User:Egm6321.f12.team3/rep6hid

=R*6.1 Determine the fifth derivative of y w.r.t x in terms of the derivatives of y w.r.t t. = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

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

Solution
In order to solve the homogeneous L2-ODE-VC:

a transformation of the independent variable from x to t was utilized. The transformation is given by the following relation:

The next step is to determine the derivatives of y with respect to the new independent variable t. The dependent variable y is a function of x, which is now a function of t after the transformation has been applied, namely $$y(x(t))$$.

The derivative of y with respect to t is as follows:

The derivative of y with respect to x can be obtained by rearranging ($$):

Differentiating ($$) with respect to t yields:

($$) can be written in shorthand as follows:

where $$y_x := \frac{dy}{dx}$$, $$y_t=\frac{dy}{dt}$$ and $$ \frac{dt}{dx} = \left(\frac{dx}{dt}\right)^{-1} = e^{-t}$$

($$) shows that the differential operator $$\frac{d}{dx}$$ can be expressed in an equivalent form $$\frac{dt}{dx}\frac{d}{dt}$$.

This can be applied to solve for the second derivative of y with respect to x:

By substituting $$\frac{dt}{dx}=e^{-t}$$ into ($$) we arrive at:

Every order derivative of y with respect to x can be determined in the same manner, so to determine $$y_{xxxxx}$$ using ($$)

=R*6.2 Determine the solution of a homogeneous L2-ODE-VC using the trial solution method. = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
1. Solve ($$) using Method 2 with trial solution $$y=x^r$$

2. Plot the solution over the interval x = 1 to x = 2.

Given
The following boundary conditions are used to solve for the two constants of the homogeneous L2-ODE-VC:

Solution
To solve ($$), take the first and second derivative of the trial solution ($$) and substitute those derivatives back into the original ODE ($$).

Substituting ($$) and ($$) into ($$) yields the following:

Factoring $$x^r$$ from ($$) produces the characteristic equation which is inside the brackets:


 * $$x^r\left[r^2-r-2r+2=0\right]$$


 * $$x^r\left[r^2-3r+2=0\right]$$


 * $$x^r\left[(r-1)(r-2)=0\right]$$

From the characteristic equation two distinct roots can be determined:

The solution to the homogeneous L2-ODE-VC has the form:

Now apply the boundary conditions to solve for the two unknown constants $$ c_1$$ and $$c_2$$.

Combining ($$) and ($$) yields the following:

Boundary Condition 1 $$\rightarrow y(x=1)=3 \Rightarrow c_1(1)+c_2(1)^2 = 3$$

Boundary Condition 2 $$ \rightarrow y(x=2)=4 \Rightarrow c_1(2)+c_2(2)^2 = 4$$

Solve ($$) and ($$) for the unknown constants:


 * $$c_1=2-2c_2$$ from $$

Substitute the above expression of $$c_1$$ in terms of $$c_2$$ into ($$) to get:


 * $$2-2c_2+c_2=3 \rightarrow 2-c_2=3 \Rightarrow c_2=-1$$

Plugging $$c_2=-1$$ into ($$) yields:


 * $$c_1+(-1)=3 \rightarrow c_1=4$$

With both constants known the solution is:

To verify the solution, plug ($$) into the original ODE ($$) and check to see if it is satisfied.

($$) is true so the solution has satisfied the original ODE.

Now to plot the solution over the interval from x=1 to x=2.

MATLAB was used to plot the solution y(x) over the interval x=1 to x=2:

The MATLAB code is shown below:



=R*6.3 Equivalence of the Two Trial Solutions for an Euler Ln-ODE-VC = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
An Euler L3-ODE-VC :

There are two methods for solving any general Euler Ln-ODE-VC of the form $$\displaystyle \sum_{i=0}^n a_i x^i y^{(i)} = 0$$.

Method 1: Use a transformation of variables to transform the Euler Ln-ODE-VC into an Euler Ln-ODE-CC

Use the transform $$\displaystyle x = e^t$$ and then use the trial solution $$\displaystyle y = e^{rt}$$ where $$r $$ is a constant.

Method 2 : Use the trial solution $$\displaystyle y = x^r$$.

Problem
Show that the trial solution for Method 2 is equivalent to the combined trial solution for Method 1.

Solution
Method 1

Transform using the trial solution:
 * $$\displaystyle x = e^t$$

We know,


 * $$y'=e^{-t}y_t$$
 * $$y^{(2)}=e^{-2t}(y_{tt}-y_t)$$
 * $$y^{(3)}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

Plugging into ($$), we get,

Further re-arrangement yields,

Now, using the trial solution, $$\displaystyle y = x^r$$ in ($$),

Thus the characteristic equation using Method 1 is :

$$\displaystyle a_3r^3+(a_2-3a_3)r^2+(a_1-a_2+2a_3)r+a_0=0$$

Method 2

Using the trial solution $$\displaystyle y = x^r$$ in ($$),


 * $$ \rightarrow a_3r(r-1)(r-2)x^r+a_2r(r-1)x^r+a_1rx^r+a_0x^r=0 $$

Rearranging the terms, we get the characteristic equation using Method 2 as:

$$\displaystyle a_3r^3+(a_2-3a_3)r^2+(a_1-a_2+2a_3)r+a_0=0$$

Thus we see that the characteristic equations attained using both Method 1 and Method 2 are the same.

