User:Egm6321.f12.team1.cla/report6

=Problem 6.1* - Higher Order Derivatives=

Link to lecture notes

Given:

 * $$x=e^t$$

Find:
The fifth derivative of y with respect to x $$y_{xxxxx}$$

Solution:
The first derivative of y with respect to x is found by taking the derivative of the fourth derivative of y.
 * $$y_{xxxxx}=\frac{dy_{xxxx}}{dx}$$

Since y is a function of x and x is a function of t the chain rule has to be applied
 * $$y_{xxxxx}=\frac{dt}{dx}\frac{d}{dt}y_{xxxx}$$

Since the 4th derivative of y with respect to x was previously found we can plug it in
 * $$y_{xxxxx}=\frac{dt}{dx}\frac{d}{dt}\left[e^{-4t}(t_{ttt}-6y_{ttt}+11y_{tt}-6_{t})\right]$$

The derivative of t with respect to x, $$\frac{dt}{dx}$$, is the reciprocal of $$\frac{dx}{dt}$$ and is equal to $$e^{-t}$$
 * $$y_{xxxxx}=e^{-t}\frac{d}{dt}\left[e^{-4t}y_{tttt}-6e^{-4t}y_{ttt}+11e^{-4t}y_{tt}-6e^{-4t}y_{t} \right ]$$


 * $$y_{xxxxx}=e^{t}\left[e^{-4t}y_{ttttt}-4e^{-4t}y_{tttt}-6e^{-4t}y_{tttt}+24e^{-4t}y_{ttt}+11e^{-4t}y_{ttt}-44e^{-4t}y_{tt}-6e^{-4t}y_{tt}+24e^{-4t}y_{t} \right ]$$


 * $$y_{xxxxx}=e^{-5t}\left[y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_{t} \right ]$$

=Problem 6.2* Find the solution assuming a trial solution =

Given
With Boundary Conditions as:

Condition 1:

Condition 2:

Condition 3:

Find
Find the solution assuming trial solution as:

$$  y=x^r $$ where r = constant and plot the solution.

Solution
$$ x^2y''-2xy'+2y=0 $$

Trial solution: $$ y=x^r $$

Taking the derivatives of the trial solution $$ y'=rx^{r-1} $$

$$ y''=r(r-1)x^{r-2} $$

Substituting the value of $$ \displaystyle y' $$ and $$ \displaystyle y'' $$ in equation ($$)

$$ x^r[r^2-3r+2]=0 $$

Now $$ x^r \neq 0 $$ therefore :

$$ r^2-3r+2=0 $$

The roots of the characteristic equation are: $$ \begin{cases}r_1=1\\r_2=2\\ \end{cases} $$

Hence the solution can be expressed as:

$$     y(x)=C_1x+C_2x^2 $$

Where $$ \displaystyle C_1 $$ and $$ \displaystyle C_2 $$ are constants need to be determined by applying boundary conditions.

Condition 1: $$ \begin{cases}C_1+C_2=3\\2C_1+4C_2=4\\ \end{cases} $$

Solving these simultaneous equations we get:

$$ \begin{cases}C_1=4\\C_2=-1\\ \end{cases} $$

The solution is thus:

$$ y(x)=-x^2+4x $$

The plot of the solution using Wolfram Alfa

Condition 2: $$ \begin{cases}C_1+C_2=-2\\2C_1+4C_2=5\\ \end{cases} $$

Solving these simultaneous equations we get:

$$ \begin{cases}C_1=-6.5\\C_2=4.5\\ \end{cases} $$

The solution is thus:

$$ y(x)=4.5x^2-6.5x $$

The plot of the solution using Wolfram Alfa

Condition 3: On applying boundary conditions we get:

$$ \begin{cases}C_1+C_2=-4\\2C_1+4C_2=7\\ \end{cases} $$

Solving these simultaneous equations we get:

$$ \begin{cases}C_1=-11.5\\C_2=7.5\\ \end{cases} $$

The solution is thus:

$$ y(x)=7.5x^2-11.5x $$

The plot of the solution using Wolfram Alfa

=Problem 6.3* - Show that the two methods to solve Euler Ln-ODE-VC are equivalent =

Given:
Euler Ln-ODE-VC

The Two methods for solving this Ln-ODE-VC are:

Method 1:


 * Stage 1: Transformation of variables


 * Stage 2: Trial solution

Method 2:Trial solution

Show:
Show that the trial solution in Method 2 is equivalent to the combined trial solution in Method 1.

Solution:
Stage 1:
 * Method 1:
 * Let $$x=e^t$$ rewrite $$y_x^{(n)}$$ in terms of $$y_t^{(n)}$$
 * $$y_x=y'=e^{-t}y_t$$
 * $$y_{xx}=y''=e^{-2t}(y_{tt}-y_t)$$
 * $$y_{xxx}=y'''=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

Stage 2:
 * Let $$y=e^{rt}$$ $$\Rightarrow y_t=r\,e^{rt}$$

Then
 * $$y_x=y'=e^{-t}y_t=e^{-t}r\,e^{rt}=r\,e^{(r-1)t}$$
 * $$y_{xx}=\frac{dy_x}{dx}=\frac{d}{dx}r\,e^{(r-1)t}=r(r-1)\,e^{(r-2)t}$$
 * $$y_{xxx}=\frac{dy_{xx}}{dx}=\frac{d}{dx}r(r-1)\,e^{(r-2)t}=r(r-1)(r-2)\,e^{(r-3)t}$$
 * $$y^{(i)}=r(r-1)(r-2)...(r-i+1)\,e^{(r-i)t} $$
 * $$y^{(i)}=r(r-1)(r-2)...(r-i+1)\,e^{(r-i)t} $$

Plugging $$y^{(i)},x=e^t $$ into Equation($$) yeilds
 * $$\sum_{i=0}^{n}a_ix^i y^{(i)}= \sum_{i=0}^{n}a_ie^{it}r(r-1)(r-2)...(r-i+1)\,e^{(r-i)t}=0$$

$$\Rightarrow \sum_{i=0}^{n}a_ie^{it}r(r-1)(r-2)...(r-i+1)\,e^{(r-i)t}=0$$ $$\Rightarrow \sum_{i=0}^{n}a_ie^{rt}r(r-1)(r-2)...(r-i+1)=0$$ The Characteristic equation of this equation is

