User:Egm6321.f09.team4/HW3

= Homework #3: Team 4 =

Problem #1:
(Lecture 13-1)


 * Given :
 * The equation $$\left ( x^ny^m \right )\left [ \sqrt{x}y^{\prime\prime}+2xy^{\prime}+3y \right ]=0$$


 * Find :
 * The values of m and n such that the equation is exact.


 * Solution :


 * The given equation can be rewritten in the following form (first condition of exactness):


 * $$ \begin{align}

f\left(x,y,p\right) &= \left(x^{n}y^{m}\right)x^{0.5} = x^{n+0.5}y^{m} \\ g\left(x,y,p\right) &= \left(x^{n}y^{m}\right)(2xp+3y) = 2x^{n+1}y^{m}p+3x^{n}y^{m+1} \\ \end{align} $$


 * To evaluate the second condition of exactness, various partial derivatives of $$f$$ and $$g$$ are required.


 * $$ \begin{align}

f_{xx} &= \frac{\partial}{\partial x}\left( \left(n+0.5\right)x^{n-0.5}y^{m}\right) = \left(n^{2}-0.25\right)x^{n-1.5}y^{m} \\ f_{xy} &= \frac{\partial}{\partial y}\left( \left(n+0.5\right)x^{n-0.5}y^{m}\right) = m\left(n+0.5\right)x^{n-0.5}y^{m-1} \\ f_y   &= mx^{n+0.5}y^{m-1} \\ f_{yy} &= m\left(m-1\right)x^{n+0.5}y^{m-2} \\ f_{xp} &= f_{yp} = 0.0 \\ \\  g_{xp} &= \frac{\partial}{\partial x}\left( 2x^{n+1}y^{m} \right) = 2\left(n+1\right)x^{n}y^{m} \\ g_{yp} &= \frac{\partial}{\partial y}\left( 2x^{n+1}y^{m} \right) = 2mx^{n+1}y^{m-1} \\ g_{pp} &= 0.0 \\ g_{y} &= 2mx^{n+1}y^{m-1}p + 3\left(m+1\right)x^{n}y^{m} \end{align} $$


 * The second condition of exactness requires satisfying two equations. The first equation (eq. 2A) is evaluated below:


 * $$ \begin{align}

&f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \quad \text{(2a)} \end{align} $$


 * Plugging in the values obtained above for the various partial derivatives:


 * $$ \begin{align}

&\left(n^2-0.25\right)x^{n-1.5}y^{m} + 2pm\left(n+0.5\right)x^{n-0.5}y^{m-1} + p^{2}m\left(m-1\right)x^{n+0.5}y^{m-2} = 2\left(n+1\right)x^{n}y^{m} + \cancel{2mpx^{n+1}y^{m-1}} - \cancel{2mpx^{n+1}y^{m-1}}- 3\left(m+1\right)x^{n}y^{m} \\ &\left(n^2-0.25\right)x^{n-1.5}y^{m} + 2pm\left(n+0.5\right)x^{n-0.5}y^{m-1} + p^{2}m\left(m-1\right)x^{n+0.5}y^{m-2} = \left[ 2\left(n+1\right)-3\left(m+1\right)\right]x^{n}y^{m} \end{align} $$


 * Dividing both sides by $$x^{n}y^{m}$$:


 * $$ \begin{align}

&\left(n^2-0.25\right)x^{-1.5} + 2pm\left(n+0.5\right)x^{-0.5}y^{-1} + p^{2}m\left(m-1\right)x^{0.5}y^{-2} = 2n - 3m - 1 \\ \end{align} $$


 * The other equation for the second exactness condition (eq. 2B) is given as follows:


 * $$ \begin{align}

&f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \\ &0.0 + p\left(0.0\right) + 2mx^{n+0.5}y^{m-1} = 0.0 \\ &2mx^{n+0.5}y^{m-1} = 0 \end{align} $$


 * The only way this equation can be satisfied for arbitrary $$x,y$$ is if $$m = 0$$. Plugging $$m=0$$ back into eq. 2A:


 * $$ \begin{align}

&\left(n^2-0.25\right)x^{-1.5} = 2n - 1 \end{align} $$


 * This equation can only be satisfied by $$n = 0.5$$. Therefore, the values of $$m,n$$ which make the original equation exact are:


 * $$ \begin{align}

m &= 0.0 \\ n &= 0.5 \end{align} $$  Good write up. Nice work. --Egm6321.f09.TA 04:13, 15 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 19:53, 7 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 11:55, 7 October 2009 (UTC)

Problem #2:
(Lecture 13-1)


 * Given :
 * The equation $$\Phi(x,y,p)=xy'+(2x^{3/2}-1)y+k_1=k_2$$ where $$k_1$$ and $$k_2$$ are constants