This makes sense because if we substitute $$\displaystyle (x = e^t)$$ in the trial solution for Method 1 $$\displaystyle (y = e^{rt})$$, we obtain $$\displaystyle y = x^r$$, the trial solution for Method 2.

=R*6.4 Complete solution for an Euler L2-ODE-VC and an Euler L2-ODE-CC = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Part 1)

1.1) Find $$a_2$$, $$a_1$$ and $$a_0$$ such that ($$) is the characteristic equation of ($$)

1.2) Find the 1st homogeneous solution $$y_1(x)=x^r$$

1.3) Find the complete solution: find $$c(x)$$ such that $$y(x)=c(x)y_1(x)$$

1.4) Find the 2nd homogeneous solution $$y_2(x)$$

Part 2)

2.1) Find $$b_2$$, $$b_1$$ and $$b_0$$ such that ($$) is the characteristic equation of ($$)

2.2) Find the 1st homogeneous solution $$y_1(x)=x^r$$

2.3) Find the complete solution: find $$c(x)$$ such that $$y(x)=c(x)y_1(x)$$

2.4) Find the second homogeneous solution $$y_2(x)$$

1.1
The characteristic equation ($$) can be expanded as follows:

and has these roots:

Assume a trial solution of $$y(x)=x^r$$, then substitute the first and second derivative of the trial solution into the original ODE ($$).

Substituting ($$) and ($$) into ($$) yields:


 * $$a_2x^2r(r-1)x^{r-2}+a_1xr^{r-1}+a_0x^r=0$$

Factoring out the trial solution $$x^r$$ from ($$) will yield the characteristic equation which is the equation inside the brackets:

Rearranging ($$) into three groups: terms multiplied $$ r^2, r^1$$ and $$ r^0 $$ yields:

By comparing ($$) with ($$), it is apparent that:


 * $$ a_2=1$$


 * $$ (a_1-a_2)=-10 \Rightarrow a_1-1=-10$$ $$ \Rightarrow a_1=-9 $$


 * $$ a_0=25$$

1.2
The first homogeneous solution is simply determined by $$y_1(x)=x^r$$ where $$r$$ is the root of the characteristic equation ($$).

1.3
To find the complete solution $$y(x)=c(x)y_1(x)$$, the unknown function $$c(x)$$ must be determined by the method of variation of parameters.

The first step is to take the first and second derivative of $$y(x)$$ and substitute them into the original ODE ($$)


 * $$y(x) = c(x)y_1(x) + c'(x)y'_1(x)+c'(x)y'_1(x)+c(x)y''_1(x)$$

Substituting ($$) and ($$) into the original ODE ($$) yields:

Substitution of the first homogeneous solution $$y_1(x)=x^5$$ yields:

Distributing ($$) and rearranging the terms into three groups: terms multiplied by $$c''(x)$$, $$c'(x)$$ and $$c(x)$$ yields:

Substituting the values of the constants $$a_2$$, $$a_1$$ and $$a_0$$ yields:

Factoring $$x^6$$ from ($$) produces the following:

By defining $$ p(x)=c'(x) $$ the order of the ODE is reduced from 2 to 1

($$) can be rearranged into standard form $$y'(x)+P(x)y(x)=0$$ by dividing through by $$x$$

The term in front of $$p(x)$$ can be identified as $$P(x)$$ and the integrating factor required to make ($$) exact is expressed as $$h(x)=e^{\int P(x)dx+k}$$


 * $$h(x)=e^{\int \frac{1}{x}dx + k} \rightarrow h(x)=e^{ln(x)+k} \rightarrow h(x)=e^{ln(x)}e^k \rightarrow h(x)=xe^k$$


 * $$h(x)\left[p'(x)+\frac{1}{x}p(x)=0\right]$$


 * $$xe^kp'(x)+xe^k\frac{1}{x}p(x)=0$$


 * $$xe^kp'(x)+e^kp(x)=0$$

Factor out and divide through by the constant $$e^k$$


 * $$e^k\left[xp'(x)+ p(x)=0\right]$$


 * $$xp'(x)+ p(x)=0$$

The left hand side of the equation can be expressed as $$\frac{d}{dx}\left(xp(x)\right)$$:


 * $$ \frac{d}{dx}\left(xp(x)\right)=0$$


 * $$ d \left(xp(x)\right)=0$$


 * $$ \int d \left(xp(x)\right) = 0 \rightarrow xp(x) = k_1 \rightarrow p(x) = \frac{k_1}{x}$$

Now that $$p(x)$$ is known, the unknown function $$c(x)$$ can be determined as follows:

$$c(x)=\int p(x)dx+ k_2 \rightarrow c(x)= \int \frac{k_1}{x}dx+k_2 \rightarrow c(x)=k_1ln(x)+k_2$$

where $$k_1$$ and $$k_2$$ are constants of integration.

