User:Egm6321.f09.team3/HW3

 It would be nice to have a title for each problem so that the table of contents is more informative, making it easier to search for the desired problem; see Team 1.

See further comments below. Egm6321.f09 15:40, 15 October 2009 (UTC)

= Problem 1 =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ x^{m+ \frac{1}{2}}y^ny'' + 2x^{1+m}y^ny' + 3x^my^{n+1} $$

Find
m & n such that the ODE is exact.

Solution
$$f(x,y,p) = x^{m+ \frac{1}{2}}y^n$$

$$g(x,y,p) = 2x^{1+m}y^ny' + 3x^my^{n+1}\! $$

which satisfies the first condition of exactness.

Condition 2 is met by satifaction of the following equations:

$$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y \qquad{...(i)}$$

$$ f_{xp} + pf_{yp} + 2f_y = g_{pp} \qquad{...(ii)}$$

The partial derivatives are given below:

$$ \begin{align} f_x &= \left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^n \\ f_{xx} &= \left(m- \frac{1}{2}\right) (m+ \frac{1}{2})x^{m- \frac{3}{2}}y^n \\ f_{xy} &= n \left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^{n-1} \\ f_y &= nx^{m+ \frac{1}{2}}y^{n-1} \\ f_{yy} &= n(n-1)x^{m+ \frac{1}{2}}y^{n-2} \\ f_p &= f_{xp} = f_{yp} = f_{pp} = 0 \\ g_x &= 2(m+1)x^my^np + 3mx^{m-1}y^{n+1} \\ g_{xx} &= 2m(m+1)x^{m-1}y^np + 3m(m-1)x^{m-2}y^{n+1} \\ g_{xy} &= 2(m+1)nx^my^{n-1}p + 3m(n+1)x^{m-1}y^n \\ g_y &= 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n \\ g_{yy} &= 2n(n-1)2nx^{m+1}y^{n-2}p + 3n(n+1)x^my^{n-1} \\ g_p &= 2x^{m+1}y^n \\ g_{px} &= 2(m+1)x^my^n \\ g_{py} &= 2nx^{m+1}y^{n-1} \\ g_{pp} &= 0 \\ \end{align} $$

Substitute into i. to get:

$$ \left(m- \frac{1}{2}\right) \left(m+ \frac{1}{2}\right)x^{m- \frac{3}{2}}y^n + 2p\cdot n\left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^{n-1} + p^2 n(n-1)x^{m+ \frac{1}{2}}y^{n-2} = 2(m+1)x^my^n + p \cdot 2nx^{m+1}y^{n-1} - 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n $$

Substitute into ii. to get:

$$ 0 + p \cdot 0 + 2 \cdot nx^{m+ \frac{1}{2}}y^{n-1} = 0 $$

Equation ii. suggests that n must be zero in order for the equation to be satisfied.

$$ n = 0 \!$$

Simplify equation i to get:

$$ \left(m- \frac{1}{2}\right) \left(m+ \frac{1}{2}\right)x^{m- \frac{3}{2}} + 0 + 0 = 2(m+1)x^m + 0 - 0 + 3x^m $$

Note the powers of x need to be the same to simplify, which suggests $$ m = \pm \frac{1}{2} $$.

$$ m = \frac{1}{2} $$ satisfies equation i.

Therefore,

$$ n = 0 \!$$

and

$$ m = \frac{1}{2} $$

The final ODE is

$$ xy'' + 2x^{\frac{3}{2}}y' + 3x^{\frac{1}{2}}my $$

or,

$$ y'' + 2x^{\frac{1}{2}}y' + 3x^{\frac{-1}{2}}my $$

 Why does $$m$$ appear in your final equation? You found $$m=\frac{1}{2}$$ so it should not appear. --Egm6321.f09.TA 04:07, 15 October 2009 (UTC)
 * }

= Problem 2 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ xp - (2x^{\frac {3}{2}} - 1)y + k_1 = k_2 $$

Find
$$y(x)\!$$

Solution
The ODE is

$$ y' - \frac{2x^{\frac{3}{2}}-1}{x}y = k_2 - k_1 $$

 $$y- \frac{2x^{\frac{3}{2}}-1}{x}y = \frac{k_2 - k_1}{x}$$. Don't forget to divide both sides by $$x$$. - Egm6321.f09.TA 23:54, 27 October 2009 (UTC)

where

$$ a_0(x) = - \frac{2x^{\frac{3}{2}}-1}{x} \qquad{\text{and}} $$

$$ b(x) = k_2 - k_1 $$

This is a linear, first order ODE with varying coefficients.