Let $$y=x^r$$ Then
 * Method 2:
 * $$y_x=y'=r\,x^{r-1}$$
 * $$y_{xx}=r(r-1)\,x^{r-2}$$
 * $$y^{(i)}=r(r-1)(r-2)...(r-i+1)\,x^{r-i} $$
 * $$y^{(i)}=r(r-1)(r-2)...(r-i+1)\,x^{r-i} $$

Plugging $$y^{(i)}$$ into Equation($$) yeilds
 * $$\sum_{i=0}^{n}a_ix^iy^{(i)}= \sum_{i=0}^{n}a_ix^ir(r-1)(r-2)...(r-i+1)\,x^{r-i}=0$$

$$\Rightarrow \sum_{i=0}^{n}a_ix^iy^{(i)}= \sum_{i=0}^{n}a_ix^rr(r-1)(r-2)...(r-i+1)=0$$ The Characteristic equation of this equation is

To conclude, comparing Equation ($$)and Equation($$), the characteristic equation of both Method 1 and 2 are exactly the same. Therefore, their solution to the Euler Ln-ODE-VC ($$) will be equivalent.

=Problem 6.4* - Solve Euler L2-ODE-VC =

Given:
The characteristic equation

with $$\lambda = 5$$

and Euler L2-ODE-VC

Find:
Part 1:

1. $$a_2, a_1, a_0$$ such that ($$) is the characteristics equation of ($$)

2. 1st homogeneous solution $$y_1(x) = x^{\lambda}$$

3. Complete solution: $$c(x)$$, such that
 * $$y(x) = c(x) y_1(x)$$

4. Find the 2nd homogeneous solution $$y_2(x)$$

Part 2:

Repeat all of the steps for Euler L2-ODE-CC

Part 1.1
Assuming trial solution
 * $$y = x^r$$ where $$r$$ is a constant
 * $$\Rightarrow y' = r x^{r-1}$$
 * $$\Rightarrow y'' = r(r-1) x^{r-2}$$

Substituting into ($$)
 * $$a_0 r (r-1) x^r + a_1 r x^r + a_2 = 0$$
 * $$\Rightarrow \left[ a_0 r^2 + (a_1 - a_0) r + a_2 \right]x^r = 0$$

Since $$x^r \neq 0$$
 * $$\Rightarrow a_0 r^2 + (a_1 - a_0) r + a_2 = 0$$

Expanding ($$)

Comparing coefficients of ($$) and ($$)
 * $$ \frac{a_2}{a_0} = 25 $$ and $$ \frac{a_1}{a_0} - 1 = -10$$

Taking $$a_0 = k$$, we have

Part 1.2
Since the characteristics equation is given below
 * $$(r - 5)^2 = 0$$

The roots of the characteristics equation are at $$r=5$$

So the homogeneous solution is given by
 * $$y_1(x) = x^r = x^5$$

Part 1.3
Assuming $$y(x) = c(x)y_1(x) = c(x)x^5$$
 * $$\Rightarrow y'(x) = c'(x)x^5 + 5 c(x)x^{5-1} = c'(x)x^5 + 5 c(x)x^4$$
 * $$\Rightarrow y(x) = c(x)x^5 + 10 c'(x)x^4 + 20 c(x)x^3$$

Substituting ($$) into ($$)
 * $$k x^2 y'' - 9k x y' + 25k y = 0$$
 * $$\Rightarrow x^2 y'' - 9 x y' + 25 y = 0$$

Using the expressions for $$y''(x), y'(x), y(x)$$


 * $$ c''(x)x^7 + 10 c'(x)x^6 + \cancel{20 c(x)x^5} - 9 c'(x)x^6 - \cancel{45 c(x)x^5} + \cancel{25 c(x)x^5} = 0$$
 * $$\Rightarrow c''(x)x^7 + c'(x)x^6 = 0$$

Substituting $$g(x) = c'(x)$$ and considering the fact that $$x^6 \neq 0$$
 * $$\Rightarrow g'(x)x + g(x) = 0$$
 * $$\Rightarrow \frac {g'(x)} {g(x)} = -\frac {1} {x}$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow log(g(x)) = -log(x)+k_1$$
 * $$\Rightarrow log(g(x)) + log(x) = k_1$$
 * $$\Rightarrow g(x)x = e^{k_1} = k_2$$
 * $$\Rightarrow g(x) = \frac {k_2}{x}$$
 * $$\Rightarrow c'(x) = \frac {k_2}{x}$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_2 \log(x) + k_3$$


 * $$\Rightarrow y(x) = (k_2 \log(x) + k_3) x^5$$

Part 1.4
The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = x^5$$ and $$\log(x) x^5$$

So the second homogeneous solution is $$y_2(x) = \log(x) x^5$$

Part 2.1
Let us assume the trial solution
 * $$y(x) = e^{rx}$$
 * $$\Rightarrow y'(x) = re^{rx}$$
 * $$\Rightarrow y''(x) = r^2 e^{rx}$$

Substituting into ($$)
 * $$b_0 r^2 e^{rx} + b_1 re^{rx} + b_2 e^{rx} = 0$$
 * $$\Rightarrow \left(b_0 r^2 + b_1 r+ b_2 \right) e^{rx} = 0$$

Since $$e^{rx} \neq 0$$
 * $$\Rightarrow b_0 r^2 + b_1 r+ b_2 = 0$$

Expanding ($$)

Comparing coefficients of ($$) and ($$)
 * $$ \frac{b_2}{b_0} = 25 $$ and $$ \frac{b_1}{b_0} = -10$$

Taking $$b_0 = k$$, we have

Part 2.2
Since the characteristics equation is given below
 * $$(r - 5)^2 = 0$$

The roots of the characteristics equation are at $$r=5$$

So the homogeneous solution is given by
 * $$y_1(x) = e^{rx} = e^{5x}$$

Part 2.3
Assuming $$y(x) = c(x)y_1(x) = c(x)e^{5x}$$
 * $$\Rightarrow y'(x) = c'(x)e^{5x} + 5 c(x)e^{5x} $$
 * $$\Rightarrow y(x) = c(x)e^{5x} + 10 c'(x)e^{5x} + 25 c(x)e^{5x}$$