 * Find :
 * Solve for y(x) by applying the integrating factor method


 * Solution :
 * This is a first-order ODE, which can be rearranged in the following form:


 * $$ \begin{align}

y' + p\left(x\right)y = g\left(x\right), \quad \text{where:} \quad p\left(x\right)&=2x^{0.5}-x^{-1} \\ g\left(x\right)&=x^{-1}\left(k_2-k_1\right) \end{align} $$


 * Let $$h\left(x\right)$$ be the integrating factor:


 * $$ \begin{align}

&h\left(x\right)y' + h'\left(x\right)y = h\left(x\right)g\left(x\right) \end{align} $$


 * Note: if we find $$h\left(x\right)$$ such that $$h\left(x\right)p\left(x\right)=h'\left(x\right)$$, the equation becomes the following:


 * $$ \begin{align}

&h\left(x\right)y' + h'\left(x\right)y = h\left(x\right)g\left(x\right) \\ &\frac{d}{dx}\left(h\left(x\right)y\right) = h\left(x\right)g\left(x\right) \quad \text{(product rule)} \\ &h\left(x\right)y+c = \int h\left(x\right)g\left(x\right)dx \\ &y\left(x\right) = \frac{\int h\left(x\right)g\left(x\right)dx - C_{1}}{h\left(x\right)} \end{align} $$


 * Then, to solve this, all we need is $$h\left(x\right)$$, which can be found by solving the constraint:


 * $$ \begin{align}

&h\left(x\right)p\left(x\right) = h'\left(x\right) \\ &h\left(x\right)\left(2x^{0.5}-x^{-1}\right) = h'\left(x\right) \\ &\frac{h'\left(x\right)}{h\left(x\right)} = 2x^{0.5} - x^{-1} \\ &\frac{d}{dx}\left\{\text{ln}\left[h\left(x\right)\right]\right\} = 2x^{0.5} - x^{-1} \\ &\text{ln}\left[h\left(x\right)\right] = \frac{4}{3}x^{1.5} - \text{ln}\left(x\right) + c \\ &h\left(x\right) = \text{exp}\left[ \frac{4}{3}x^{1.5}-\text{ln}\left(x\right)+c\right] \\ &h\left(x\right) = C_{2} x^{-1} \text{exp}\left(\frac{4}{3}x^{1.5}\right) \end{align} $$


 * Finally, $$y\left(x\right)$$ is found by plugging in $$h\left(x\right)$$:


 * $$ \begin{align}

y\left(x\right) &= \frac{\int\left[ C_{2}x^{-1}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)\cdot x^{-1}\left(k_{2}-k_{1}\right)\right]dx - C_{1}}{C_{2}x^{-1}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)} \\ y\left(x\right) &= x\text{ exp}\left(\frac{-4}{3}x^{1.5}\right)\left[ C_{2}\left(k_2-k_1\right)\int x^{-2}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)dx - \frac{C_1}{C_2}\right] \end{align} $$





You have an extra power of $$x$$ in the denominator of the integrand. Egm6321.f09.TA 01:24, 28 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 16:55, 6 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 11:58, 7 October 2009 (UTC)

Problem #3:
(Lecture 13-1)


 * Given :
 * A class of ODE's: linear, 2nd order ODE's with varying coefficients (L2_ODE_VC).


 * Find :
 * The mathematical structure of $$\Phi\left(x,y,p\right)$$ that yields the above class.


 * Solution :


 * Consider the following equation:


 * $$ \begin{align}

\Phi\left(x,y,p\right) = C, \quad \text{(where C is a constant)} \end{align} $$


 * Take the total derivative of this equation.


 * $$ \begin{align}

&\frac{d}{dx}\left[ \Phi\left(x,y,p\right) \right] = 0 \\ &\Phi_{x}dx + \Phi_{y}dy + \Phi_{p}dp = 0 \\ &\Phi_{x} + \Phi_{y}\frac{dy}{dx} + \Phi_{p}\frac{dp}{dx} = 0 \\ &\Phi_{x} + \Phi_{y}y' + \Phi_{p}y'' = 0 \quad \text{(since } p=y' \text{)} \end{align} $$


 * For any $$\Phi\left(x,y,p\right)$$, this equation will produce an exact, second-order ODE with varying coefficients. However, we were given the additional restriction that the ODE must be linear; i.e., the equation must be linear in both $$y$$ and $$y'$$, and there should be no terms containing $$y$$, $$y'$$, or $$y''$$ multiplied together. These requirements lead to three conditions on the partial derivatives of $$\Phi$$:
 * $$\Phi_x$$ cannot contain powers of $$y$$ or $$p$$ greater than 1. If both are present, they cannot be multiplied together.
 * $$\Phi_y$$ cannot contain $$y$$ or $$p$$.
 * $$\Phi_p$$ cannot contain $$y$$ or $$p$$.