In standard form, $$ N = 1 $$ and $$ M = a_0 \cdot y $$.

This makes

$$ \begin{align} h(x) &= \exp \left( \int a_0(s) ds\right) \\ &= \exp \left( \int \left(-2 \sqrt{s} + s^{-1}\right) ds\right) \\ &= x \cdot \exp \left(- \frac{4}{3} x^{\frac {3}{2}}\right) \end{align} $$



I think your error above lead to an incorrect integrating factor. Your process is correct however. Egm6321.f09.TA 23:54, 27 October 2009 (UTC)

Multiply the ODE by the integrating factor to get:

$$ (hy)' = hb $$

Integrate to get:

$$ y(x) = \frac{1}{h(x)} \int h(s) b(s) ds $$

$$ y(x) = \frac{1}{x \cdot \exp \left(- \frac{4}{3} x^{\frac {3}{2}}\right)} \int s \cdot \exp \left(- \frac{4}{3} s^{\frac {3}{2}}\right) \cdot (k_2 - k_1) ds $$


 * }

= Problem 3 =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A class of exact linear second order ODEs with varying coefficients

Find
The mathematical structure of $$\phi$$

Solution
Since $$ F = \frac {d \phi}{dx} = \phi_x(x,y,p) + \phi_y(x,y,p) \cdot y' + \phi_p(x,y,p) \cdot y' $$

and $$ F = P(x) y'' + Q(x) y' + R(x) y \!$$

This suggests

i. $$ P(x) = \phi_p \!$$ ii. $$ Q(x) = \phi_y\! $$

iii. $$ R(x)y = \phi_x \!$$

From i.,

$$ \phi = \int P(x) dp = P(x) \cdot p + C(x,y) $$

From ii.,

$$ \phi = \int Q(x) dy = Q(x) \cdot y + C(x,p) $$

This suggests $$ \phi $$ has a structure of:

$$ \phi = P(x) \cdot p + Q(x) \cdot y + C(x,y) $$

Take the parital partial derivative with respect to x,

$$ \frac{\partial{\phi}}{dx} = \frac{\partial{P}}{dx} \cdot p + \frac{\partial{Q}}{dx} \cdot y + \frac{\partial{C}}{dx} $$

and compare to iii, which suggests C is not a function of x!

The final form is:

$$ \phi = P(x) \cdot p + T(x) \cdot y + k $$

where P(x) and T(x) are functions, and k is a constant.


 * }

Good result. very concise. Egm6321.f09.TA 01:58, 28 October 2009 (UTC)

= Problem 4 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The conditions for exactness:

$$ \begin{align} &F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n \\ &f_i = \dfrac {\partial F}{\partial y^{(i)}} \qquad{\text{such that}}\\ &f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 \end{align} $$

Find
the n= 1 case

Solution
$$ f_1 = \frac{\partial{F}}{dy'} = \frac{\partial{}}{dy'}(\phi_x + \phi_y y') = \phi_y $$

Therefore,

$$ f_0 - \frac{df_1}{dx} = \frac{\partial {F}}{dy} - \frac{d}{dy}(\phi_y) = \frac{\partial{(\phi_x + \phi_y y')}}{dy'} - \frac{d}{dx}(\phi_y) $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy}y' - \phi_yx - \phi_{yy}y' $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy}  - \phi_yx $$$ = 0 $. The zero is what allows you to make the statement $\phi_{xy}=\phi_{yx}$

Therefore,

$$ f_0 - \frac{df_1}{dx} = 0 $$ is equivalent to $$ \phi_{xy}  = \phi_yx $$, which is the condition of exactness for the linear first order ODE!