Substituting ($$) into ($$)
 * $$k y'' - 10 k y' + 25k y = 0$$
 * $$\Rightarrow y'' - 10 y' + 25 y = 0$$

Using the expressions for $$y''(x), y'(x), y(x)$$


 * $$ c''(x)e^{5x} + \cancel{10 c'(x)e^{5x}} + \cancel{25 c(x)e^{5x}} - \cancel{10 c'(x)e^{5x}} - \cancel{50 c(x)e^{5x}} + \cancel{25 c(x)e^{5x}} = 0$$
 * $$\Rightarrow c''(x)e^{5x} = 0$$

Since $$e^{5x} \neq 0$$
 * $$\Rightarrow c''(x) = 0$$

Integrating twice w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_1x + k_2$$


 * $$\Rightarrow y(x) = (k_1x + k_2) e^{5x}$$

Part 2.4
The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = e^{5x}$$ and $$xe^{5x}$$

So the second homogeneous solution is $$y_2(x) = xe^{5x}$$

Pavel Bhowmik (talk) 21:30, 13 November 2012 (UTC)

=R* 6.5 - Find the particular solution =

Find:
Use the variation of constants method to find the particular solution, $$y_{p}(x)$$, by finding the unknown $$A(x)$$

Solution:
Since it is the goal to find A(x) we need to set up an equation that can be used to find this function. This can be achieved by using ($$) and substituting it into ($$).

The first step is to find $$y'(x)$$


 * $$y'(x)=A'(x)y_{H}(x)+A(x)y'_{H}(x)$$

Since we have know the homogeneous solution we can take the derivative of it and substitute.


 * $$y'_{H}(x)=-a_{0}(x)\exp\left(-\int a_{0}(x)dx\right)$$


 * $$y'(x)=A'(x)\exp\left(-\int a_{0}(x)dx\right)+A(x)\left(-a_{0}(x)\exp\left(-\int a_{0}(x)dx\right)\right)$$


 * $$y'(x)=\exp\left(-\int a_{0}(x)dx\right)\left(A'(x)-a_{0}(x)A(x)\right)$$

Now that we have an expression for $$y'(x)$$ we can substitute this into ($$).


 * $$\exp\left(-\int a_{0}(x)dx\right)\left(A'(x)-a_{0}(x)A(x)\right)+a_{0}(x)A(x)\exp\left(-\int a_{0}(x)dx\right)=Q(x)$$

Simplifying


 * $$\exp\left(-\int a_{0}(x)dx\right)\left(A'(x)-a_{0}(x)A(x)+a_{0}(x)A(x)\right)=Q(x)$$
 * $$\exp\left(-\int a_{0}(x)dx\right)A'(x)=Q(x)$$

Now we have an expression in which we can solve for $$A(x)$$ and the solution is shown below
 * $$A(x)=\int Q(x)\exp\left(\int a_{0}(x)dx\right)dx$$

Now we can substitute this into ($$) to find the particular solution


 * $$y(x)=\int Q(x)\exp\left(\int a_{0}(x)dx\right)dx\left[\exp\left(-\int a_{0}(x)dx\right)\right]$$

=Problem 6.6* - Special IFM =

Link to lecture notes ,

Given:
The 2nd exactness condition

Find:
The partial differential equations that govern the integrating factor $$h(t,y)$$ where x is replaced with t because $$f(t)$$ is a function of t.

Solution:
First need to multiply the integrating factor $$ h(t,y)$$ through ($$) yielding,
 * $$ h(t,y)a_{2}y''+ h(t,y)a_{1}y'+ h(t,y)a_{0}y= h(t,y)f(t)$$

and then we need to subtract the left hand side of the equation to the right side to set the equation equal to zero and satisfy the first exactness condition,
 * $$ h(t,y)a_{2}y''+ h(t,y)a_{1}y'+ h(t,y)a_{0}y-h(t,y)f(t)=0$$

From here we can find f and g to find the partial differential equations governing the integrating factor $$ h(t,y)$$
 * $$ f=a_{2}h(t,y)$$
 * $$ g=h(t,y)a_{1}y'+ h(t,y)a_{0}y-h(t,y)f(t)$$

Now finding the partial derivatives for equations ($$) and ($$)
 * $$f_{t}=a_{2}h_{t}$$
 * $$f_{tt}=a_{2}h_{tt}$$
 * $$f_{ty}=a_{2}h_{ty}$$
 * $$f_{yy}=a_{2}h_{yy}$$
 * $$g_{t}=a_{1}ph_{t}+a_{0}yh_{t}-f(t)h_{t}-hf(t)'$$
 * $$g_{tp}=a_{1}h_{tp}$$
 * $$g_{y}=a_{1}ph_{y}+a_{0}yh_{y}+a_{0}h-f(t)h_{y}$$
 * $$g_{yp}=a_{1}h_{y}$$

Referring back to ($$) we know,
 * $$ f_{tt}+2pf_{ty}+p^2*f_{yy}=g_{tp}+pg_{yp}-g_{y}$$

Plugging in our partial derivatives,
 * $$a_{2}h_{tt}+2pa_{2}h_{ty}+p^2*a_{2}h_{yy}=a_{1}h_{tp}+pa_{1}h_{y}-a_{1}ph_{y}+a_{0}yh_{y}-f(t)h_{y}$$

Canceling terms we get,

Now we need ($$) for further solving and the terms are below,
 * $$f_{tp}=0$$
 * $$f_{yp}=0$$
 * $$f_{y}=a_{2}h_{y}$$
 * $$g_{pp}=0$$

Plugging the terms into ($$) we get,
 * $$0+p(0)+2a_{2}h_{y}=0$$
 * $$2a_{2}h_{y}=0$$

And the only way this above equation is true is if,
 * $$h_{y}=0$$

Now we can use $$h_{y}=0$$ in ($$) for further simplfication,
 * $$a_{2}h_{tt}+2pa_{2}h_{ty}+p^2*a_{2}h_{yy}=a_{1}h_{tp}+a_{0}yh_{y}-f(t)h_{y}$$
 * $$a_{2}h_{tt}+2pa_{2}(0)+p^2*a_{2}(0)=a_{1}h_{tp}+a_{0}y(0)+a_{0}h-f(t)(0)$$
 * $$a_{2}h_{tt}=a_{1}h_{tp}+a_{0}h$$
 * $$a_{2}h_{tt}-a_{1}h_{tp}-a_{0}h=0$$

Therefore we can solve these partial differential equations for the integrating factor.

