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Problem 1: Proof of exponential of transpose matrix
Report problem 5.1 from.

Find: Prove
Prove ($$)

Solution

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The exponential of a matrix is given by

the transpose of a matrix has the following properties ,

Thus,

Thus,

This proved the ($$)

Problem 2: Proof of exponential of diagonal matrix
Report problem 5.2 from.

Given: The formula of exponential of diagonal matrix
A diagonal matrix

where $$\mathbb{C}^{n\times n}$$ represents the set of $$n\times n$$ matrices with complex coefficients

The exponential of the diagonal matrix is given by

Find: Prove
Prove ($$)

Solution

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The exponential of a scalar is

The exponential of a matrix is

It is then easy to find out

from ($$),

This proved ($$)

Problem 3: Prove property of eigenvalue matrix
Report problem 5.3 from section 20-6

Given: The expression of eigenvalue matrix, the diagonal matrix and exponentiation of matrix
The exponentiation of matrix is given by,

For a matrix$$A$$,

The eigenvalue matrix is made of matrix of eigenvector as shown below. Suppose we have a matrix $$\mathbf{A}$$. The eigenvector is given by,

$$\phi$$ is eigenvector, If there are n linearly independent eigenvectors, we will have,

So the the equation can be written as,

Where,

The matrix can be diagonalizable, which is given by,

Find: The expression of exponentiation of matrix can also be diagonalizable
$$\exp \mathbf A =\mathbf \phi \, \text{Diag}[e^{\lambda_1}\cdots e^{\lambda_n}]\, \mathbf \phi ^{-1}$$

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. Use equation ($$) into equation($$), we have,
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Since $$\mathbf \phi \, \mathbf \phi^{-1} =1$$, we have,

According to associative property, which is given by,

where $$\mathbf A, \mathbf B ,\mathbf C \in \mathbf E^{n\times n}$$. So the equation turns into,

In the problem 2, we have already prove that,

So the equation turns into,

Problem 4: Showing the Decomposition of a Matrix
Based on lecture notes section 20

Given: A Matrix
Also, according to section 20-4, a matrix can be decomposed as follows:

where the right-hand side of the equation consists of the eigenvector matrix, a diagonal eigenvalue matrix and the inverse of the eigenvector matrix, respectively.

Lastly,

Solution

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To start off, we must find the eigenvalues of the matrix above

Taking the determinant and setting it equal to zero, we can see that we get

Using these eigenvalues, we get the following eigenvectors:

Next, using the method of Determinants to find the inverse of a matrix, which states for a matrix:

The inverse is found by:

Employing this technique on ($$) :

Lastly, our diagonal eigenvalue matrix is seen to be:

Therefore, the matrix can be written, in decomposed form, as:

which is identical to ($$)

Now for the second part, from ($$), we can see that the exponential of our matrix can be expressed as:

The diagonal matrix can be re-written using Euler's formula from (5) in section 27-6 to arrive at the final answer:

Problem 5: First Integral for a class of exact L2-ODE-VC
Problem 5 from lecture notes section 21-5.

Given: A class of exact L2-ODE-VC
Given a class of linear, 2nd order, ordinary differential equations with varying coefficients of the form:

Find: A general form of the first integral
Show that the first integral of the form given below generates a class of exact L2-ODE-VC of the form given in ($$).

where

Solution: Derive the first integral

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Integrating ($$) with respect to p yields:

The partial derivatives with respect to x and y are given by

Substituting back into the general form of L2-ODE-VC ($$)

where

Integrating ($$) results in

where

Substituting into ($$) produces

Taking the partial derivative with respect to y and comparing to Q(x) (since $$\phi_y = Q(x)$$ from ($$) and ($$))

Therefore, $$k_1 '$$ cannot be a function of y, and $$k_1$$ must be a constant. This means

Problem 6: Solve certain L2-ODE-VC
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Given: A L2-ODE-VC
The linear 2nd-order ordinary differential equation with varying coefficients(L2-ODE-VC):

Find: Show exactness and solve through integration

 * 1) Show the equation is exact
 * 2) Find first integration $$\phi$$
 * 3) Solve for $$y(x)$$

Solution of the L2-ODE-VC

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This solution was produced without referring to previous solutions. 1, Show exactness:
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First write ($$) into:

Obiviously, ($$) represents that the L2-ODE-VC satisfies the 1st exactness conditin.