 You have an operator $$\frac{\partial}{dy'}$$ which does not make sense. You cannot mix partials derivatives and total derivatives. Furthermore the $$y'$$ in the denominator should be $$y$$. --Egm6321.f09.TA 14:14, 15 October 2009 (UTC)
 * }

Also, be careful of the typo: $$\displaystyle \phi_yx$$ is not the same as $$\displaystyle \phi_{yx}$$; you need to use the curly brackets around the subscripts "$$\displaystyle xy$$" if you want to write $$\displaystyle \phi_{yx}$$. Egm6321.f09 15:40, 15 October 2009 (UTC)

= Problem 5 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The conditions for exactness:

$$ \begin{align} &F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n \\ &f_i = \dfrac {\partial F}{\partial y^{(i)}} \qquad{\text{such that}}\\ &f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 \end{align} $$

Find
the n=2 case

Solution
$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

Integrate with respect to x to get:

$$ \int f_0 dx - f_1 + \frac{df_2}{dx} $$

where $$ f_0 = \frac{\partial{F}}{dy} = \phi_{xy} + \phi_{yy} y' + \phi_{y'y} y'' $$

Therefore, $$ \int f_0 dx = \phi_y $$

Now, take the derivative with respect to x to get the original equation,

$$ \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy} y' + \phi_{yy'} y'' $$

which is, of course, $$ f_0 \!$$

The final equations is validated:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

 These steps are not logically connected and do not prove the expressions in the problem statement. --Egm6321.f09.TA 01:32, 16 October 2009 (UTC)
 * }

= Problem 6 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The Legendre linear, second order ODE with varying coefficients: $$ F = (1-x^2)y'' - 2xy' + n(n+1)y = 0\! $$

Find
Verify the exactness of the equation using:

a. equations 4 and 5 from 10-2

b. equation 5 from 14-1

Solution
a. Using the following 2 equations:

$$ \begin{align} &\text{(i)} \quad f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y \\ &\text{(ii)} \quad f_{xp} + pf_{yp} + 2 f_y = g_{pp} \qquad{\text{where}} \\ &F= f(x,y,p)y'' + g(x,y,p) \qquad{\text{and}} \\ &f(x,y,p) = 1-x^2 \\ &g(x,y,p) = -2xp + n(n-1) y \end{align} $$

The partial derivatives are:

$$f_x = -2x\!$$

$$f_{xx} = -2\!$$

$$f_{xy} = f_y = f_{yy} = f_{xp} = f_{yp} = f_p = f_{pp} = 0\!$$

$$g_x = -2p\!$$

$$g_{xp} = -2\!$$

$$g_y = n(n+1)\!$$

$$g_p = -2x\!$$

$$ g_{xx} = g_{xy} = g_{yy} = g_{yp} = g_{pp} = 0\! $$

Substitute to get:

i. $$ -2 + 2p \cdot 0 + p^2 \cdot 0 = -2 + p \cdot 0 - n(n+1) $$

ii. $$ 0 + p \cdot 0 + 2 \cdot 0 = 0 $$

The ODE is not exact because the equation i is not satisfied.

b. Using the following equation:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} = 0 $$

where

$$ f_0 = \frac{\partial{F}}{dy} = n(n+1) $$

$$ f_1 = \frac{\partial{F}}{dy'} = -2x $$

$$ f_2 = \frac{\partial{F}}{dy''} = 1-x^2 $$

Combining,

$$n(n+1) - (-2) + (-2) = 0\!$$

Which is exact for the solution of

$$ n = 0 \! \qquad{\text{(trivial)}} \!$$

or

$$ n=-1 \!$$

However, the ODE can be made exact using the integrating factors method:

$$h(x,y) = x^my^n\!$$

Multiply by the ODEto get:

$$ x^my^n [ (1-x^2)y'' - 2xy' + n(n+1) y = 0 ] \!$$

which makes

$$ \begin{align} f(x,y,p) &= x^my^n - x^{m+2}y^n \qquad{\text{and}} \\ g(x,y,p) &= - 2x^{m+1}y^np + \alpha(\alpha+1)x^m y^{n+1} \end{align} $$