Given:
Now consider the trial solution $$ h_{t}=exp(\alpha t)$$ for the integrating factor and multiplying it by ($$),

where the right hand side of the equation becomes,

In ($$) $$\bar a_{2}y$$ cancels to avoid making a $$ y' $$

Find:

 * $$\bar a_{1}$$ and $$\bar a_{0}$$ in terms of $$a_{2},a_{1},a_{0}$$

Solution:
First we need to differentiate ($$)with respect to $$ t $$ using product rule,
 * $$ \alpha exp(\alpha t)[\bar a_{1}y'+\bar a_{0}y]+exp(\alpha t)[\bar a_{1}y''+\bar a_{0}y']$$
 * $$ exp(\alpha t)[\alpha\bar a_{1}y'+\alpha\bar a_{0}y+[\bar a_{1}y''+\bar a_{0}y']$$

Combining terms of the same power of $$ y $$ yields,
 * $$ exp(\alpha t)[\bar a_{1}y''+(\bar a_{0}+\alpha\bar a_{1})y'+\alpha\bar a_{0}y$$

Now we set this equal to the integrand on left had side of ($$),
 * $$exp(\alpha t)[a_{2}y+a_{1}y'+a_{0}y]=exp(\alpha t)[\bar a_{1}y+(\bar a_{0}+\alpha\bar a_{1})y'+\alpha\bar a_{0}y$$

Canceling $$exp(\alpha t)$$ from both sides of the equation,
 * $$[a_{2}y+a_{1}y'+a_{0}y]=[\bar a_{1}y+(\bar a_{0}+\alpha\bar a_{1})y'+\alpha\bar a_{0}y$$

From here we can produce several equations given this relationship. First off we know $$a_{2}$$has to equal $$\bar a_{1}$$ because it is the only terms on both sides multiplied by $$ y''$$. This gives us $$\bar a_{1}$$ in terms of $$a_{2}$$,

We also know $$(\bar a_{0}+\alpha\bar a_{1})$$ must equal $$ a_{1})$$,

Finally we can get $$ \bar a_{0} $$ in terms of $$  a_{0} $$ using the y terms on the r.h.s and l.h.s
 * $$\alpha\bar a_{0}=a_{0}$$

Find:
The quadratic function of $$\alpha$$

Solution:
Solve ($$) for a quadratic function of $$ \alpha $$ in terms of $$ a_{2},a_{1},a_{0}$$
 * $$\bar a_{0}=a_{1}-\alpha a_{2}=\frac{a_{0}}{\alpha}$$
 * $$a_{1}-\alpha a_{2}-\frac{a_{0}}{\alpha}=0$$

Now multiply through the equation to get $$ \alpha $$,
 * $$\alpha a_{1}-(\alpha^2) a_{2}-\alpha \frac{a_{0}}{\alpha}=0$$

Canceling and rearranging terms we get the final answer for the quadratic solution for $$ \alpha $$,
 * $$ (\alpha^2) a_{2}-\alpha a_{1}+a_{0}=0$$

Given:
The reduced form,

Find:
Show that it can be solved using the intergating factor method (IFM).

Solution:
In order for ($$) to be solved using IFM it has to be in the following form,

We can get ($$) into this form with the following steps,
 * $$[\bar a_{1}y'+\bar a_{0}]=exp(-\alpha t)\int exp(\alpha t)f(t)dt$$

Then dividing both sides of the equation by $$ \bar a_{1} $$
 * $$\bar a_{1}y'+\bar a_{0}=exp(-\alpha t)\int exp(\alpha t)f(t)dt$$

Now it is the correct form and can be solved using IFM.

Given:
The correct form of ($$).

Find:
The solution $$ y(t) $$ for the excitation $$ f(t) $$ using the IFM

where $$ b(s) $$ is the left hand side of ($$)

Solution:
In ($$) we can denote $$\frac {\bar a_{1}}{\bar a_{0}}$$ to be $$ \beta $$ which simplifies the integral to,
 * $$\bar h(t)=exp[\int^t \beta dt]$$
 * $$\bar h(t)=exp[\beta t]$$

Now that we have our integrating factor we can plug this into ($$)
 * $$ y(t)=\frac {1}{exp[\beta t]}[\int^s exp[\beta s]\frac{exp(-\alpha s)\int exp(\alpha s)f(s)ds}{\bar a_{1}}+k_{2}]$$

This gives the most general form of the solution $$ y(t) $$

Given:
Our relationships from part 6.6.2.1

Show:

 * $$\alpha \beta=\frac{a_{0}}{a_{2}} $$
 * $$\alpha + \beta=\frac{a_{1}}{a_{2}} $$

Solution:
For,
 * $$\alpha \beta=\frac{a_{0}}{a_{2}} $$

lets first find $$\alpha$$ using ($$)
 * $$\bar a_{0}=\frac{a_{0}}{\alpha}$$

Solving for $$ \alpha $$ we get,
 * $$\alpha=\frac{ a_{0}}{\bar a_{0}}$$

And we know from the previous problem, $$ \beta=\frac{\bar a_{0}}{a_{1}}$$ Plugging in these terms we get,
 * $$\alpha \beta=\frac{ a_{0}}{\bar a_{0}}\frac{\bar a_{0}}{\bar a_{1}}$$

From ($$) we know,
 * $$\bar a_{1}=a_{2}$$

And substituting in $$a_{2}$$ we get,
 * $$\alpha \beta=\frac{ a_{0}}{\bar a_{0}}\frac{\bar a_{0}}{ a_{2}}$$