Then, as for the 2nd exactness condition, we need to check if ($$) satisfies the following expressions:

we have: $$LHS1=RHS1=-cosx$$, $$LHS2=RHS2=0$$

Thus, ($$)satisfies the 1st and 2nd exactness conditions i.e. it is exact.

2, Find $$\phi$$

As for ($$), we know that there exist a $$\phi$$ which satisfies:

According to ($$), we have:

Moreover, we obtain:

{{NumBlk|:|$$\left\{\begin{matrix} \phi _y=h_y\\ \phi _x=-psinx+h_x\end{matrix}\right.$$|$$}}

Institute ($$) into ($$) and make integration, we obtain:

Where $$k_1$$ is a constant.

Finally, institute ($$) into ($$), we have:

Problem 7: Show the Equivalence
Report problem 5.7 from lecture notes section 22-4.

Given: 2 forms of 2nd exactness condition of N2-ODE
One form of 2nd exactness condition of N2-ODE shown as below

where

The other form of 2nd exactness condition of N2-ODE shown as below

where

Find: show the equivalence of these two forms
Show this equivalence to symmetry of mixed 2nd partial derivatives of first integral $$\phi$$

Solution: expand 1st form to show that it is equal to the 2nd form

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As the definition

The first integral $$\phi$$ is a function of $$(x,y,y')$$ that are considered as 3 independent variables. Thus $$ g_i$$for $$ i=0,1,2 $$ can be expanded as

To make it easier to get the solution, I just expand the derivatives not too far toward the detal(bottom) level; instead, just stay at the more general(top) level as much as possible as shown below.

Thus, Eq.(7.1) can be expanded as

p and q are not generally equal to 0, thus in order to make the Eqn to be equal to 0, the terms of this Eqn must be satiesfied for these conditions as below

Which means

Problem 8: Show the 2nd Exactness Condition
Report problem 5.8 from lecture notes section 22-6.

Given: General Equation of N2-ODE
One general Equation of N2-ODE with the form of 1st exactness condition shown as below

which can be also expressed as below

where

Define

Thus Eq.(8.2) can be rewritten as

Find: show the 2nd Exactness Condition
Since $$1$$ and $$q$$ are lineraly independent ($$q=y''$$ is in general not a constant), we must have

In other words, show the 2nd Exactness about N2-ODE

Solution: work with coefficients in 2nd exactness condition

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One form of 2nd exactness condition of N2-ODE shown as below

where

In Eq.(8.1), $$g$$ and $$f$$ are all functions of $$(x,y,y')$$ that are considered as 3 independent variables. Thus $$ g_i$$ for $$ i=0,1,2 $$ can be expanded as

Next

Thus, Eq.(8.8) can be expanded as

q are not generally equal to 0, thus in order to make the Eqn to be equal to 0, the terms of this Eqn must be satiesfied for these conditions as below

which means

This is 2nd Exactness Condition

Problem 9: Use Maclaurin series to expand equations
Report problem 5.9 from section 64

Find: Maclaurin series expansion and definition of hypergeometric function
The Maclaurin series is an Taylor series expansion of function at x about 0. Which is given by,

Where, $$f^n(x) = \frac{d^n f(x)}{dx^n} $$

Two functions are given by,

$$(1-x)^{-a}$$

$$\frac{1}{x}\arctan (1+x)$$

The notation of hypergeometric function is given by,

Where $$-1< x < 1$$,

$$\begin{cases} a(0):=1\\a(k)=a(a+1)\cdots(a+k-1)\end{cases}$$

Solution

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First function
SInce $$1-x=1$$. the expansion of this function is given by,

According to the notation,when $$ b = c=d$$, the equation satisfied, so the expression is given by,

Second function
The directly expansion of this function is wrong, since the $$\frac{1}{x}$$ is undefined at the zero point and direct derivation is complicated. But we can expand the $$\arctan (x+1)$$ first, then divide it by x.

