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Problem 1: Find fifth-order derivative of y with respect to x
Report problem 6.1 from.

Given: transformation of variables
y(x) is a function of variable x. Transformation of variable x gives new variable t,

Find:$$y_{xxxxx}$$
Find $$y_{xxxxx}$$ in terms of the derivatives of y with respect to t.

Solution

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From ($$),

Thus,

Problem 2: Solving a L2-ODE Using Method 2
Based on lecture notes Section 31

Given: A L2-ODE With Boundary Conditions
Boundary Conditions:

Trial Solution:

Find: Solve the L2-ODE Using Trial Solution
Solve the given homogeneous L2-ODE using the trial solution ($$)

Solution

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To start off, we must find the derivatives of the trial solution, which can be easily shown to be:

Now substituting these derivatives and the trial solution in to ($$)

Since $$\displaystyle x^r $$ can not equal zero for all values of r,

Factoring the expression, we get:

Thus, the general solution will look like:

Using the boundary conditions from ($$) and ($$),

Solving the system of equations using substitution, we get:

Finally, we substitute the constants back in to ($$)

Problem 3: Show the Equivalance
Report problem 6.3 from lecture notes section 31-5.

Given: Trial solution in Method 2 and the combined trial solutions in Method 1
One general Equation of Euler Ln-ODE-VC shown as below

which can be also expressed as a compact form below

Two methods for sloving Euler Ln-ODE-VC Method 1: Stage 1: transformation of variables

Stage 2: Trial solution

Method 2: Trial solution

Find: show the Equivalance
show that the trial solution in Method 2 is equivalent ot the combined trial solutions in Method 1

Solution: work with Method 1 and Method 2

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Stage 1
first, we know that $$x=e^t$$, $$\frac{dt}{dx}=e^{-t}$$, thus the general form of Ln-ODE-VC can be written as below

Stage 2
$$\cdot\cdot\cdot$$

Thus, the Ln-ODE-VC can be written as below

To make it more compact, I eliminating$$e^{rt}$$ which is independent with i. thus it can be rewritten as

Method 2
$$\cdot\cdot\cdot$$

Thus, Ln-ODE-VC can be rewritten as

To make it more compact, I eliminating the $$ x^r$$ which is independant with the i.

Apparently, Eq.(3.12) which is generated by Method 1 is equivalant with Eq.(3.19) which is generated by Method 2

Problem 4: Method of Variation of Parameter
Report problem 6.4 from section 32.

Given: The Characteristic Equation
The characteristic equation is given by,

Where $$ \lambda = 5$$ is given. The Euler-L2-ODE-VC is given by,

The Euler-L2-ODE-CC is given by,

Find: The first and complete solution of equations
1. Find $$ a_2,a_1, a_0$$ such that the characteristic function is ($$).

2. Find the first homogeneous solution.

3. Find the complete solution.

4. Find the second homogeneous solution.

5. Repeat the above steps to equation

Solution

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Figure out the coefficient of the equations
So we use the trial solution, which is given by ,

So the derivation of y is given by,

So drive the equation($$)into equation($$), we have,

The characteristic equation is given by,

According to the linear-independence, we can have,

The Euler-L2-ODE-VC then changes into,

Find out the first homogenous solution
The first homogenous solution is given by,

Figure out the complete solution
Use the Variation of Parameters, we assume that the complete solution is given by,

So drive equation ($$)into equation ($$), we have,

Suppose Z=U', we can reduce the order of equation, which is given by,

So now the equation is L1-ODE-VC.When $$ x^5 \neq 0$$, both sides are divided by $$x^5$$, we will have,

Use Integration Factor Method, we have,

So the solution of L1-ODE-VC is given by,

So according to equation ($$), we will have,

So the complete solution is given by,

Figure out the second homogeneous solution
The second homogeneous solution is given by,

Solve the other Euler L2-ODE-CC
So use trial solution, we have,

Drive equation i ($$)into equation ($$), we have,

According to equation and property of linear independence, we have,

So the Rulor-ODE-CC turns into,

Use Method of Variation of Parameters, we assume that the complete solution is given by,

So drive equation ($$) into equation ($$), we have,

So the solution of U is given by,

So the complete solution is given by,

SO the second homogeneous solution is, $$u_2= xe^{5x}$$

Problem 5: Find the particular solution using method of variation of parameters
Report problem 5 from.