Canceling $$\bar a_{0}$$,
 * $$\alpha \beta=\frac{ a_{0}}{a_{2}}$$

which was the relationship we were trying to prove. Now to show that,
 * $$\alpha + \beta=\frac{a_{1}}{a_{2}} $$

are equivalent.
 * $$ \beta= \frac{\bar a_{0}}{a_{1}}$$

We are going to use ($$) to solve for $$ \alpha$$
 * $$a_{1}=\bar a_{0}+\alpha a_{2}$$
 * $$a_{1}-\bar a_{0}=\alpha a_{2}$$
 * $$\alpha=\frac{a_{1}-\bar a_{0}}{a_{2}}$$

where $${a_{2}}=\bar a_{1}$$ therefore,
 * $$\alpha + \beta=\frac{a_{1}-\bar a_{0}}{\bar a_{1}}+ \frac{\bar a_{0}}{a_{1}}$$
 * $$\alpha + \beta=\frac{a_{1}-\bar a_{0}+\bar a_{0}}{\bar a_{1}}$$
 * $$\alpha + \beta=\frac{a_{1}}{\bar a_{1}}$$

where $${a_{2}}=\bar a_{1}$$
 * $$\alpha + \beta=\frac{a_{1}}{ a_{2}}$$

Thereby showing the second condition and proving $$ \alpha,\beta $$ to be roots of the quadratic function,
 * $$(\lambda-\alpha)(\lambda-\beta)=\lambda^2-(\alpha+\beta)\lambda+\alpha\beta=0$$

Given:
The general solution $$ y(t) $$ from part 6.6.2.4,

Find:
The particular solution $$ y_{P}t $$

Solution:
First we start with the solution ($$) with the goal of breaking it up into the homogeneous and particular solutions.
 * $$y(t)=\frac {1}{exp[\beta t]}[\int^s exp[\beta s]\frac{exp(-\alpha s)\int exp(\alpha s)f(s)ds}{\bar a_{1}}+k_{2}]$$

Using an exponential rule we get
 * $$ y(t)=\frac {1}{exp[\beta t]}[\int^s \frac{exp[\beta -\alpha ]s}{\bar a_{1}}\int exp(\alpha s)f(s)ds+k_{2}]$$

Next we can pull $$ \bar a_{1} $$ out of the first integral because it is a constant.
 * $$ y(t)=\frac {1}{exp[\beta t \bar a_{1}]}[\int^s exp[\beta -\alpha ]s \int exp(\alpha s)f(s)ds+k_{2}]$$

Now we can move $$ exp[\beta t] $$ to the numerator but making it $$ exp[-\beta t] $$
 * $$ y(t)=\frac {exp{[-\beta t}]}{ \bar a_{1}}[\int^s exp[\beta -\alpha ]s \int exp(\alpha s)f(s)ds+\int^t exp(\beta-\alpha)k_{1}dt]$$

In the next step of breaking apart the solutions we can remove the dummy variable $$ s $$ and replace it with $$ t $$
 * $$ y(t)=\frac {exp{[-\beta t]}}{ \bar a_{1}}[\int^t exp[\beta -\alpha ]t \int exp(\alpha t)f(t)dt+\int^t exp(\beta-\alpha)k_{1}dt]$$

Solving the final integral in the equation gives us,
 * $$ y(t)=\frac {exp{[-\beta t]}}{ \bar a_{1}}[\int^t exp[\beta -\alpha ]t [\int exp(\alpha t)f(t)]dt+ exp(\beta-\alpha)t\frac{k_{1}}{\beta-\alpha}$$

This leaves us with the homogeneous part of the solution $$ y_{H}(t) $$ and the particular part of the solution $$ y_{P}(t) $$
 * $$y_{H}(t)=exp(\beta-\alpha)t\frac{k_{1}}{\beta-\alpha}$$
 * $$y_{P}(t)=\frac {exp{[-\beta t]}}{ \bar a_{1}}[\int^t exp[\beta -\alpha ]t [\int exp(\alpha t)f(t)]dt $$

Given:
The particular solution from the part 6.6.2.6 above,

where,

Find:
Verify that the particular solution found ($$) is the same as particular solutions found in tables.

Solution:
From = we find that the particular solution for ($$) should be of the form,
 * $$ Cx^m exp(ax)=(A_{0}+A_{1}x+...+A_{m}x^m)exp(ax)$$

Equation ($$) below,
 * $$ f(t)=t exp(bt)$$

tells us that $$ t=C=m$$ for our differential equation and after further simplification of $$ y_{P}(t) $$ or ($$) it should have the form,
 * $$ Cx^m exp(ax)=(A_{0}+A_{1}x+...+A_{m}x^m)exp(ax)$$

Given:
The two characteristic differential equations below,

Find:
Solve the general solution ($$) repeated below,
 * $$ y(t)=\frac {1}{exp[\beta t]}[\int^s exp[\beta s]\frac{exp(-\alpha s)\int exp(\alpha s)f(s)ds}{\bar a_{1}}+k_{2}]$$

for both characteristic differential equations ($$)and ($$)with defined hyperbolic function below,
 * $$ f(t)=tanh(t)$$

Solution:
From ($$) we know, $$ r=-1, r=2 $$ to be the roots with $$ \alpha=-1 $$, $$ \beta=2 $$ we can now plug these values into the general solution,
 * $$ y(t)=\frac {1}{exp[2t]}[\int^s exp[2s]\frac{exp(1s)\int exp(-1s)f(s)ds}{\bar a_{1}}+k_{2}]$$

Then we substitute in the exciting function $$ f(t) $$
 * $$ y(t)=\frac {1}{exp[2t]}[\int^s exp[2s]\frac{exp(1s)\int exp(-1s)tanh(s)ds}{\bar a_{1}}+k_{2}]$$

This gives our general solution for the first characteristic equation ($$) We can find the general solution for the second characteristic equation ($$) using the same method. The only difference this is r is a repeated root or $$ r=4 $$ twice. Therefore,
 * $$ \beta=\alpha=4 $$

but the exciting function $$ f(t) $$ remains the same. Plugging these terms into the general solution $$ y(t) $$ yields,
 * $$ y(t)=\frac {1}{exp[4t]}[\int^s exp[4s]\frac{exp(-4s)\int exp(4s)tanh(s)ds}{\bar a_{1}}+k_{2}]$$

This is the general solution for the characteristic differential equation ($$) with the hyperbolic function $$ f(t)$$.