 * To meet these conditions, we define $$\Phi$$ to have the following structure:


 * $$ \begin{align}

\Phi\left(x,y,p\right) = A\left(x\right) + B\left(x\right)y + C\left(x\right)p + D, \quad \text{(where D is a constant)} \end{align} $$


 * This yields the following values for the partial derivatives:


 * $$ \begin{align}

\Phi_x &= A'\left(x\right) + B'\left(x\right) y + C'\left(x\right) p \quad &\text{(condition 1 met)} \\ \Phi_y &= B\left(x\right) \quad &\text{(condition 2 met)}\\ \Phi_p &= C\left(x\right) \quad &\text{(condition 3 met)} \end{align} $$



It is not clear that you have arrived at the proper result. To finish this you need to equate your terms to those in the problem statement. Egm6321.f09.TA 02:01, 28 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 18:18, 6 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 12:00, 7 October 2009 (UTC)

Problem #4:
(Lecture 13-3)


 * Given :
 * A nonlinear 1st order ODE $$F\left(x,y,y'\right)=0=\frac{d}{dx}\Phi\left(x,y\right)$$, and the exactness condition $$f_o-\frac{df_1}{dx}=0$$.


 * Find :
 * Show that this exactness condition is equivalent to the condition $$\Phi_{xy}=\Phi_{yx}$$, derived earlier in the course. Hint: $$f_1=\Phi_y$$.


 * Solution :


 * Start by expanding the definition of our ODE:


 * $$ \begin{align}

F\left(x,y,y'\right) &= \frac{d}{dx}\left[\Phi\left(x,y\right)\right] = 0 \\ &= \Phi_{x}\left(x,y\right)dx + \Phi_{y}\left(x,y\right) dy \\ &= \Phi_{x}\left(x,y\right) + \Phi_{y}\left(x,y\right)\frac{dy}{dx} \\ &= \Phi_{x}\left(x,y\right) + \Phi_{y}\left(x,y\right) p \quad \text{(where } p=y' \text{)} \end{align} $$


 * By definition, we know that:
 * $$ \begin{align}

f_{i} &= \frac{\partial F}{\partial y^{\left(i\right)}} \end{align} $$


 * Expanding this definition for $$i=0, i=1$$ yields:


 * $$ \begin{align}

f_{0} &= \frac{\partial F}{\partial y^{\left(0\right)}} = \frac{\partial F}{\partial y} \\ &= \Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p \\ \\ f_{1} &= \frac{\partial F}{\partial y^{\left(1\right)}} = \frac{\partial F}{\partial p} \\ &= \Phi_{y}\left(x,y\right) \end{align} $$


 * Inserting these two values into the exactness condition:


 * $$ \begin{align}

&f_{0} - \frac{d}{dx} f_{1} = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \frac{d}{dx}\left[\Phi_{y}\left(x,y\right)\right] = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \left[\Phi_{yx}\left(x,y\right)\frac{d}{dx}\left(x\right) + \Phi_{yy}\left(x,y\right)\frac{d}{dx}y\right] = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \left[ \Phi_{yx}\left(x,y\right) + \Phi_{yy}p \right] = 0 \\ &\Phi_{xy}\left(x,y\right) = \Phi_{yx}\left(x,y\right) + \Phi_{yy}p - \Phi_{yy}p \\ &\Phi_{xy}\left(x,y\right) = \Phi_{yx}\left(x,y\right) \end{align} $$

 Good write up- nice work. This is very clear. --Egm6321.f09.TA 14:24, 15 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 16:15, 7 October 2009 (UTC)

Problem #5:
(Lecture 14-1)
 * Given :
 * An exact nonlinear 2nd order ODE $$F\left(x,y,y',y''\right)=0=\frac{d}{dx}\Phi\left(x,y,y'\right)$$, and the exactness condition $$f_0-\frac{df_1}{dx}+\frac{d^{2}f_2}{dx^2}=0$$.


 * Find :
 * Show that $$f_1=\frac{d}{dx}f_2+\Phi_y$$
 * Show that $$\frac{d}{dx}(\Phi_y)=f_0$$
 * Show that $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$
 * Relate the condition $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ to equations 4 and 5 on page 10-2.