Given: A non-homogeneous L1-ODE-VC
The linear 1st-order ordinary differential equation with varying coefficients(L1-ODE-VC):

The homogeneous solution to ($$):

Find: The particular solution to the non-homogeneous L1-ODE-VC
Find the particular solution: $$y_P(x)$$

Solution: Using method of variation of parameters

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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. Consider the following trial solution:
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where $$A(x)$$ is the unknown to be found.

Instituting ($$) into ($$) yeilds:

i.e.

Solve the above equation, we have:

where $$C$$ is an integration constant.

Thus, the solution to ($$) works out to be:

From the solution above, the particular solution could be found:

where $$y_H(x)=exp[-\int_{}^{x}a_0(s)ds]$$ as ($$) already shown.

Problem 6: Special IFM to solve nonhomogeneous L2-ODE-VC con't
Report problem 6.6 from lecture notes section 33-1.

Given: Nonhomogeneous L2-ODE-CC cont'd
One general Equation of Euler L2-ODE-VC shown as below

Find: some specific Questions
1. Find the PDEs that govern the integrating factor h(x,y) for the Eq.(6.1) Recall the 2 relations in the 2nd exactness condition for N2-ODEs. solve these PDEs for h(x,y) 2. Trial solution for the integrating factor $$ h(t)= e^{\alpha t}$$, which is similar to the trial solution for the Euler L2-ODE-CC (a homogeneous Ln-ODE-CC), where $$ \alpha$$ is unknown to be determined.

Because of the integrating factor in exponential form, assume the l.h.s. take the form:

Clearly $$\bar a_2=0$$ to avoid having $$y'''$$ when differentiating the r.h.s; an advantage is to reduce the order of the resulting ODE.

2.1. Find $$(\bar a_1,\bar a_0)$$ in terms of $$(a_0,a_1,a_2)$$ 2.2. Find the quadratic eqution for $$\alpha $$ Use the Eq. shown as below

2.3. Reduced-order equation: Eq.(6.3) and Eq.(6.4) lead to

a L1-ODE-CC, thus easily solvable by the IFM

2.4. Use the IFM to solve

2.5. show that

thus $$(\alpha,\beta)$$are roots of the quadratic equation:

which is the same as Eq.(6.6)

2.6. deduce the particular solution $$y_P(t)$$ for general excitation $$f(t)$$

2.7. verify result with table of particular solutions for:

2.8. solve the nonhomogeneous L2-ODE-CC with the following excitation: Gaussian distribution:

for the coefficients $$ (a_0,a_1,a_2) $$, consider two different characteristic equations: 2.8.1.

2.8.2

2.9. for each case in 2.8.1 and 2.8.2, determine the fundamental period of undamped free vibration. Plot the homogeneous soln $$y_H(t)$$ for about 5 periods, the partucular solution $$y_P(t)$$ for the excitation for the same time interval, and the complete solution $$y(t)$$, assuming ero initial conditions.

Solution: solve the problems

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Part 1
Using the IFM to solve the problem, thus the general form of Ln-ODE-VC can be written as below

Because $$ h(t,y) $$ is a function only with respect of y. Thus $$ h(t,y) $$ can be rewritten as $$ h(t) $$ Eq.(6.15) can be rewritten as

which is satiesfied the 1st exactness condition:

where

Then calculate that

Then examine that with the 2nd exactness condition

which can be specified

solve the 2 eqn, then yeild

Thus this problem can be solved by solving Eq.(6.23)

2.1
Given that

Substituting Eq.(6.24) into Eq.(6.23)

Substituting Eq.(6.24) into Eq.(6.3)

Because $$ a_2=o$$, thus Eq. (6.26) can be rewritten as

differentiate the right side of the Eq.(6.27)

Solving for $$(\bar a_1,\bar a_0)$$ in terms of $$(a_0,a_1,a_2)$$ as below

2.2
Equate Eq.(6.30) and Eq.(6.31) yeild

which can be rearranged as below

2.3
Thus

2.4
Solve the problem by IFM

By definition

Then the integration of the h(t) can be rewritten as below

Thus

where

Substitute the $$ \bar h(t) B(t)$$ into the Eq.(6.41) yeilds

2.5
Substituting Eq.(6.29)(6.30)(6.31) into the Eq. shown below

Thus

2.6
By Euler L2-ODE-CC, I add the integrating constants $$k_1, k_2$$ to make Eq.(6.42) more general

where

2.7
substitute $$ f(t)=te^{bt}$$ into the Eq.(6.45) yeild

use Wolfram-Alpha integrating Eq.(6.46) yeilds

Which can be rewritten as

2.8.1
Given that

two solution are$$ r_1=-1,r_2=2 $$,in other words,$$ \alpha=-1,\beta=2 $$ Thus, two homogeneous solution for the original function are

As Gaussian distribution

use Wolfram-Alpha integrating Eq.(6.46) yeilds

Thus,

2.8.2
Given that

two solution are$$ r_1=r_2=4 $$,in other words,$$ \alpha=\beta=4 $$ Thus, two homogeneous solution for the original function are

As Gaussian distribution

use Wolfram-Alpha integrating Eq.(6.46) yeilds

Thus

Problem 7: Show Equivalence between two expressions
Report problem 6.7 from section 35.