So the complete solution is:

1.4
The second homogeneous solution $$y_2(x)$$ can be determined by examining the complete solution $$ y(x) $$ which is the linear summation of the two homogeneous solutions.

Distributing the complete solution in ($$) yields:

It is apparent that the second homogeneous solution is:

2.1
Assume a trial solution of the form:

Take the first and second derivative of the trial solution and substitute those derivatives into the original ODE ($$):

Substituting ($$) and ($$) into ($$) yields:

After factoring out the trial solution $$e^{rx}$$ from ($$) we arrive at:

{{NumBlk|:|$$e^{rx}\left[b_2r^2+b_1r+b_0=0\right]$$|$$

By comparing ($$) with ($$) the unknown coefficients can be readily determined:

$$ b_2 = 1$$

$$ b_1=-10$$

$$ b_0=25$$

2.2
The first homogeneous solution $$ y_1(x)=e^{rx}$$ where $$r$$ is the root of the characteristic equation ($$)

2.3
Assuming a complete solution of the form:

Take the first and second derivative of ($$) and substitute those derivatives back into ($$)

Substituting ($$) and ($$) into the original ODE ($$) yields:

Substituting the values of the coefficients $$b_2$$, $$b_1$$ and $$b_0$$ and the first homogeneous solution $$y_1(x)=e^{5x}$$ yields:

Rearranging ($$) into three groups: terms multiplied by $$c''(x)$$, $$c'(x)$$ and $$c(x)$$ yields:

($$) reduces to:

$$c''(x)=0 \rightarrow c(x)=k_1x+k_2$$

So the complete solution is:

2.4
The second homogeneous solution $$y_2(x)$$ can be determined by inspecting the complete solution in ($$)

It is apparent that the second homogeneous solution is:

=R*6.5 Particular solution using variation of parameters method = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Determine the particular solution using variation of constants method

Given
The particular solution is of the form

where $$y_H(x) $$ is the homogeneous solution of the L1-ODE of the form $$\displaystyle y' + P(x)y = Q(x) $$.

Solution
Taking the derivative of ($$),

Taking the derivative of ($$),

From ($$), ($$) and ($$), we can write:

Substituting ($$), ($$) and ($$) in $$\displaystyle y' + P(x)y = Q(x) $$,


 * $$\displaystyle exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)] + P(x)A(x)\cdot exp[-\int P(x)dx]= Q(x) $$


 * $$\displaystyle \rightarrow exp[-\int P(x)dx](A(x)'-P(x)A(x)+A(x)P(x))=Q(x)$$


 * $$\displaystyle \rightarrow exp[-\int P(x)dx]\cdot A(x)' = Q(x)$$


 * $$\displaystyle \rightarrow A(x)' = \frac {Q(x)}{exp[-\int P(x)dx]} = Q(x)\cdot exp[\int P(x)dx] $$


 * $$\displaystyle \rightarrow A(x) = \int Q(x) \cdot exp[\int P(x)dx]dx $$

Thus, paricular solution can be expressed as :

$$\displaystyle y(x) = \int Q(x) \cdot exp[\int P(x)dx]dx \cdot exp[-\int P(x)dx]$$

=R*6.6 Solving of Special IFM=

Part 1
In this problem, y is a function of independent variable t, thus the integral factor $$ h(x,y) $$ here could be written as a function of variables$$ y $$ and$$ t $$ instead of$$ x $$, that is, $$ h(t,y) $$.

First exactness condition,

Thus

Recall the second exactness condition,

Substituting ($$) and ($$) into ($$) and ($$) yields

From ($$), we have

That is, $$ h(t,y) $$ is function of t only, thus $$ h(t,y)=h(t) $$.

For $$ h(t,y)=h(t) $$, we have,

Substituting ($$) into ($$)   yields

Part 2.1
Where $$C_1 $$ is integral constant. Differentiating both sides of ($$)  with respect to t yields

The functions $$y'', y', y$$ are linear independent with each other, thus we have,

Thus

Part 2.2
Because,

We have

Part 2.3
Substituting ($$) into ($$)  yields

Thus

Obviously that ($$) is a L1-ODE-CC which can be solved by the IFM.