Using the wolframe Alpha, we can get the derivatives of $$\arctan(1+x)$$, which are given by,

Where,

$$f^{(i)}= \frac{d^i f}{(dx)^i}$$

The derivatives seems like no path. But when we plug $$ x=0$$ into the equations, we will have,

So obviously at x=0, the constant doesn't fit the expressions given in the lecture note.

Personally I think when we use x=-1, we could have,

So we can get the expansion of $$ \arctan(x+1)$$, which is given by,

equation($$) is divided by x, then we get,

So the condition of the problem should be changed, then the answer is satisfied.

Problem 10: plot the local maximum and show equations
Report problem 5.10 from

Find: plot and proof

 * 1) Use MATLAB to plot $$F(5,-10;1,x)$$ near $$x=0$$ to display the local maximum (or maxima) in this region.
 * 2) Show that $$

Solution

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1. As we know

in which

$$\begin{cases}(a)_0:=1\\(a)_k:=a(a+1)(a+2)\cdots(a+k-1)\end{cases}$$

The MATLAB codes for this function called is as below: ( it is not possible to get the $$k=\infty$$, so I set $$ n=k$$ as a number large enough.)

The main program is

The graph is showed below, in which $$x\in [-0.1,0.8]$$. And the red circle indicate the local maximum in the neighbourhood of 0.

In order to get the local maximum of $$F(a,b;c,x)$$, we can find the points of it whose differential value is 0 and make further judgment. Using the codes below to get the specific value in which the differential value is 0 in the neighbourhood of$$x=0$$.

We get the local maximum at $$x_1=0.0717,x_2=   0.2288  ,x_3=  0.4442,x_4=    0.6834 $$ near $$x=0$$, according to $$.

2. If we expand the right hand side of $$. we get

According to $$, the left hand side of $$could be written as

So the LHS=DHS.

proof end

Problem 11: plot and find
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Find: plot and proof

 * 1) For each value of time t, solve for altitude z(t), then plot z(t) versus t.
 * 2) Find the time when the projectile returns to the ground.

Solution

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Substitute the condition $$into the $$, we get

If we use the $$derived previously to get the equation of $$t$$, the answer and graph of it are identical.

MATLAB is unable to get the inverse function of $$z=F^{-1}(t)$$, but we can draw the plot of t when $$z(t)\in[-5,5]$$ as below with MATLAB:

So as we can see, $$t\in(-0.2,0.72)$$. If we exchange the axes of X and Y, then we get:

Unfortunately,It is hard for me to say at which point the projectile return to the ground because the physical meaning of this graph is unclear.

Problem 12: analyze the hypergeometric differential equation and verify the hypergeometric fuction to be a solution of it
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Given: the hypergeometric differential equation and the hypergeometric function
The hypergeometric differential equation:

The hypergeometric function:

where: $$k=1,2,\cdots $$, $$(a)_k=a(a+1)(a+2)\cdots (a+k+1)$$, and $$(b)_k, (c)_k$$ have the same expressions as $$(a)_k$$.

Find: Show exactness and verify solution
1, Is ($$) exact?

2, Is ($$) in the power form

3, Verify that ($$) is indeed a solution of ($$)

Solution

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This solution was produced without referring to previous solutions.
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1, Show exactness:

First write ($$) into:

Where $$p=y'$$.

Obiviously, ($$) represents that the L2-ODE-VC satisfies the 1st exactness conditin.

Then, as for the 2nd exactness condition, we need to check if ($$) satisfies the following expressions:

As for ($$), we have: $$\left\{\begin{matrix}LHS1=1\\ RHS1=ab-a-b-1\end{matrix}\right. \Rightarrow LHS1\neq RHS1$$ and $$LHS2=RHS2=0$$

Thus, ($$)satisfies the 1st but not satisfies the 2nd exactness conditions, i.e. it is not exact.

2, Power form?

We could see the power form as following:

Comparing ($$) with ($$), we know:

$$\alpha $$ and $$r$$ can not be found to satisfy $$ax^r=x(1-x)$$.