The partial derivatives are:

$$f_x = mx^{m-1}y^n - (m+2)x^{m+1}y^n\!$$

$$f_{xx} = m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n\!$$

$$f_{xy} = mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}\!$$

$$f_y = nx^my^{n-1} - nx^{m+2}y^{n-1}\! $$

$$ f_{yy} = n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2}\! $$

$$ f_{xp} = f_{yp} = f_p = f_{pp} = 0 \!$$

$$ g_x = -2(m+1)x^my^np + m\alpha (\alpha + 1)x^{m-1}y^{n+1} \!$$

$$ g_{xx} = -2m(m+1)x^{m-1}y^np + m(m-1)\alpha (\alpha + 1)x^{m-2}y^{n+1} \!$$

$$ g_{xy} = -2(m+1)nx^my^{n-1}p + m\alpha (\alpha + 1)(n+1)x^{m-1}y^{n} \!$$

$$ g_y = - 2nx^{m+1}y^{n-1}p + (n+1)\alpha(\alpha+1)x^m y^{n} \!$$

$$ g_{yy} = - 2n(n-1)x^{m+1}y^{n-2}p + n(n+1)\alpha(\alpha+1)x^m y^{n-1} \!$$

$$ g_p = -2x^{m+1}y^n\! $$

$$ g_{yp} = -2nx^{m+1}y^{n-1} \!$$

$$ g_{xp} = -2(m+1)x^my^n\! $$

$$ g_{pp} = 0 \!$$

Substituting into equation i:

$$ m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n + 2p \cdot (mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}) + p^2 \cdot (n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2}) = -2(m+1)x^my^n + p \cdot (-2nx^{m+1}y^{n-1}) - (- 2nx^{m+1}y^{n-1}p) + (n+1)\alpha(\alpha+1)x^m y^{n}) $$

and into equation ii:

$$ 0 + p \cdot 0 + 2 \cdot (nx^my^{n-1} - nx^{m+2}y^{n-1}) = 0 $$

Equation ii can only be solved by $$ n = 0 $$

which leaves equation i as:

$$ m(m-1)x^{m-2} - (m+2)(m+1)x^{m} + 2p \cdot (0 + 0) + p^2 \cdot (0 - 0) = -2(m+1)x^m + p \cdot (0) - (-0) + \alpha(\alpha+1)x^m) $$

Which may be satisfied for $$ m = 1 $$

(Incidentally, the solution requires $$ \alpha = 1 $$ to be true.)


 * }

 Avoid using the symbol $$\displaystyle n$$ in the integrating factor $$\displaystyle h(x,y) = x^m y^n$$ since $$\displaystyle n$$ may be confused with the order $$\displaystyle n$$ in the Legendre differential equation; use $$\displaystyle h(x,y) = x^a y^b$$ instead. It is not clear how $$\displaystyle m = 1$$ can be a solution to make the Legendre differential equation exact. See Team 1. Egm6321.f09 15:40, 15 October 2009 (UTC)

= Problem 7 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The definitions of linearity:

$$ \begin{align} &\text{(a)} \quad L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) \\ &\text{(b)} \quad \text{(i)} \quad L(u + v) = L(u) + L(v) \qquad{\text{and}} \qquad \text{(ii)} \quad L( \lambda u ) = \lambda L(u) \end{align} $$

Find
Show the two definitions are true.

Solution
Apply b. ii to get:

$$ L( \alpha u ) = \alpha L(u)$$

and

$$ L( \beta v ) = \beta L(v) $$

Combine using b. i to get:

$$ L( \alpha u + \beta v) = \alpha L(u) + \beta L(v) $$

which matches the definition of linearity given in a.

 clear explanation. --Egm6321.f09.TA 04:03, 16 October 2009 (UTC)
 * }

= Problem 8 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
Knowing the plots of the other shape functions

Find
Plot the shape function $$N_{j+1}^{2}(x)$$.