Part 2.4
Rearranging ($$) yields

Obviously the right hand side of ($$)is a function of t only, this equation could be divided by $$ \bar a_1 $$ on both sides to be the form of ($$), that is,

Recall ($$), which presents one form of general non-homogeneous L1-ODE-VC and replace the independent variable x with t in this problem,

Comparing ($$) with ($$)yields,

Where $$k_1=\frac{C_1}{\bar a_1} $$ is a constant. According to ($$), the integral factor $$ \bar h(t) $$ for ($$)is

$$ \beta $$ is introduced in this problem to represent $$ \bar a_0/\bar a_1 $$. It had been proved that the integral constant $$ k_0 $$ here is not necessary in previous report. Thus we have,

Therefore, we can find the solution for general excitation $$ f(t) $$ with the help of ($$),

Part 2.5
Definition of $$ \beta $$ in this problem is,

Recall the ($$)and ($$), then we have,

For $$ \alpha + \beta$$,

Thus, $$ \alpha, \beta$$ are roots of the quadratic equation,

Part 2.6
Part I if  $$ \alpha \ne \beta $$,

$$\displaystyle y(t)=\frac{1}{ e^{\beta t }}\left[\int^t \frac{e^{(\beta -\alpha) s}}{\bar a_1}\int e^{\alpha \tau }f(\tau )d\tau ds+k_1 \int^t e^{(\beta-\alpha) s}ds+k_2 \right] $$

$$\displaystyle\Rightarrow y(t)=\frac{1}{ e^{\beta t }}\int^t \frac{e^{(\beta -\alpha) s}}{\bar a_1}\int e^{\alpha \tau }f(\tau )d\tau ds+\frac{k_1}{ e^{\beta t }}\int^t e^{(\beta-\alpha) s}ds+\frac{k_2}{ e^{\beta t }}$$

$$\displaystyle \Rightarrow y(t)=\frac{1}{ e^{\beta t }} \int^t \frac{e^{(\beta -\alpha) s}}{\bar a_1}\int e^{\alpha \tau }f(\tau )d\tau ds+ \frac{k_1}{ e^{\beta t }}(\frac{e^{(\beta-\alpha) t}}{\beta-\alpha}-1)+\frac{k_2}{ e^{\beta t }}$$

$$\displaystyle \Rightarrow y(t)=\frac{1}{ e^{\beta t }} \int^t \frac{e^{(\beta -\alpha) s}}{\bar a_1}\int e^{\alpha \tau }f(\tau )d\tau ds+ \frac{k_1e^{-\alpha t}}{\beta-\alpha}-\frac{k_1}{ e^{\beta t }}+\frac{k_2}{ e^{\beta t }}$$

$$\displaystyle \Rightarrow y(t)=\frac{1}{ e^{\beta t }} \int^t \frac{e^{(\beta -\alpha) s}}{\bar a_1}\int e^{\alpha \tau }f(\tau )d\tau ds+ \frac{k_1}{\beta-\alpha}e^{-\alpha t}+(k_2-k_1) e^{-\beta t}$$

Where,

The characteristic equation for the homogenous counterpart of ($$) is,

Substituting ($$)and ($$)into ($$)yields

Thus the roots of the characteristic equation are $$ \lambda_1=-\alpha,  \lambda_2=-\beta $$

Therefore the homogenous solution for ($$) can be written as,

Comparing ($$)with ($$), it is clear that the last two terms on the right hand side of ($$) form the homogenous solution if we replace the arbitrary constant $$ k_3 $$ with another constant $$ C^1_H $$,and $$ k_4 $$ with $$ C^2_H $$, that is,

Because $$ y(t)=y_P(t)+ y_H(t)$$

Thus we reach the conclusion that

Part II if  $$ \alpha = \beta $$

The roots of the characteristic equation in this case are $$ \lambda_1=\lambda_2=-\alpha $$

Therefore the homogenous solution for ($$) can be written as,

Comparing ($$)with ($$), it is clear that the last two terms on the right hand side of ($$) form the homogenous solution if we replace the arbitrary constant $$ k_2 $$ with another constant $$ C^1_H $$,and $$ k_1 $$ with $$ C^2_H $$, that is,

Because $$ y(t)=y_P(t)+ y_H(t)$$

Thus we reach the conclusion that

Part 2.7
For

If $$ \alpha \ne \beta $$

We have the particular solution as follows

If $$ \alpha = \beta $$

We have the particular solution as follows

Because

Thus, for $$ \alpha \ne \beta $$, substituting ($$) into ($$) yields

Where the integral can be,thus,

Where

For $$ \alpha = \beta $$, substituting ($$) into ($$) yields

Where

Comparing

Part 2.8.1
The characteristic equation is,

Comparing ($$) with ($$) yields

Since $$ \alpha \ne \beta $$, for $$f(t)=e^{-t^2} $$, we have the particular solution as follows

Because

Where $$\phi= \tau-1/2$$.

We have the error function, which could be expressed as,

Thus, clearly we have,

Substituting ($$) into ($$) yields

Substituting ($$) into ($$) yields

Because

$$\displaystyle \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}e^{-3t}erf(t-1/2)+\frac{1}{3}\int^t e^{-3s}erf'(s-1/2) ds$$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}e^{-3t}erf(t-1/2)+\frac{1}{3}\int^{t-1/2} e^{-3(\theta+1/2)}erf'(\theta) d\theta$$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}e^{-3t}erf(t-1/2)+\frac{1}{3}\frac{2}{\sqrt{\pi}}\int^{t-1/2} e^{-3(\theta+1/2)}e^{-\theta^2}d\theta $$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}e^{-3t}erf(t-1/2)+\frac{1}{3}\frac{2}{\sqrt{\pi}}\int^{t-1/2} e^{-(\theta^2+3\theta+9/4)}e^{3/4}d\theta $$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}e^{-3t}erf(t-1/2)+\frac{e^{3/4}}{3}\frac{2}{\sqrt{\pi}}\int^{t-1/2} e^{-(\theta^2+3\theta+9/4)}d\theta $$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}[e^{-3t}erf(t-1/2)-\frac{2e^{3/4}}{\sqrt{\pi}}\int^{t-1/2} e^{-(\theta+3/2)^2}d\theta] $$

