can conclude that the expression (5.12.0.3) is a solution of the equation (5.12.0.1).

Author and References

 * Solved and Typed by -- Rui Che

Problem R*5.13 – Exactness of Classical Special Functions
sec27-2

Statement
For the Legendre and Hermite equations,

1. Verify the exactness. For the second exactness condition, use two methods. Method 1 can be found at (1) and (2) sec16-5. Method 2 can be found at (1) sec22-3.

2. If the Hermite equation is not exact, check if it obeys the power form and see if it can be made exact via the Integrating Factor Method.

3. Verify the following are homogeneous solutions of the Hermite Differential Equation: a. $$ H_0 = 1 $$ b. $$ H_1 = 2x $$ c. $$ H_2 = 4x^2-2 $$

Given
The Legendre Differential Equation is written as
 * {| style="width:100%" border="0"

$$ (1-x^2)y'' - 2xy' + n(n+1)y=0 $$     (5.13.1)
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 * }

The Hermite Differential Equation is written as
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$$ y''-2xy'+2ny = 0 $$     (5.13.2)
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 * }

The first exactness condition states that $$ \phi (y',y,x) = k $$ exists such that
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$$ G(y,y',y,x) = \frac{d \phi (y',y,x)}{dx} = g(x,y,p)+f(x,y,p) y = 0 $$     (5.13.3)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }

where one defines $$ p=y' $$. The ODE must be able to be written in the form shown in equation 5.13.1.

The second exactness condition can be stated in two methods. Method 1, as shown above in the problem statement, is written as
 * {| style="width:100%" border="0"

$$ f_{xx}+2Pf_{xy}+P^2f_{yy}=g_{xp}+Pg_{yp}-g_y $$     (5.13.4a) and
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 * }
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$$ f_{xp}+Pf_{yp}+2f_y=g_{pp} $$     (5.13.4b)
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 * }

such that $$ f_{xx}=\frac{\partial }{\partial x}\frac{\partial f}{\partial x} $$, etc.

Method 2, as shown above in the problem statement, is written as
 * {| style="width:100%" border="0"

$$ g_0 - \frac{dg_1}{dx} + \frac{d^2 g_2}{dx^2}=0 $$     (5.13.5) such that $$ g_i = \frac{\partial G}{\partial y^{(i)}} $$.
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 * <p style="text-align:right">
 * }

The power form for a L2-ODE is
 * {| style="width:100%" border="0"

$$ \alpha x^ry'' + \beta x^sy' + \gamma x^ty $$     (5.13.6) and should the equation not be exact, the Integrating Factor is
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 * <p style="text-align:right">
 * }
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$$ h(x,y) = x^my^n $$. (5.13.7)
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 * <p style="text-align:right">
 * }

Part A: The Legendre Differential Equation
The Legendre Differential Equation is given by equation 5.13.?. One can define $$ g(x,y,p) $$ and $$ f(x,y,p) $$ as follows:
 * {| style="width:100%" border="0"

$$ g(x,y,p):= -2xy' + n(n+1)y = -2xp + n(n+1)y $$     (5.13.8)
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 * <p style="text-align:right">
 * }


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$$ f(x,y,p):= 1-x^2 $$     (5.13.9) Thus $$ g(x,y,p) + f(x,y,p)y'' = 0 $$. Since this is of the form expressed in equation 5.13.3, the first exactness condition is satisfied.
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 * <p style="text-align:right">
 * }

From equations 5.13.4a and 5.13.4b, one needs to take the various derivatives of $$ f $$ and $$ g $$ to show the second exactness condition for method 1. They are calculated as follows:
 * {| style="width:100%" border="0"

$$ f_{xx}=\frac{\partial }{\partial x} \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}[-2x]=-2 $$     (5.13.10)
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 * <p style="text-align:right">
 * }


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

$$ f_{xy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial x}=\frac{\partial }{\partial y}[-2x]=0 $$     (5.13.11)
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 * <p style="text-align:right">
 * }


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

$$ f_{yy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}[0]=0 $$     (5.13.12)
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 * <p style="text-align:right">
 * }


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$$ g_{xp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial x}=\frac{\partial }{\partial P}[-2p]=-2 $$     (5.13.13)
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 * }


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$$ g_{yp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial y}=\frac{\partial }{\partial P}[n(n+1)]=0 $$     (5.13.14)
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 * <p style="text-align:right">
 * }


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$$ g_{y}=\frac{\partial g}{\partial y}=n(n+1) $$     (5.13.15)
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 * }