Given: Two expressions of paticular solution of y
From lecture note, the L2-ODE-VC is given by,

The particular solution is given by

Where the $$ u_1(x) $$ is the 1st homogeneous solution and h(x) is integration factor.

In King's book, for the same equation, the particular solution is given by,

Where $$ u_1, u_2$$ are the 1st and 2nd homogeneous solution, and W(s) is wroskian function, which is given by,

Some choices of variation of parameters,

Find: equivalence of two expressions and the feasibility
1. Show the equivalence of two expressions of particular solution 2 Test the feasibility of the other choices of variation of parameters

Solution

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Show the equivelance
According to lecture note, the expression of the 2nd homogeneous solution is given by,

So we can derive the below expression,

So the derivative of equation can be expressed by,

The expression of King's book can be reorganized into below,

So for equation($$), use integrations by parts , we will have,

So the equivalence has been prooved.

check the feasibility
For the equation, we just consider the$$'+'$$, since the other situation is the same. So we have,

So the equation turns into.

So we just circle back to the original equation and nothing can be done to reduce the order.

For the second choice, we have,

So drive equation back to equation, the coefficient of U is given by,

Obviously, there is no way to make it equal to zero, so we can not reduce the order of equation.

For the third choice,we have

We can see that this choice has make the order of $$U$$ even higher, which strongly implies that the equation cannot be solved by this choice.

Problem 8: Roots of a Trial Solution
Based on lecture notes Section 35

Given: Homogeneous L2-ODE
For the above equation, using a trial solution of:

and a characteristic equation of:

We are given the two roots of the equation as:

Find: Validity of the Root and Solution
Explain why ($$) is not a valid root, i.e., $$\displaystyle u_2(x) = e^{xr_2(x)} $$ is not a valid solution.

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.
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To start off, we must find the derivatives of the trial solution

Now substituting these derivatives and the trial solution in to ($$)

From here it is quite obvious that:

And thus, ($$) is not a valid root.

Given:
1.L2-ODE-VC:

2.Trial solution:

3.Characteristic equation:

the roots are:

Find:

 * 1) select a valid homogeneous solution, and call it $$u_1$$.
 * 2) Find the 2nd homogeneous solution $$u_2(x)$$ by variation of parameters, and compare to $$e^{xr_2(x)}$$.

Solution

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1.select 1st homogeneous solution: The valid root is $$r_1=1$$, so the valid 1st homogenous solution is $$e^{xr_1}=e^x$$.

2.Find 2nd homogeneous solution by variation of parameters. We get the operation of variation of parameters as follows.

Now substitute the terms in $$with that from $$:

follow the operations from$$to $$, then we get:

So the 2nd homogeneous solution is $$u_2(x)=x$$.

Comparing to $$e^{xr_2}=e^{\frac{x}{x-1}}$$ reflects that $$r_2(x)$$ is not a valid root.

Given:
The schematic of the motion of a particle in the air :

Derive the equations of motion:

Consider the case $$k\neq 0$$ and $$v_{x0}=0$$: ($$) turns into

Which can be written as :

Separation of variables

For all values of $$n$$ :

Give particular values of some parameters :

Initial vertical velocity:$$z(t):=v_y(0)=50$$

Find:

 * For each value of $$n$$, find the vertical velocity $$z(t)$$ vs. time t, plot this function.
 * For each value of $$n$$, find the altitude $$y(t)$$ vs. time t, find the time when the projectile returns to the ground.

Use numerical methods to find the value of each given value of time t if the explicit expression cannot be obtained.