$$\displaystyle \Rightarrow \int^t e^{-3s}erf(s-1/2) ds=-\frac{1}{3}[e^{-3t}erf(t-1/2)-\frac{2e^{3/4}}{\sqrt{\pi}}\int^{t+1} e^{-\omega^2}d\omega]$$

Where $$ \theta=s-1/2, \omega=\theta+3/2 $$.

Substituting ($$) into ($$) yields

Since $$ \bar a_1 =a_2 $$, we have the particular solution for nonhomogeneous L2-ODE-CC ($$)  as follows

Where $$ A_1=\frac{-e^{1/4}\sqrt{\pi}}{6\bar a_1} $$.

The homogeneous solution for ($$) whose characteristic equation is $$ (r+1)(r-2)=0 $$ is

Therefore we have the solution for ($$),

Part 2.8.2
The characteristic equation is,

Comparing ($$) with ($$) yields

Since $$ \alpha = \beta $$, for $$f(t)=e^{-t^2} $$, we have the particular solution as follows,

The integral forming ($$) has the similar form of the one forming ($$), thus we can process this integral by same method.

$$\displaystyle y_P(t)= \frac{1}{\bar a_1 e^{-4 t }}\int^t \int^s e^{-(\tau^2+4 \tau+4)}e^4d\tau ds$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4}{\bar a_1 e^{-4 t }}\int^t \int^s e^{-(\tau+2)^2}d\tau ds $$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}\int^t erf(s+2)ds$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}[erf(t+2)t-\frac{2}{\sqrt{\pi}} \int^{t+2} (\sigma-2)\cdot e^{-\sigma^2}d\sigma]$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}[erf(t+2)t-\frac{2}{\sqrt{\pi}} \int^{t+2} \sigma\cdot e^{-\sigma^2}d\sigma+\frac{4}{\sqrt{\pi}} \int^{t+2} e^{-\sigma^2}d\sigma]$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}[erf(t+2)t-\frac{2}{\sqrt{\pi}} \int^{t+2} \sigma\cdot e^{-\sigma^2}d\sigma+2erf(t+2)]$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}[erf(t+2)t-\frac{1}{\sqrt{\pi}} \int^{t+2}e^{-\sigma^2}d\sigma^2+2erf(t+2)]$$

$$\displaystyle \Rightarrow y_P(t)= \frac{e^4\sqrt{\pi}}{2\bar a_1 e^{-4 t }}[(t+2)erf(t+2)+\frac{1}{\sqrt{\pi}} \int^{(t+2)^2}e^{-\varphi}d\varphi]$$

Where $$ \sigma=s+2, \varphi=\sigma^2 $$.

Since $$ \bar a_1 =a_2 $$, we have the particular solution for nonhomogeneous L2-ODE-CC ($$)  as follows

Where $$ A_2 = \frac{e^4\sqrt{\pi}}{2a_2 }$$.

The homogeneous solution for ($$)  whose characteristic equation is $$ (r-4)^2=0 $$ is

Therefore we have the solution for ($$)  ,

Part 2.9
For case in Part 2.8.1, The characteristic equation is,

Thus

The solution for for($$)  is,

Then we could conclude

Where $$ A_1=\frac{-e^{1/4}\sqrt{\pi}}{6\bar a_1} $$.

For the zero initial conditions, we have,

Assume the parameter $$\bar a_1$$ to be 1, evaluating both equations above yields

Values of $$C^1_H$$ and $$ C^2_H$$ can be determined from both equations above as follows.

Thus the solution for ($$) is,



For case in Part 2.8.2, The characteristic equation is,

Thus

The solution for ($$) is,

Then we could conclude

Where $$ A_2 = \frac{e^4\sqrt{\pi}}{2a_2 } $$.

For the zero initial conditions, we have,

Assume the parameter $$\bar a_1$$ to be 1, evaluating both equations above yields

Values of $$C^1_H$$ and $$ C^2_H$$ can be determined from both equations above as follows.

Thus the solution for ($$) is,



=R*6.7 Show that (1) p.34-6 agrees with King 2003 p.8(1.6) = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Part 1
show agrees with

Part 1
Starting from Eq(1) p.34-6

Integrating both side Eq(6.2) becomes.