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

$$ f_{xp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial x}=\frac{\partial }{\partial P}[-2x]=0 $$     (5.13.16)
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 * <p style="text-align:right">
 * }


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$$ f_{yp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial y}=\frac{\partial }{\partial P}[0]=0 $$     (5.13.17)
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 * }


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$$ f_y=\frac{\partial f}{\partial y} =0 $$     (5.13.18)
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 * }


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$$ g_{pp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial P}=\frac{\partial }{\partial P}[-2x]=0 $$     (5.13.19)
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Plugging equations 5.13.10-5.13.19 into equations 5.13.? and 5.13.?, one can see that
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$$ -2+2p(0)+p^2(0)=-2+p(0)-n(n+1) $$     (5.13.20)
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and


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$$ 0+p(0)+2(0)=0 $$     (5.13.21)
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 * }

From equation 5.13.20, it can be seen that the Legendre Differential Equation is only exact if $$ n=0,-1 $$. Thus for $$ n\ne 0,-1 $$, the equation is not exact and the IFM would need to be used to solve the equation.

For the second method, one can see that
 * {| style="width:100%" border="0"

$$ g_0 = \frac{\partial G}{\partial y^{(0)}}=n(n+1) $$     (5.13.21)
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 * <p style="text-align:right">
 * }


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$$ g_1 = \frac{\partial G}{\partial y^{(1)}}=-2x $$     (5.13.22)
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 * }


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$$ g_2 = \frac{\partial G}{\partial y^{(2)}}=1-x^2 $$     (5.13.23)
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 * }


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$$ \frac{dg_1}{dx} = -2 $$     (5.13.24)
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 * <p style="text-align:right">
 * }


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$$ \frac{d^2 g_2}{dx^2} = -2 $$     (5.13.25)
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 * }

Plugging equations 5.13.21-5.13.25 into equation 5.13.5, one can see that
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$$ n(n+1)-(-2)+(-2)=0 $$     (5.13.26)
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 * }

Thus the Legendre Differential Equation is only exact if $$ n=0,-1 $$. Thus for $$ n\ne 0,-1 $$, the equation is not exact and the IFM would need to be used to solve the equation. This is consistent with the result of Method 1.

Part B: The Hermite Differential Equation.
The Hermite Differential Equation is given by equation 5.13.2. One can define $$ g(x,y,p) $$ and $$ f(x,y,p) $$ as follows:
 * {| style="width:100%" border="0"

$$ g(x,y,p):= -2xy'+2ny=-2xp+2ny $$     (5.13.27)
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 * }


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

$$ f(x,y,p):= 1 $$     (5.13.28) Thus $$ g(x,y,p) + f(x,y,p)y'' = 0 $$. Since this is of the form expressed in equation 5.13.3, the first exactness condition is satisfied.
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 * <p style="text-align:right">
 * }

From equations 5.13.?a and 5.13.?b, one needs to take the various derivatives of $$ f $$ and $$ g $$ to show the second exactness condition for method 1. They are calculated as follows:
 * {| style="width:100%" border="0"

$$ f_{xx}=\frac{\partial }{\partial x} \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}[0]=0 $$     (5.13.29)
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 * <p style="text-align:right">
 * }


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$$ f_{xy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial x}=\frac{\partial }{\partial y}[0]=0 $$     (5.13.30)
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 * <p style="text-align:right">
 * }


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$$ f_{yy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}[0]=0 $$     (5.13.31)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }


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

$$ g_{xp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial x}=\frac{\partial }{\partial P}[-2p]=-2 $$     (5.13.32)
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 * <p style="text-align:right">
 * }


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

$$ g_{yp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial y}=\frac{\partial }{\partial P}[2n]=0 $$     (5.13.33)
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 * <p style="text-align:right">
 * }


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$$ g_{y}=\frac{\partial g}{\partial y}=2n $$     (5.13.34)
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 * <p style="text-align:right">
 * }


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

$$ f_{xp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial x}=\frac{\partial }{\partial P}[0]=0 $$     (5.13.35)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }


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

$$ f_{yp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial y}=\frac{\partial }{\partial P}[0]=0 $$     (5.13.36)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ f_y=\frac{\partial f}{\partial y} =0 $$     (5.13.37)
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 * <p style="text-align:right">
 * }


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

$$ g_{pp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial P}=\frac{\partial }{\partial P}[-2x]=0 $$     (5.13.38)
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 * <p style="text-align:right">
 * }

Plugging equations 5.13.29-5.13.38 into equations 5.13.4a and 5.13.4b, one can see that
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$$ 0+2p(0)+p^2(0)=-2+p(0)-2n $$     (5.13.39)
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 * }

and


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$$ 0+p(0)+2(0)=0 $$     (5.13.40)
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 * <p style="text-align:right">
 * }

From equation 5.13.39, it can be seen that the Hermite Differential Equation is only exact if $$ n=-1 $$. Thus for $$ n\ne -1 $$, the equation is not exact and the IFM would need to be used to solve the equation.