=R*6.7 - Show the solution agrees with the form in King 2003=

Given
Hint:

To Show
Show that the above equation agrees with the form in King, i.e.


 * $$ \displaystyle y_P(x)=\int^{x}f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds $$

with


 * $$ \displaystyle W:=u_1 u_2'-u_2 u_1'=\begin{vmatrix}u_1&u_2\\u_1'& u_2'\end{vmatrix}$$

Solution
First let us consider the term
 * $$\int^{x}\frac{W(x)}dx = \int^{x}\frac{W(x)} \frac{u_2(x)}{u_1(x)}dx$$

Considering


 * $$\int^{x}\frac{W(x)}dx = \int^{x}p(x) q(x) dx$$

Integrating by parts


 * $$\int^{x}\frac{W(x)}dx = \left [ \int^{x} p(x) dx \right] q(x) - \int^{x} \left[ \frac{\mathrm{d}}{\mathrm{d}x} q(x) \right]  \left[ \int^{x} p(x) dx  \right] dx$$

Substituting form ($$) and ($$)
 * $$\int^{x}\frac{W(x)}dx = \left [ \int^{x} \frac{W(x)} dx \right] \frac{u_2(x)}{u_1(x)}  - \int^{x} \left( \frac{u_2(x)}{u_1(x)}  \right)'  \left[ \int^{x} \frac{W(x)} dx  \right] dx$$

Using ($$)
 * $$\int^{x}\frac{W(x)}dx = \left [ \int^{x} \frac{W(x)} dx \right] \frac{u_2(x)}{u_1(x)}  - \int^{x}  \frac{1}{h(x)}    \left[ \int^{x} \frac{W(x)} dx  \right] dx$$

Expanding the differentiation on ($$) we get:

Using ($$) and ($$)


 * $$y_P(x)=u_1(x)\int{\frac{1}{h(x)}}\left[\int \frac{{f(x)}u_1(x)}{W(x)}\,dx \right]\,dx $$

Using ($$)
 * $$y_P(x)=u_1(x)\left[\left(\int\frac{W(x)}dx \right)\frac{u_2(x)}{u_1(x)}-\int\frac{W(x)}dx \right]$$

Changing the dummy integration variables from $$x$$ to $$s$$,


 * $$y_P(x)=u_1(x)\left[\left(\int^{x}\frac{W(s)}ds \right)\frac{u_2(x)}{u_1(x)}-\int^{x}\frac{W(s)}ds \right]$$
 * $$\Rightarrow y_P(x)=u_2(x)\int^{x}\frac{W(s)}ds - u_1(x)\int^{x}\frac{W(s)}ds$$

Moving $$ u_1(x), u_2(x)$$ inside integral and rearranging,


 * $$y_P(x)=\int^{x}\frac{W(s)}ds - \int^{x}\frac{W(s)}ds$$
 * $$\Rightarrow y_P(x)=\int^{x}f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds\qquad \blacksquare$$

Pavel Bhowmik (talk) 15:24, 13 November 2012 (UTC)

=Problem 6.8* - Invalid Root =

Given:
First homogeneous solution:

Trail solution:

Characteristic equation:

WA:

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

Solution:
Let $$ y=u_2(x) = e^{xr_{2}(x)}=e^{\frac{x}{x-1}} $$ then
 * $$y'= e^{\frac{x}{x-1}}\,( \frac{x}{x-1})' $$

$$\Rightarrow y'= e^{\frac{x}{x-1}}\, \frac{(x-1)-x}{(x-1)^2} $$ $$\Rightarrow y'= e^{\frac{x}{x-1}}\, \frac{-1}{(x-1)^2}  $$ and
 * $$ y'' = \frac{2(x-1)}{(x-1)^4}e^{\frac{x}{x-1}} - \frac{1}{(x-1)^2}\frac{-1}{(x-1)^2}e^{\frac{x}{x-1}}$$

$$\Rightarrow y'' = \frac{2(x-1)+1}{(x-1)^4} e^{\frac{x}{x-1}} $$ $$\Rightarrow y'' = \frac{2x-1}{(x-1)^4}e^{\frac{x}{x-1}} $$ Plug $$\displaystyle y, y^{'}, y^{''}$$ into the left hand side of Equation ($$) yeilds
 * $$ (x-1)y^{''}-xy^'+y $$

$$= \frac{2x-1}{(x-1)^3}e^{\frac{x}{x-1}} + \frac{x}{(x-1)^2}e^{\frac{x}{x-1}} + e^{\frac{x}{x-1}}  $$ $$= e^{\frac{x}{x-1}}\, \left[ \frac{2x-1}{(x-1)^3} +\frac{x}{(x-1)^2}+1 \right] $$ $$= e^{\frac{x}{x-1}}\, \left[ \frac{2x-1+x^2-x+x^3-3x^2+3x-1}{(x-1)^3} \right] $$ $$= e^{\frac{x}{x-1}}\, \left[ \frac{x^3-2x^2+4x-2}{(x-1)^3} \right] $$ $$\neq 0$$ Therefore $$\displaystyle u_2(x) = e^{xr_{2}(x)}$$ is not a valid solution to the Equation ($$).

=Problem 6.9* - Find homogeneous solution to L2-ODE-VC =

Given:
The following L2-ODE-VC

Find:
A valid homogeneous solution and call it $$ u_1(x) $$.

Find the 2nd homogeneous solution $$ u_2(x) $$ using variation of parameters method and compare it to $$ e^{xr_2(x)} $$.