Recall Integration by parts

Lets evaluate the RHS of Eq(6.7.8) separately

plug Eq(6.7.10) & Eq(6.7.12) into Eq(6.7.8)

u_1(x) cancels out

plug Eq(6.7.1) into Eq(6.715)

Then we arrive to King 2003 p.8(1.6)

case 1
assume that we know u_1(x) as 1st homogenous solution, and by plugging in the case 1 equation we get two unknown and one equation which is difficult to solve.

case 2 & case 3
the only difference is that in Eq 2 p.34-2 we multiply by u_1(x), in this case we multiply U(x) by 1\ u_1(x)(case 2). and we multiply u_1(x) by  1\ U(x) clearly the product rule is easier to handle than the quotient rule.

=R*6.8 Invalid Roots of a Characteristic Equation= On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
The L2-ODE-VC

Trial solution

Correspoding characteristic equation

The two roots of the characteristic equation above are

Problem
Explain why $$ r_2(x)$$ is not a valid root, i.e., $$ u_2(x) = e^{xr_{2}(x)}$$ is not a valid solution.

Solution
It is assumed that $$ r$$ in trial solution is constant, thus root $$ r$$ of characteristic equation should not be a function of x, that is, $$ r \ne r(x)$$.

Actually, assuming there exists a solution for ($$) which equals the trial solution associated with $$ r_2 = \frac{1}{x-1}$$, that is, $$ u_2=e^{\frac{x}{x-1}} $$, substituting this so called solution into ($$) yields conclusion easily that $$ (x-1)u_2''-xu_2'+u_2 \ne 0$$, thus $$ u_2=e^{\frac{x}{x-1}} $$ is not a valid solution.

Consider, if, $$ r$$ in trial solution is not assumed to be constant, but a function of x in general, thus the corresponding trial solution is $$ y=e^{r(x)x} $$, then we have,

Substituting ($$) together with ($$) and expression of y into ($$) yields

Obviously $$ r(x)=\frac{1}{x-1}$$ would satisfy the equation above if it is a valid solution for ($$). Now check the validity of $$ r(x)=\frac{1}{x-1}$$ for satisfying ($$), we have,

Substituting ($$),($$)together with ($$) into left hand side of ($$) yields,

Where $$ f_1(x) $$ and $$ f_2(x) $$ are both polynomial of x in which the highest power of x is less than 3 and 5 respectively, while$$ f_3(x) $$ is polynomial in terms of x in which the highest power of x is less than 3. Obviously LHS could not equal to -1 since greatest power of x in denominator is 3 while 5 in numerator. Thus we have

Therefore $$ r(x)=\frac{1}{x-1}$$ does not satisfy ($$), it is not a valid solution for ($$). =R*6.9 Find second Homogeneous Solution Using Variation of Parameters Method =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
For the L2-ODE-VC (Linear Second Order Differential Equation with Variable Coefficients)

Select a valid homogeneous solution and call it $$\displaystyle u_1$$. Find the second homogeneous solution $$\displaystyle u_2$$ by method of variation of parameters and compare it to $$\displaystyle e^{xr_2(x)}$$

Solution
Let us assume a trial solution $$\displaystyle u_1 = e^{rx} = y$$, where r is a constant.

$$\displaystyle y' = re^{rx}$$

$$\displaystyle y'' = r^2e^{rx}$$

Substituting in ($$)

$$\displaystyle (x-1)(r^2e^{rx}) - x(re^{rx}) + e^{rx} = 0$$

$$\displaystyle e^{rx}((x-1)r^2 - xr +1) = 0$$

$$\displaystyle e^{rx}(xr(r - 1) - (r^2 - 1)) = 0$$

Let us analyze this for all x.

At x=0, for the equation to be zero, $$\displaystyle r = \pm 1$$

For all other x,

$$\displaystyle r(r-1) = 0$$ AND $$\displaystyle r^2-1 = 0$$

From the first condtion, $$\displaystyle r=0,1$$

From the second condition $$\displaystyle r = \pm 1$$

Thus combining all the cases, we get r = 1, for the equation to satisfy.

Thus : $$\displaystyle u_1 = e^x$$

Rewriting ($$)

$$\displaystyle y'' - \frac{x}{(x-1)}y' + \frac{1}{(x-1)}y = 0$$

$$\therefore \displaystyle a_1 = - \frac{x}{(x-1)}$$

From Lecture notes ($$),

$$\displaystyle u_2(x) = u_1(x) \int \frac{1}{{u_1(x)}^2} exp \left[ -\int a_1(x)dx \right]$$

$$\displaystyle u_2 = e^x \int e^{-2x} exp \left[ -\int - \frac{x}{(x-1)}dx \right]$$

Integrating Using Wolfram Alpha ,

$$\displaystyle u_2 = e^x \int e^{-2x}. e^x (x-1) dx$$

$$\displaystyle u_2 = e^x \int e^{-x}(x-1) dx$$

Using Wolfram Alpha to integrate again

$$\displaystyle u_2 = e^x (-e^{-x}x)$$

$$\displaystyle u_2 = -x$$

= R 6.10 Calculation of Time Taken by a projectile to hit the Ground =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
Where $$ _2F_1 (a_1,a_2;b_1;x)$$ is a Hypergeometric Function.