For the second method, one can see that
 * {| style="width:100%" border="0"

$$ g_0 = \frac{\partial G}{\partial y^{(0)}}=2n $$     (5.13.41)
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 * <p style="text-align:right">
 * }


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$$ g_1 = \frac{\partial G}{\partial y^{(1)}}=-2x $$     (5.13.42)
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 * <p style="text-align:right">
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$$ g_2 = \frac{\partial G}{\partial y^{(2)}}=1 $$     (5.13.43)
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$$ \frac{dg_1}{dx} = -2 $$     (5.13.44)
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$$ \frac{d^2 g_2}{dx^2} = 0 $$     (5.13.45)
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Plugging equations 5.13.41-5.13.45 into equation 5.13.5, one can see that
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$$ 2n-(-2)+0=0 $$     (5.13.46)
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Thus the Hermite Differential Equation is only exact if $$ n=-1 $$. Thus for $$ n\ne -1 $$, the equation is not exact and the IFM would need to be used to solve the equation. This is consistent with the result of Method 1.

Part 2. If the Hermite equation is not exact, check if it obeys the power form and see if it can be made exact via the Integrating Factor Method.
The Hermite Equation is not exact if $$ n\ne -1 $$. For this case, one needs to see if the equation is of the power form. The power form was given in equation 5.13.?. Therefore if $$ \alpha = 1 $$, $$ r = o $$, $$ \beta =-2 $$, $$ s=1 $$, $$ \gamma =2n $$, and $$ t = o $$, the Hermite Differential Equation is of the power form $$ \forall n \in \mathbb {R} $$. As such, the IFM can be used such that the integrating factor is given in equation 5.13.?. The calculations are as follows:


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$$ x^my^n[y-2xy'+2ky]=x^my^ny-2x^{m+1}y^ny'+2kx^my^{n+1}=0 $$     (5.13.47)
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The $$ n $$ in the Hermite Differential Equation was redefined as $$ k $$ to avoid confusion with the Integrating Factor. From 5.13.47, one can define $$ g(x,y,p) $$ and $$ f(x,y,p) $$ as follows:
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$$ g(x,y,p):= -2x^{m+1}y^ny' + 2kx^my^{n+1} $$     (5.13.48)
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$$ f(x,y,p):= x^my^n $$     (5.13.49)
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From equations 5.13.4a and 5.13.4b, one needs to take the various derivatives of $$ f $$ and $$ g $$ to solve for $$ m,n $$. They are calculated as follows:
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$$ f_{xx}=\frac{\partial }{\partial x} \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}[mx^{m-1}y^n]=m(m-1)x^{m-2}y^n $$     (5.13.50)
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$$ f_{xy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial x}=\frac{\partial }{\partial y}[mx^{m-1}y^n]=mnx^{m-1}y^{n-1} $$     (5.13.51)
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$$ f_{yy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}[nx^my^{n-1}]=n(n-1)x^my^{n-1} $$     (5.13.52)
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$$ g_{xp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial x}=\frac{\partial }{\partial P}[-2(m+1)x^my^np+2kmx^{m-1}y^{n+1}]=-2(m+1)x^my^n $$     (5.13.53)
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$$ g_{yp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial y}=\frac{\partial }{\partial P}[-2(m+1)x^my^np+2kmx^{m-1}y^{n+1}]=-2nx^{m+1}y^{n-1} $$     (5.13.54)
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$$ g_{y}=\frac{\partial g}{\partial y}=-2nx^{m+1}y^{n-1}p+2k(n+1)x^my^n $$     (5.13.55)
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$$ f_{xp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial x}=\frac{\partial }{\partial P}[mx^{m-1}y^n]=0 $$     (5.13.56)
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$$ f_{yp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial y}=\frac{\partial }{\partial P}[nx^my^{n-1}]=0 $$     (5.13.57)
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$$ f_y=\frac{\partial f}{\partial y} = nx^my^{n-1} $$     (5.13.58)
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$$ g_{pp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial P}=\frac{\partial }{\partial P}[-2x^{m+1}y^n]=0 $$     (5.13.59)
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Plugging equations 5.13.50-5.13.59 into equations 5.13.4a and 5.13.4b, one can see that
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$$ m(m-1)x^{m-2}y^n+2pmnx^{m-1}y^{n-1}+p^2n(n-1)x^my^{n-2}=-2(m+1)x^my^n+p(-2)nx^{m+1}y^{n-1}+2nx^{m+1}y^{n-1}p-2k(n+1)x^my^n $$     (5.13.60)
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and