Therefore, ($$) is not in the power form.

3, Verify that ($$) is indeed a solution of ($$)

If the statement that "($$) is indeed a solution of ($$)" is true, we could obtain:

where: $$k=1,2,\cdots $$, $$(a)_k=a(a+1)(a+2)\cdots (a+k+1)$$, and $$(b)_k, (c)_k$$ have the same expressions as $$(a)_k$$;

if we institute ($$), ($$) and ($$) into the left hand side of ($$)(which we call $$LHS_{hde}$$), we could have:

Next we will verify if ($$) is true.

Unify each expression to the form of "$$\sum_{k=0}^{\infty} [\cdots \cdots ] \frac{x^{k+1}}{k!}$$":

Sum all the five expressions above, we obtain:

considering the following expression of $$(a)_k$$(Similar expressions could be derived as for $$(b)_{k}$$ and $$(c)_{k}$$.):

We have:

thus, ($$) is true.

Therefore, ($$) is one solution of ($$).

Problem: Verify exactness and Solutions of L2-ODE-VC
Problem R*5.13 from lecture notes section 27.

Given: 2 L2-ODE-VC
Given the following two linear, 2nd order, ordinary differential equations with varying coefficients:

Find: Exactness, Integrating factors and Solution
Verify the exactness of the two equations. If ($$) is not exact, check whether it is in power form and can be made exact using IFM with $$h(x,y) = x^m y^n$$. Also, verify the first few Hermite polynomials as shown below are homogenous solutions to ($$).

Solution

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Setting:

then it is apparent that ($$) satisfies the first exactness condition, i.e., that it is of the form:

Similarly, setting

makes it apparent that ($$) satisfies the first exactness condition.

The second exactness condition is checked using relations (1) and (2) in lecture notes section 16-5 and also by relation (1) in lecture notes section 22-3.

The partial derivatives of ($$) and ($$) are given below.

Plugging into (1) and (2) of 16-5

The two sides of ($$) are only equal if $$n = 0$$.

For relation (1) of 22-3,

Again, the relation is only true for $$n = 0$$. Therefore ($$) is only exact for $$n = 0$$.

The partial derivatives of ($$) and ($$) are given below.

Plugging into (1) and (2) of 16-5

Since ($$) is never true, ($$) is not exact.

For relation (1) of 22-3,

Again, the relation not true and therefore ($$) is not exact.

($$) can be considered to be in the power form given in (3) of 21-2 if

Therefore, an integrating factor of the form $$h(x,y) = x^m y^n$$ can be used. This results in, where $$n$$ in the original equation is replaced with $$s$$ to avoid confusion:

The power coefficients, $$m$$ and $$n$$ must now be determined. The partial derivatives are now given by:

Relation (2) of 16-5 becomes

Therefore, $$n=0$$. Relation (1) of 16-5 now becomes

This implies

Which is satisfied when $$m = 1, s = -2$$. This means the integrating factor that makes the original equation exact is:

Another solution is the first Hermite polynomial:

Which implies $$n=0$$. Therefore

And the differential equation is satisfied.

For $$n=1$$, the Hermite polynomial solution is:

Therefore,

The equation is satisfied.

For $$n=2$$, the Hermite polynomial solution is:

Therefore,

Which indeed is a homogenous solution.

Problem 14: Method of Trial Solution for Euler-Bernoulli Beam
Based on lecture notes Section 30

Given: Undetermined Coefficient, Root of a Characteristic Equation
The Euler-Bernoulli beam equation is given as:

After separation of variables, the X function becomes:

The solution is given to be in the form of:

Find: Solutions to X
Find the expressions of X(x) in terms of $$\displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x $$

Solution

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The general solution takes on the form:

We will start with the imaginary parts and use Euler's formula from (5) in section 27-6

Now, the hyperbolic sine and hyperbolic cosine are defined as follows:

For this problem, we note that

Thus,

Therefore, the solutions have been successfully expressed in terms of $$\displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x$$ as seen from ($$), ($$), ($$), ($$).