Problem
Consider the integral in (3) Pg.63-8 and ($$)

$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$

a= k/m = 2 and b = g = 10

For each value of n, solve for vertical velocity z(t), plot z(t) vs t, find the altitude y(t) vs time t,and find the time when projectile returns to ground.

Case I: n=2

Case II: n=3

Case I: For n=2
Substituting the values of n,a and b in ($$), we get:

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

$$ = \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{2}; \frac{3}{2};\frac{-2z^2}{10}\right) $$

Or we can directly integrate the given expression,

$$ \int \frac{dz}{2z^2+10} = \frac {1}{10} \int \frac{dz}{\frac{z^2}{5} + 1} $$

Let $$ u = \frac {z}{\sqrt 5}, \therefore u^2 = \frac{z^2}{5}, dz = \sqrt 5 du $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {\sqrt 5}{10} \int \frac{du}{u^2 + 1} $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {1}{2 \sqrt 5} tan^{-1}(u) + k $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {1}{2 \sqrt 5} tan^{-1}(\frac {z}{\sqrt 5}) + k = -t$$

We have initial condition as at t = 0, z = 50

Putting it in the above integral get $$k = -\frac {1}{2 \sqrt 5} tan^{-1}(\frac {50}{\sqrt 5})$$

Thus we have,

$$\frac {1}{2 \sqrt 5} tan^{-1}(\frac {z}{\sqrt 5}) = \frac {1}{2 \sqrt 5} tan^{-1}(\frac {50}{\sqrt 5})- t $$

$$tan^{-1}(\frac {z}{\sqrt 5}) = tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t $$

$$\frac {z}{\sqrt 5} = tan(tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t)$$

Now,

$$y(t) = \int z(t) = \sqrt 5 \int tan(tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t)$$

Integrating using Wolfram Alpha we get,

$$y(t) = \frac{1}{2\sqrt 5} log(cos(2 \sqrt 5 t - tan^{-1}(\frac {50}{\sqrt 5}))$$

The plot for y(t) and z(t) vs t are given below.

From the plot, we see that the projectile hits the ground at t = 0.6825 sec. It may noted that the Plot of Altitude y(t) has been translated to (0,0) for convenience of calculation of Time t when the projectile reaches the ground.

The MATLAB Code for the above plot is as follows:

Case II: For n=3
Substituting the values of n,a and b in ($$), we get:

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^3+10} = \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{3};1+ \frac{1}{3};-2\frac{z^3}{10}\right) + k $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^3+10} = \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{3}; \frac{4}{3};\frac{-2z^3}{10}\right) - \frac{1}{10} 50 _2F_1 \left(1,\frac{1}{3}; \frac{4}{3};\frac{-2(50)^3}{10}\right) = -t$$

Now using Wolfram alpha, we find

$$k = 5 \times 0.041334 = 0.2067$$

$$\therefore \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{3}; \frac{4}{3};\frac{-2z^3}{10}\right) = -t + .2067$$

=R*6.11 Reduction of Order: Method 2=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Given Legendre Equation for n = 2,

$$\displaystyle (1-x^2)y''-2xy'+2(2+1)y=0 $$

$$\displaystyle (1-x^2)y''-2xy'+6y=0 $$

and the first homogeneous solution $$\displaystyle P_2 = \frac{1}{2}(2x^2-1)$$

Find the second homogeneous solution using the Variation of Parameters, Reduction of Order Method 2

Nomenclature
$$\displaystyle P_n(x), n = 0,1,2..$$ is the first homogeneous solution for Legendre Equation of nth order.

$$\displaystyle Q_n(x), n = 0,1,2..$$ is the second homogeneous solution for Legendre Equation of nth order.

Solution
$$\displaystyle y(x)$$ can be represented as $$\displaystyle U(x)u_1(x)$$ where $$\displaystyle U(x)$$ is an unknown function to be found out and $$\displaystyle u_1(x)$$ is a homogeneous solution for $$\displaystyle y(x)$$

In this case, $$\displaystyle u_1(x) = P_2(x) = \frac{1}{2}(2x^2-1)$$

A L-2 ODE-VC can be written as

Comparing the Legendre Equation for n=2,

we have

and

$$\displaystyle y = Uu_1$$

$$\displaystyle y' = U'u_1 + U{u'}_1$$

$$\displaystyle y = Uu_1 + U'{u'}_1 + U'{u'}_1 + U{u}_1 = Uu_1 + 2U'{u'}_1 + U{u''}_1$$

$$\displaystyle y + a_1(x)y' + a_0(x)y = U(a_0u_1 + a_1u'_1 + u_1) + U'(a_1u_1 + 2u'_1) + U''(u_1) = 0$$

But $$\displaystyle u_1$$ is the homogeneous solution for ($$) thus $$\displaystyle a_0u_1 + a_1u'_1 + u''_1 = 0$$

is a L-2 ODE with a missing dependent variable.