Solution


[[media:HW3-8.jpg|Graph]]


 * }



correct. Egm6321.f09.TA 03:08, 28 October 2009 (UTC)

= Problem 9 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$x^2 y'' - 2xy' + 2y = 0\!$$

Find
$$y_{xxx} \qquad{\text{and}} \!$$

$$y_{xxxx}\!$$

Solution
Recall that

$$ \frac{d}{dt} y = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{dy}{dx} \cdot e^t $$

Which yields

$$ y_t = y_x \cdot e^t $$

Rearrange to get:

$$ y_x = y_t \cdot e^{-t} $$

Take the third derivative:

$$ \begin{align} \frac{d^3}{dx^3}y =& \frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(y\right)\right)\right) \\ =&\frac{dt}{dx}\frac{d}{dt}\left(\frac{dt}{dx}\frac{d}{dt}\left(\frac{dt}{dx}\frac{d}{dt}\left(y\right)\right)\right) \\ =& e^{-t}\frac{d}{dt}\left(e^{-t}\frac{d}{dt}\left(e^{-t}\frac{d}{dt}\left(y\right)\right)\right) \\ =& e^{-t}\frac{d}{dt}\left(e^{-t}\frac{d}{dt}\left(e^{-t} y_t\right)\right) \\ =& e^{-t}\frac{d}{dt}\left(e^{-t}\left(-e^{-t} y_t + e^{-t}y_{tt}\right)\right) \\ =& e^{-t}\frac{d}{dt}(-e^{-2t} y_t + e^{-2t}y_{tt}) \\ =& e^{-t}(2e^{-2t} y_t - e^{-2t}y_{tt} - 2e^{-2t}y_{tt} + e^{-2t}y_{ttt}) \\ =& 2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt} \\ y_{xxx} =& e^{-3t} (2y_t - 3y_{tt} + y_{ttt}) \end{align} $$

To find $$ y_{xxxx} $$ take an additional derivative:

$$ y_{xxxx} = \frac{dt}{dx}\frac{d}{dt}(2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt}) $$

$$ y_{xxxx} = e^{-t} (-3 e^{-3t}y_{ttt} + e^{-3t}y_{tttt} + 9 e^{-3t}y_{tt} - 3 e^{-3t}y_{ttt} - 6 e^{-3t}y_t + 2e^{-3t}y_{tt} $$

$$ y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$


 * }



good solution. Cculd be expressed a little cleaner. Egm6321.f09.TA 03:22, 28 October 2009 (UTC)

= Problem 10 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
Solve equation 1 on p.16-1,

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

using the method of trial solution

$$ y=e^{rx}$$

directly for the boundary conditions

$$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$

Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Find
The solution using the method of trial solution $$ y = e^{rx} $$

Part 1
Assume a trial solution of the given ODE to be $$ y = e^{rx} $$, where r is a constant and x is the independent variable. By differentiating the trial solution we get the equations:

$$ y = e^{rx} $$

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

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

Therefore when we plug the trial solution into the given equation we get the solution

$$ {x}^{2}{r}^{2}e^{rx} - 2xre^{rx} + 2e^{rx} = 0 $$

Simplifying we can eliminate the $$ e^{rx} $$ because it appears in every term, leaving us with:

$$ {x}^{2}{r}^{2} - 2xr + 2 = 0 $$

Rearranging

$$ {r}^{2}{x}^{2} - 2xr + 2 = 0 $$

This solution of r must a function of x that cannot be solved directly, and thereby violating our initial assumptions. This departure from the assumption in the solution means that we must try another solution to find agreement. Therefore we should chose another guess and continue. The next most likely trial solution to try would be $$ y = {x}^{r} $$, which as shown in the note produces a viable solution.

Part 2
Assume the solution

$$ y = {x}^{r} $$

, where r is a constant, and the full solution is

$$ y(x) = {C}_{1}{x}^{{r}_{1}} + {C}_{2}{x}^{{r}_{2}} $$

$$ y = {x}^{r} $$

$$ y' = r{x}^{r-1} $$

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

Plugging in these values

$$ {x}^{2}r(r-1){x}^{r-2} - 2xr{x}^{r-1} + 2{x}^{r} = 0 $$

Simplifying with standard algebraic techniques we get the solution

$$ {r}^{2} - 3r + 2 = 0 $$, which determines $$ {r}_{1} = 1 $$ and $$ {r}_{2} = 2 $$.

Therefore we have our solution

$$ y(x) = {C}_{1}{x} + {C}_{2}{x}^{2} $$

using the boundary conditions we obtain:

$$ 3 = {C}_{1} + {C}_{2} $$

$$ 4 = 2{C}_{1} + 4{C}_{2} $$

And solving for the constants:

$$ {C}_{1} = 4 $$

$$ {C}_{2} = -1 $$

The solutions is therefore: $$ y(x) = 4x - {x}^{2} $$

The plot of this solution is shown below.