Solution:
Let us assume the trial solution
 * $$y(x) = e^{rx}$$
 * $$\Rightarrow y'(x) = re^{rx}$$
 * $$\Rightarrow y''(x) = r^2 e^{rx}$$

Substituting into ($$)
 * $$(x-1) r^2 e^{rx} - x re^{rx} + e^{rx} = 0$$
 * $$\Rightarrow \left( (x-1) r^2 - x r + 1 \right) e^{rx} = 0$$

Since $$e^{rx} \neq 0$$
 * $$\Rightarrow (x-1) r^2 - x r + 1 = 0$$
 * $$\Rightarrow (x-1) r^2 - (x - 1) r - r + 1  = 0$$
 * $$\Rightarrow (x - 1) r (r - 1) - (r - 1 ) = 0$$
 * $$\Rightarrow ((x - 1) r - 1 )(r - 1 ) = 0$$

The roots of $$r$$ are
 * $$r=1, r=\frac{1}{x-1}$$

Taking the root $$r=1$$, the homogeneous solution is $$y_1(x) = e^x$$

Assuming $$y(x) = c(x)y_1(x) = c(x)e^x$$
 * $$\Rightarrow y'(x) = c'(x)e^x + c(x)e^x $$
 * $$\Rightarrow y(x) = c(x)e^x + 2c'(x)e^x + c(x)e^x $$

Substituting into ($$)
 * $$(x - 1) (c''(x)e^x + 2c'(x)e^x + c(x)e^x) - x (c'(x)e^x + c(x)e^x) + c(x)e^x = 0$$

Since $$e^x \neq 0$$
 * $$(x - 1) (c''(x) + 2c'(x) + c(x)) - x (c'(x) + c(x)) + c(x) = 0$$
 * $$\Rightarrow xc(x) + 2xc'(x) + \cancel{xc(x)} - c(x) - 2c'(x) - \cancel{c(x)} - x c'(x) - \cancel{x c(x)} + \cancel{c(x)} = 0 $$

Substituting $$ c'(x) = p(x) $$
 * $$\Rightarrow c''(x) = p'(x) $$


 * $$\Rightarrow xp'(x) + xp(x) - p'(x) - 2p(x) = 0 $$
 * $$\Rightarrow xp'(x) - p'(x) = 2p(x) - xp(x) $$
 * $$\Rightarrow \frac {p'(x)}{p(x)} = \frac {1} {x - 1} - 1 $$

Integrating w.r.t. $$x$$
 * $$\Rightarrow \log (p(x)) = \log (x - 1) - x + k_1$$
 * $$\Rightarrow p(x) = (x - 1) e^{- x + k_1} = (x - 1) e^{- x }e^{ k_1} = k_2(x - 1) e^{- x }$$
 * $$\Rightarrow c'(x) = k_2 x e^{- x } - k_2 e^{- x }$$
 * $$\Rightarrow c'(x) = k_2 x e^{- x } - k_2 e^{- x }$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_2 \int^x x e^{- x } \mathrm{d} x+ k_2 e^{- x } + k_3$$

Integrating by parts
 * $$\Rightarrow c(x) = k_2 x \int^x e^{- x } \mathrm{d} x - k_2 \int^x \left[ \frac{\mathrm{d}}{\mathrm{d}x}x \int^x e^{- x }\mathrm{d} x \right]\mathrm{d} x + k_2 e^{- x } + k_3$$


 * $$\Rightarrow c(x) = - k_2 x e^{- x } + k_2 \int^x e^{- x } \mathrm{d} x + k_2 e^{- x } + k_3$$


 * $$\Rightarrow c(x) = - k_2 x e^{- x } - \cancel{k_2 e^{- x }} + \cancel{k_2 e^{- x }} + k_3 = - k_2 x e^{- x } + k_3$$


 * $$\Rightarrow y(x) = c(x) e^x = (- k_2 x e^{- x } + k_3)e^x$$

The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = e^{x}$$ and $$x$$

So the second homogeneous solution is $$y_2(x) = x$$, which is equal to $$e^{xr(x)}$$ only if
 * $$r(x) = \frac {\log{x}} {x}$$

Which is clearly not a root of the characteristic equatiob. Pavel Bhowmik (talk) 05:28, 14 November 2012 (UTC)

=Problem 6.10* - Plot the altitude vs time =

Given:

 * $$z(t):=v_{y}(t)$$
 * $$n=2,3, a=k/m=2, b=mg=10$$
 * $$z(0)=v_{y}(0)=50$$

Find:
Plot $$z(t)$$ versus $$t$$ for each value of n. Also plot the altitude $$y(t)$$ vs time and find the time the projectile hits the ground

Solution:
For n=2 we can integrate ($$) by hand using the following integration rule.

Plugging in the parameter values given into ($$) and taking the derivative involves the following steps


 * $$\frac{1}{10}\int \frac{1}{\frac{z^2}{5}+1}du$$

To get this equation into the form of ($$) we can say $$u=\frac{z}{\sqrt{5}}$$ which would make $$du=\frac{1}{\sqrt{5}}dz$$. Substituting these values in gives us...


 * $$\frac{\sqrt{5}}{10}\int \frac{1}{u^2+1}du=\frac{\sqrt{5}}{10}\tan^{-1}\left(u\right)+c$$

We can now plug $$ u$$ back in.


 * $$-t=\frac{\sqrt{5}}{10}\int \frac{1}{u^2+1}du=\frac{\sqrt{5}}{10}\tan^{-1}\left(\frac{z}{\sqrt{5}}\right)+c$$

The same solution was found using Wolfram Alpha and can be found at this link

Solve for z:
 * $$z=\sqrt{5}tan\left(-2\sqrt{5}t+c2\sqrt{5}\right)$$

Once again we know that at $$t=0 z=50$$ so we can find the constant
 * $$c=-1.78e14$$

Since we cannot solve for z we will use Matlab to plot z(t) vs t

We know that at $$t=0; z=50$$ so we can solve for the constant.
 * $$c=\frac{\sqrt{5}}{10}\tan^{-1}(10\sqrt{5})=1.706$$

Therfore...
 * $$z=\sqrt{5}tan\left(-2\sqrt{5}t+1.526\right)$$

To find the altitude as a function time we can just take the integral of the velocity with respect to time
 * $$y(t)=\int\left[\sqrt{5}tan\left(-2\sqrt{5}t+1.526\right)\right]+c$$

Using matlab to solve symbolically we obtain...
 * $$y(t)=\frac{1}{2}\log\left(\cos\left(2\sqrt{5}t-\frac{763}{500}\right)\right)+c$$

Assuming $$y=0$$ when $$t=0$$ we can solve for c...
 * $$c=-\frac{1}{2}\log\left[\cos\left(\frac{-763}{500}\right)\right]$$

And...
 * $$y(t)=\frac{1}{2}\log\left(\cos\left(2\sqrt{5}t-\frac{763}{500}\right)\right)-\frac{1}{2}\log\left[\cos\left(\frac{-763}{500}\right)\right]$$

The particle returns to the ground at approximately 0.68 seconds

For n=3 an explicit solution is not obvious and int his case wolfram alpha is used to solve the problem using the hypergeometric function


 * $$\frac{1}{10} \left(\frac{\sqrt[3]{5} \log \left(\frac{z}{\sqrt[3]{5}}+1\right)}{3 z}-\frac{\sqrt[3]{-5} \log \left(1-\frac{e^{-\frac{i \pi }{3}}z}{\sqrt[3]{5}}\right)}{3 z}+\frac{(-1)^{2/3} \sqrt[3]{5} \log \left(1-\frac{e^{\frac{i \pi }{3}} z}{\sqrt[3]{5}}\right)}{3 z}\right)$$

This is extremely hard to plot so we will use an approximation based on the following series...