We define $$\displaystyle Z = U'$$

Substituting in ($$) we get

$$\displaystyle Z' + Z\frac{a_1u_1 + 2u'_1}{u_1} = 0$$

From ($$)

$$\displaystyle h(x) = exp \left[ \int \left\{a_1(x) + \frac{2u'_1(x)}{u_1(x)}\right\} dx \right]$$

As shown in Section 34

Substituting $$\displaystyle a_1$$ from ($$) and $$\displaystyle u_1 = P_2$$

$$\displaystyle h(x) = \frac{1}{4} (2x^2-1)^2 exp \left[ \int \frac {-2x}{1-x^2}dx \right]$$

$$\displaystyle h(x) = \frac{1}{4} (2x^2-1)^2 exp [ log(x^2-1)]$$

$$\displaystyle h(x) = \frac{1}{4} (2x^2-1)^2 (x^2-1)$$

$$\displaystyle u_2(x) = u_1(x)\int \frac{1}{h(x)}dx $$

$$\displaystyle u_2(x) = \frac {1}{2} (2x^2-1) \int \frac{4}{(2x^2-1)^2(x^2-1)}dx$$

Integrating using Wolframalpha we get

$$\displaystyle u_2(x) = \frac {1}{2} (2x^2-1)\left[\frac{4x}{2x^2-1} + \sqrt 2 log \left(\frac {(2x + \sqrt 2)}{\sqrt 2 -2x}\right) + 2 log \left( \frac{1 - x}{x + 1}\right) \right]$$

This equation does not match with the equation that has to be proved. Thus something is amiss. But the procedure to solve for $$\displaystyle Q_2$$ is exactly as mentioned.

In the equation to be proved, we see a term $$\displaystyle \frac {1}{2} (3x^2-1)$$ instead of $$\displaystyle \frac {1}{2} (2x^2-1)$$, so $$\displaystyle P_2$$ is probably equal to $$\displaystyle \frac {1}{2} (3x^2-1)$$

Substituting $$\displaystyle P_2 = u_1 = \frac {1}{2} (3x^2-1)$$ in ($$)

$$\displaystyle h(x) = \frac{1}{4} (3x^2-1)^2 (x^2-1)$$

We know $$\displaystyle u_2(x) = u_1(x)\int \frac{1}{h(x)}dx $$

$$\displaystyle u_2(x) = \frac {1}{2} (3x^2-1) \int \frac{4}{(3x^2-1)^2(x^2-1)}dx$$

Integrating using Wolframalpha we get

$$\displaystyle u_2(x) = \frac {1}{2} (3x^2-1)\left[\frac{1}{2} \left(\frac{6x}{3x^2-1} + log \left( \frac{1 - x}{x + 1}\right)\right) \right]$$

$$\displaystyle u_2(x) = \frac{3x}{2} + \frac {1}{4}(3x^2-1)log \left( \frac{1 - x}{x + 1}\right) $$

Which is the required solution, so there indeed was an error in printing.

=R*6.12 Solution of a Non-homogeneous Legendre Equation (L2-ODE-VC) by Variation of Parameters=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Given the non-homogeneous Legendre Equation (n = 2),

where $$\displaystyle f(x)=1$$ and the first homogeneous solution $$\displaystyle u_1(x)=P_1(x)=x$$, FIND the final solution $$y(x)$$ by Variation of Parameters

Nomenclature
Let $$\displaystyle u_1(x)$$ be the first homogeneous solution and $$\displaystyle u_2(x)$$ be the second homogeneous solution for Legendre Equation of 2nd order.

Solution
Let us first rearrange the given ODE into the form a general L2-ODE-VC, $$y''+a_1(x)y'+a_0(x)y=f(x)$$

For simplicity, let us assume the final solution in terms of the known first homogeneous solution $$u_1(x)$$,

To avoid repetition and lengthy derivations, Let us make use of the results provided in R*6.11 - ($$) and Lecture Notes Section 34, p.34-4 and p.34-5

Let us first calculate $$h(x)$$

$$\displaystyle h(x)=x^2 exp[\int \frac{-2x}{1-x^2}dx]$$

Now, plugging ($$) in ($$), we get,

$$\displaystyle Z(x)=\frac{1}{x^2(1-x^2)}[k_2+\int x^2(1-x^2) \frac{1}{x(1-x^2)}dx]$$

Using ($$) in ($$), we get,

Finding the integral using Wolfram Alpha

$$\displaystyle U(x)=k_1+\frac{-(2k_2x+x)log(1-x)+(2k_2+x)log(x+1)-4k_2}{4x}$$

Now, plugging ($$) in ($$), we get,

$$\displaystyle y(x)=[k_1-\frac{k_2}{x}+\frac{k_2}{2}log\frac{1+x}{1-x}+\frac{1}{4}log\frac{1+x}{1-x}](x)$$

$$\displaystyle y(x)=[k_1x-k_2+\frac{k_2}{2}x log\frac{1+x}{1-x}+\frac{x}{4}log\frac{1+x}{1-x}]$$

($$) gives the final solution for $$y(x)$$ by Variation of Parameters.

=Contributing Members=

= References =