 * 1) $$n=2$$
 * 2) $$n=3$$

Solution

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According to $$, $$and$$, the uniform expression can be written as:

The inverse funciton of $$can be show as below in terms of $$t$$:

So we get:

When n=2
The $$-b come to be:

With the initial value$$-c:

The accurate number of $$k$$ is $$\frac{1536841641999493}{4503599627370496}$$ byWolframApha and WolframApha, and the form of $$can be written as:

So we can plot the figure of z vs t as follow:

The inverse function of $$can be obtained by Matlab command $$ g = finverse(f,var) $$as follows :

So we can plot the figure of t vs z as follow:

Integrate as $$shows we get:

when the projectile reach the ground, y, the altitude, reach 0. So we need to solve the t in this equation below:

UseWolframAlpha to get the answer. The answer is

We can see that the expression of y is a  periodic function, the positive value which is nearest to 0 is $$0.6825$$, at which time the projectile reach the ground.

So can we solve this problem by numerical methods. The program and plot obtained by MATLAB are showed below

When n=3
The $$-b come to be:

With the initial value$$-c:

The number of $$k\doteq0.20667022059582652141$$ by WolframAlpha, so we just write the expression as below :

So we can plot the figure of z vs t as follow:

The inverse function of $$cannot be obtained by MATLAB, So numerical methods is needed.

Discard the part that contains imaginary number for these have no physical meaning, then exchange the axes of z vs t to t vs z, get the graph like this:

Then integrate z to get y. When the projectile reach ground, the value of y should reach 0, which means at the t in which the area of blue and red are equal in $$.The plot and MATLAB programs are showed below.

The answer, according to the $$, is $$0.5024$$, at which time the projectile reach the ground.

*When n=1
The $$-b come to be:

With the initial value$$-c:

The number of $$k=\frac{\log{11}}{2}$$ by WolframAlpha, so we just write the expression as below :

So we can plot the figure of z vs t as follow. Discard the part that contains imaginary number afterwards for these have no physical meaning:

The inverse function of $$showed below can be obtained by MATLAB, and so can numerical methods.

Then integrate z to get y. When the projectile reach ground, the value of y should reach 0, which means at the t in which the area of blue and red are equal in $$.

We can get the answer is$$5.49991$$, at which time the projectile reach the ground, from WolframAlpha. And so can we get it from numerical methods.The plot and MATLAB programs are showed below.

Problem 11: Find the 2nd homogeneous solution of the Legendre equation
Problem R*6.11 from lecture notes section 37.

Given: The 1st homogeneous solution
For Legendre's equation, given by:

The first homogeneous solution when $$n=2$$ is

Find: The 2nd homogeneous solution
Show that

is the second homogeneous solution using variation of parameters.

Solution: Reduction of order formula

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First, note the typo in the problem definition. The 1st homogeneous solution should be

The variation of parameters method is only used to find the particular solution to a non-homogeneous differential equation once the homogeneous solutions are already known. However, we are trying to find the second homogeneous solution to the homogeneous Legendre equation, therefore the reduction of order formula will be used.

First, Legendre's equation is rearranged.

Now we can evaluate the inner integral of the reduction of order formula. Setting $$$$

From this it follows

Therefore the reduction of order formula now becomes

Using the WolframAlpha integrator, the right hand side evaluates to

After rearranging and simplifying, the 2nd homogeneous equation as shown in the problem statement is found.

Problem 12: Find the final solution by variation of parameters
Report problem 12 from.

Given: A non-homogeneous Legendre equation and the 1st homogeneous solution
The non-homogeneous Legendre equation(L2-ODE-VC)

Given the 1st homogeneous solution being

Find: The final solution to the non-homogeneous Legendre equation
Find the final solution $$y(x)$$

Solution: Using method of variation of parameters

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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. Consider the following trial solution
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where $$A(x)$$ is the unknown to be found.

Instituting ($$) into ($$) yeilds

Rearrange ($$)

Assume that

then ($$) turns to be

Note that ($$) is a L1-ODE-VC which could be solved directly or by IFM.

Noticing that

{{NumBlk|:|$$ \left\{\begin{matrix} \phi_{Bx}=1-3x^2 \\ \phi_{xB}=2(1-2x^2) \end{matrix}\right.\Rightarrow \phi_{Bx}\neq \phi_{xB}$$|$$}}

i.e. ($$) is not exact and IFM should be used to solve it

Multiply ($$) with certain integration factor $$h(x)$$

Recall the 2nd exactness condition $$\bar{\phi}_{Bx}=\bar{\phi}_{xB}$$, we have:

i.e.

where $$k_1$$ is an integrating constant.

Moreover we could get the solution to ($$)

where $$k_2$$ is an integrating constant.

Integrating ($$) yeilds

where $$k_3$$ is another integrating constant.

Plug ($$) back into ($$), we get the final solution to ($$):

where $$C_1$$ and $$C_2$$ are two arbitrary constants.