 * }



very good. Egm6321.f09.TA 03:47, 28 October 2009 (UTC)

= Problem 11 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ u_1(x) z' + [a_1(x)u_1(x) + 2u_1'(x)]z = 0 $$

Find
Solution using the integrating factor method.

Solution
$$ z' + \left[a_1(x) + 2 \frac {u_1'(x)}{u_1(x)}\right]z = 0 $$

$$ a_0 = \left[a_1 + 2 \frac {u_1'}{u_1}\right] \qquad{\text{and}}$$

$$ b = 0 \!$$

The integrating factor h(x) is:

$$ \begin{align} h(x) =& \exp \left( \int a_0(s) ds \right) \\ =& \exp \left( \int a1(s) + 2\frac {u_1'(s)}{u_1(s)} ds\right) \\ =& \exp \left( \int a1(s) ds + 2 ln(x)\right) \end{align} $$

$$ (hz)' = 0 $$

Integrate to solve:

$$ hz = \text{constant} $$

Rearrange to solve for z:

$$ z = \frac {\text{constant}}{(u_1(x))^2} \cdot \exp \left[- \int a_1(s) ds\right] $$
 * }



very good. Egm6321.f09.TA 04:03, 28 October 2009 (UTC)

= Problem 12 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The reduction of order method 2 using $$ y(x) = U(x)u_1(x) $$

Find
Alternate definitions of y(x) that create reduction of order method 2

$$\text{(a.)} \quad y(x) = U(x) \pm u_1(x) $$

$$\text{(b.)} \quad y(x) = \dfrac{U(x)}{u_1(x)} $$

$$ \text{(b.)} \quad y(x) = \dfrac{u_1(x)}{U(x)} $$

Solution
$$\text{(a.)}$$

Take the first and second derivatives of y(x) to get:

$$y'(x) = U' \pm u_1'$$

$$y(x) = U \pm u_1''$$

The desired form is

$$ y'' + a_1y' + a_0y = 0 $$

, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0U \pm a_0u_1 + a_0U' \pm a_1u_a' + U \pm u_1'' $$

which is

$$ 0 = a_0U \pm a_0u_1 + a_0U' \pm a_1u_a' + U \pm u_1 $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

$$\text{(b.)}$$

Take the first and second derivatives of y(x) to get:

$$y'(x) = \dfrac{U'}{u_1} + \dfrac{U}{u_1'}$$

$$y(x) = \dfrac{U}{u_1} + 2\dfrac{U'}{u_1'} + \dfrac{U}{u_1''}$$

The desired form is

$$ y'' + a_1y' + a_0y = 0 $$

, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 \dfrac{U}{u_1} + a_1 \dfrac{U'}{u_1} + a_1 \dfrac{U}{u_1'} + \dfrac{U}{u_1} + 2\dfrac{U'}{u_1'} + \dfrac{U}{u_1''} $$

which is

$$ 0 = a_0 \dfrac{U}{u_1} + a_1 \dfrac{U'}{u_1} + a_1 \dfrac{U}{u_1'} + \dfrac{U}{u_1} + 2\dfrac{U'}{u_1'} + \dfrac{U}{u_1} $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

$$\text{(c.)}$$

Take the first and second derivatives of y(x) to get:

$$y'(x) = \dfrac{u_1'}{U} + \dfrac{u_1}{U'}$$

$$y(x) = \dfrac{u_1}{U} + 2\dfrac{u_1'}{U'} + \dfrac{u_1}{U''}$$

The desired form is

$$ y'' + a_1y' + a_0y = 0 $$

, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 \dfrac{u_1}{U} + a_1 \dfrac{u_1'}{U} + a_1 \dfrac{u_1}{U'}+ \dfrac{u_1}{U} + 2\dfrac{u_1'}{U'} + \dfrac{u_1}{U''}$$

which is

$$ 0 = a_0 \dfrac{u_1}{U} + a_1 \dfrac{u_1'}{U} + a_1 \dfrac{u_1}{U'}+ \dfrac{u_1}{U} + 2\dfrac{u_1'}{U'} + \dfrac{u_1}{U} $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.
 * }



very good. Egm6321.f09.TA 04:15, 28 October 2009 (UTC)