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$$ 0+p(0)+2(nx^my^{n-1}=0 $$     (5.13.61)
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From equation 5.13.61, one can see that $$ 2nx^my^{n-1}=0 \therefore n=0 $$. Substituting this result back into equation 5.13.60, one can see that
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$$ m(m-1)x^{m-2}=-2(m+1)x^m-2kx^m $$     (5.13.62) This equation will only be valid if both sides are equal to zero. Thus it can be shown that $$ m=0 $$ when $$ k=-1 $$ and $$ m=1 $$ when $$ k=-2 $$. Since $$ n=0 $$, $$ m\ne 0 $$, because then the integrating factor would be 1. Therefore, $$ m=1 $$, and the integrating factor is $$ h(x,y)=x^1y^0=x $$.
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Part A: $$ H_0(x)=1 $$
When $$ H_0(x)=1 $$, $$ y(x)=1 $$, $$ y'(x)=0 $$, and $$ y''(x)=0 $$. Therefore the Hermite Differential Equation becomes
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$$ 0=0-2x(0)+2k(1) $$     (5.13.63) This is a valid homogeneous solution if k=0.
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Part B: $$ H_1(x)=2x $$
When $$ H_0(x)=2x $$, $$ y(x)=2x $$, $$ y'(x)=2 $$, and $$ y''(x)=0 $$. Therefore the Hermite Differential Equation becomes
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$$ 0=0-2x(2)+2k(2x) $$     (5.13.64) This is a valid homogeneous solution if k=1.
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Part C: $$ H_2(x)=4x^2-2 $$
When $$ H_0(x)=4x^2-2 $$, $$ y(x)=4x^2-2 $$, $$ y'(x)=8x $$, and $$ y''(x)=8 $$. Therefore the Hermite Differential Equation becomes
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$$ 0=8-2x(8x)+2k(4x^2-2) $$     (5.13.65) This is a valid homogeneous solution if k=2.
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Author and References

 * Solved and Typed by -- Kaitlin Harris
 * Reviewed by -- Seong Hyeon Hong

Problem R*5.14 – Using Euler's formula find the expression for X(x) in terms of cos Kx, sin Kx, cosh Kx and sinh Kx
sec27-6 sec30-8

Statement
Find the expressions for


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$$ X(x)= e^{rx} $$     (5.14.1)
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in terms of Cos Kx, Sin Kx, Cosh Kx and Sinh Kx

Solution
Euler's formula based on Eq(5) sec27-6


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$$ e^{ikx}=cos kx+ i sin kx $$ (5.14.2)
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Trigonometrical definition of $$ e^{ikx} $$ in terms of cosh kx and sinh kx,


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$$ e^{\pm ikx}=cosh(ikx)\pm sinh(ikx) $$     (5.14.3)
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From vibration analysis problem,


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$$ X^{(4)}- K^{4} x = 0 $$     (5.14.4)
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$$ r_{1,2}=\pm K, r_{3,4}=\pm iK $$ (5.14.5)
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Based on Eq(1) sec30-8


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$$ X(x)= \sum_{i=1}^{4}c_{i}e^{r_{i}x} $$     (5.14.6)
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$$ X(x)= c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+c_{4}e^{r_{4}x} $$     (5.14.7)
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Substituting from (5.14.5) for values of $$ r_{1}, r_{2}, r_{3}, r_{4} $$


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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx}+c_{3}e^{iKx}+c_{4}e^{-iKx} $$     (5.14.8)
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Substituting from Eq. (5.14.2) and (5.14.3),


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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx} + c_{3} (cos{Kx}+i\sin{Kx})+c_{4}(cosh{Kx}-i\sinh{Kx}) $$     (5.14.9)
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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx} + c_{3}\cos{Kx}+c_{4}\cosh{Kx} +i (c_{3}\sin{Kx}-c_{4}\sinh{Kx}) $$     (5.14.10)
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which separates it into real and imaginary parts.

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Pushkar Mishra

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.