 * $$1+\frac{ab}{c}\frac{x}{1!}+\frac{a(a+1)b(b+1)}{c(c+1}\frac{x^{2}}{2!}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)}\frac{x^{3}}{3!}+\cdot\cdot\cdot$$

Now we can find an expression for $$t$$


 * $$t=-\left[\frac{z}{10}+\frac{-z^4}{200}+\frac{z^7}{1750}+\frac{-z^{10}}{1250}\right]+c$$

Since we cannot solve for z we will use Matlab to plot the results



Since the velocity continues to increase it does not seem reasonable that the particle will ever return to the ground.

= R*6.11 - Using variation of parameters to find 2nd homogeneous solution =

Given
$$(1-x^2)y''-2xy'+6y=0$$ for n=2

First homogeneous solution: $$P_2(x)=\frac{1}{2}(3x^2-1)$$

To Show
the second homogeneous solution: $$Q_2(x)=\frac{1}{4}(3x^2-1)\log \left ( \frac{1+x}{1-x} \right )-\frac{3}{2}x $$

Solution
Putting the Legendre equation in the standard form,

$$y''- \underbrace {\frac{2x}{(1-x^{2})}}_{a_1(x)}y'+\underbrace {\frac{6}{(1-x^{2})}}_{a_0(x)}y=0 $$

Since the solution $$P_{2}$$ is given and we need to show that $$Q_{2}$$ is a solution take $$u_{2}(x)=Q_{2}(x),u_{1}(x)=P_{2}(x) $$

From the notes we have: $$u_{2}(x):=u_{1}(x)\int\frac{1}{u^{2}(x)}exp[-\int a_{1}(x)dx]dx $$

$$\int \frac{2x}{(1-x^{2})}dx,a=1-x^{2},da=-2xdx $$

$$ \int \frac{-1}{a}da=-\log(a)=-\log(1-x^{2}) $$

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

Using Wolfram Alfa to perform the integral

$$\int\frac{4}{(3x^{2}-1)^{2}(x^{2}-1)}dx=\frac{1}{2}(\frac{6x}{3x^{2}-1}+\log(\frac{1-x}{1+x})) $$

$$ u_{2}(x)=Q_{2}(x)=\frac{1}{4}(3x^{2}-1)\log (\frac{1+x}{1-x})-\frac{3}{2}x $$

= R*6.12 - Solving Non-homogeneous Legendre equation =

Given
Non-homogeneous Legendre equation

where $$ \displaystyle f(x)=1$$ The 1st homogeneous solution $$ \displaystyle u_1(x)=P_1(x)=x$$

Find:
The final solution $$y(x)$$ by variation of parameters

Solution
Rewrite Equation ($$) into

According to Lecture Note (4)34-5, we can find the 2 homogeneous solutions using formula $$u_{2}(x):=u_{1}(x)\int\frac{1}{u^{2}(x)}exp[-\bar{a_1}(x)]dx $$ where $$\bar{a_1}(x) = \int a_{1}(x)dx $$ $$\Rightarrow \bar{a_1}(x)= \int -\frac{2x}{1-x^{2}}dx = log(1-x^2)$$

Applying $$u_{1}(x)=x,\bar{a_1}(x) $$ into $$u_{2}(x) $$ we get
 * $$u_{2}(x)=x\int\frac{1}{x^2 }\, exp[-log(1-x^2)]dx $$

$$\Rightarrow u_{2}(x)=x\int\frac{1}{x^2 (1-x^2)}dx $$ $$\Rightarrow u_{2}(x)= \frac {x}{2}log \frac{1+x}{1-x} -1$$ as showed in lecture note (4)p.7-1

The final solution of Equation ($$) is separated by two parts: homogeneous solution $$y_H(x)$$and particular solution $$y_P(x)$$ see Lecture Note (3)34-5

where $$y_H(x)=k_1u_1(x)+k_2u_2(x)$$ and (see Lecture Note (1)34-6 ) $$y_P(x)=u_1(x) \int \frac{1}{h(x)}\left[ \int h(x) \frac{f(x)}{u_1(x)} dx \right] dx  $$ in which (see Lecture Note (4)34-4 )
 * $$h(x)=u_1^2(x)exp\left[\int a_1(x)dx \right]$$

$$\Rightarrow h(x)= x^2 exp\left[\bar{a_1}(x) \right]$$ $$\Rightarrow h(x)= x^2 exp\left[log(1-x^2) \right]$$ $$\Rightarrow h(x)= x^2(1-x^2) $$

Apply$$u_{1}(x),h(x)$$ into the particular solution yields $$y_P(x)=x \int \frac{1}{x^2(1-x^2)}\left[ \int x^2(1-x^2) \frac{1}{x} dx \right] dx  $$ Using Wolfram Alpha to solve the integrating, we get $$y_P(x)= \frac{x^2}{4} + \frac{x}{8} log \left( \frac{1+x}{1-x} \right) $$

Applying homogeneous solution $$y_H(x)$$and particular solution $$y_P(x)$$ into Equation($$), we can get the final solution which is $$y_(x)=k_1x+k_2 \left(\frac {x}{2}log \frac{1+x}{1-x} -1 \right) +\frac{x^2}{4} + \frac{x}{8} log \left( \frac{1+x}{1-x} \right)  $$

=Contribution Table=

= References =