= Problem 13 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

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

Find
u_1 and u_2 using the following two methods:

$$ y = ax^b \qquad{...(i)} $$

$$ y = e^{rx} \qquad{...(ii)}$$

and compare the two solutions and the solution using the reduction of order method 2 using the boundary conditions $$y(0) = 1$$ and $$y(1) = 2$$

Part 1
Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1)

Find solution using the trial solution

$$ y=ax^{b} $$

$$ y'=bax^{b-1} $$

$$ y''=b(b-1)ax^{b-2} $$

Substituting into Equation 1

$$ (1-x^{2}) ax^{b-2} -2x bax^{b-1} +2 ax^{b}  =0 $$  This equation does not match the line below it. Egm6321.f09.TA 04:59, 28 October 2009 (UTC) $$ (1 - {x}^{2})*(a{b}^{2} - ab)*{x}^{b-2} -2 abx^{b} +2 ax^{b}  =0 $$

Dividing through by $$ {x}^{b} $$ $$ (1 - {x}^{2})*(a{b}^{2} - ab)*{x}^{-2} -2 ab +2 a  =0 $$ This results in the following equation:

$$ ({x}^{-2} - 1)*a*({b}^{2} - b) - 2 ab + 2 a  =0 $$

We can divide through by a, leaving us with:

$$ ({x}^{-2} - 1)*({b}^{2} - b) - 2 b + 2 = 0 $$

Therefore by comparison of forms we know:

$$ {u}_{1}(x) = {x}^{-2} - 1 $$ ,and $$ {u}_{2}(x) = 1 $$

$$ {k}_{1} = {b}^{2} - b $$, and $$ {k}_{2} = -2*b + 2 $$

Therefore the solution is $$ y(x) = ({b}^{2} - b)*({x}^{-2} - 1) + (-2*b + 2)*1 $$

using the boundary conditions to solve for b we obtain:

$$ 1 = -1*({b}^{2} - b) + (-2*b + 2) $$

$$ 2 = -2*b + 2 $$

Given these parameters $$ b = 0 $$, making the trial solution $$ a*{x}^{0} $$ , this trivial solution illustrates this is not a solution to the ODE. We must therefore try another solution. I am having difficulty following what you did in your trial solution. Egm6321.f09.TA 05:01, 28 October 2009 (UTC)

Part 2
Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1)

Find solution using the trial solution

$$ y=e^{rx} $$

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

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

Substituting into Equation 1

$$ (1-x^{2}) r^{2}e^{rx} -2x re^{rx}  +2 e^{rx}  =0  $$ Assuming $$ y=er^{x}\neq0$$, divide through by $$  er^{x}$$ $$ (1-x^{2}) r^{2} -2x r  + 2 = 0 $$          Following this results we can construct our solution:

$$ {u}_{1}(x) = 1 - {x}^{2} $$ ,and $$ {u}_{2}(x) = -x*r + 1 $$

$$ {k}_{1} = {r}^{2} $$, and $$ {k}_{2} = 2 $$ Therefore the solution is $$ y(x) = {r}^{2}*(1 - {x}^{2}) + 2*(-x*r + 1) $$

using the boundary conditions to solve for r we obtain:

$$ 1 = {r}^{2} + 2 $$

Therefore $$ r = \sqrt{-3} $$

$$ 2 = -2*r + 2 $$

Therefore $$ r = 0 $$

Since one of the roots of r is zero (similar to the previous case) making the solution of the form, $$ y(x) = e^{0} $$, it is shown that this is also not the solution. 
 * }

You should be able to make one of these techniques work. Egm6321.f09.TA 05:31, 28 October 2009 (UTC)

= Contributing Team Members =

Angela Diggs Egm6321.f09.team3.diggs 01:36, 7 October 2009 (UTC) Barrett Mooney Egm6321.f09.team3.mooney 17:30, 6 October 2009 (UTC) Catherine Snow EGM6321.f09.Team3.cgs 16:27, 7 October 2009 (UTC) Francisco Kuljevan -- Egm6321.f09.team3.Kuljevan.Francisco 18:55, 7 October 2009 (UTC)