 * Solution :


 * Expanding $$F(x,y,y',y'')$$ yields:
 * $$ \begin{align}

F &=\frac{d\Phi}{dx}(x,y,p) \qquad \text{(where } p=y' \text{)}\\ &=\Phi_x\frac{dx}{dx}+\Phi_y\frac{dy}{dx}+\Phi_p\frac{dp}{dx} \\ &=\Phi_x+\Phi_y p+\Phi_p p' \end{align} $$


 * By definition:
 * $$ \begin{align}

&f_i=\frac{\partial F}{\partial y^{(i)}} \end{align} $$


 * Expanding this definition for $$i=0,i=1,i=2$$ yields:
 * $$ \begin{align}

f_0&=\frac{\partial F}{\partial y} = \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' \\ \\ f_1&=\frac{\partial F}{\partial p} = \Phi_{xp} + \Phi_{yp}p + \Phi_{y} + \Phi_{pp}p' \\ \\ f_2&=\frac{\partial F}{\partial p'} = \Phi_p \end{align} $$


 * Part 1 :


 * $$ \begin{align}

f_{1} &= \frac{df_2}{dx} + \Phi_y \\ \Phi_{xp}+\Phi_{yp}p+\Phi_{y}+\Phi_{pp}p' &= \frac{d}{dx}\left(\Phi_p\right) + \Phi_{y} \\ \Phi_{xp}+\Phi_{yp}p+\Phi_{y}+\Phi_{pp}p' &= \Phi_{px} + \Phi_{py}p + \Phi_{pp}p' + \Phi_{y} \\ \Phi_{xp}+\Phi_{yp}p &= \Phi_{px} + \Phi_{py}p \\ 0 &= 0 \quad \text{Proof. (since } \Phi_{ab}=\Phi_{ba} \text{, for all a,b)} \end{align} $$


 * Part 2 :


 * $$ \begin{align}

\frac{d}{dx}\left(\Phi_y\right) &= f_{0} \\ \Phi_{yx}+\Phi_{yy}p+\Phi_{yp}p' &= \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' \\ \Phi_{yx}+\Phi_{yp}p' &= \Phi_{xy} + \Phi_{py}p' \\ 0 &= 0 \quad \text{Proof. (since } \Phi_{ab}=\Phi_{ba} \text{, for all a,b)} \end{align} $$


 * Part 3 :


 * $$ \begin{align}

f_0 - \frac{df_1}{dx} + \frac{d^{2}f_{2}}{dx^{2}} = 0 \end{align} $$


 * From part 1, we know that:


 * $$ \begin{align}

\frac{d}{dx}\left( f_{2} \right) = f_{1} - \Phi_{y} \end{align} $$


 * Differentiating both sides of this equation:


 * $$ \begin{align}

\frac{d^2}{dx^2}\left(f_{2}\right) &= \frac{d}{dx}\left(f_{1}\right) - \frac{d}{dx}\left(\Phi_{y}\right) \\ &= \frac{d}{dx}\left(f_{1}\right) - f_{0} \quad \text{(by result from Part 2)} \end{align} $$


 * Plugging this result back into the equation we're trying to prove:


 * $$ \begin{align}

f_{0} - \frac{df_1}{dx} + \frac{d^{2}f}{dx^{2}} &= 0 \\ f_{0} - \frac{df_1}{dx} + \left[ \frac{df_1}{dx} - f_{0} \right] &= 0 \\ f_{0} - f_{0} &= 0 \\ 0&=0 \quad \text{Proof.} \end{align} $$


 * Part 4 :


 * The old second-order exactness conditions which we're trying to derive are given as follows:


 * $$ \begin{align}

\quad &f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + p g_{yp} - g_{y} \\ \quad &f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \end{align} $$


 * Using the expansion of $$F$$ at the beginning of the solution, the functions $$f\left(x,y,p\right)$$ and $$ g\left(x,y,p\right)$$ are given as follows:


 * $$ \begin{align}

\text{(2a)} \qquad &f\left(x,y,p\right) = \Phi_p\left(x,y,p\right) \\ \text{(2b)} \qquad &g\left(x,y,p\right) = \Phi_x\left(x,y,p\right) + \Phi_y\left(x,y,p\right)p \end{align} $$


 * Taking the necessary partial derivatives:


 * $$ \begin{align}

f_{y} &= \Phi_{py}, \quad &f_{yy} = \Phi_{pyy}, \qquad &f_{xy} = \Phi_{pxy} \\ f_{xx} &= \Phi_{pxx}, \quad &f_{xp} = \Phi_{pxp}, \qquad &f_{yp} = \Phi_{pyp} \end{align} $$


 * $$ \begin{align}

g_{y} &= \Phi_{xy}+\Phi_{yy}p \\ g_{yp} &= \Phi_{xyp}+\Phi_{yy}+\Phi_{yyp}p \\ g_{xp} &= \Phi_{xxp} + \Phi_{yx} + \Phi_{yxp}p \\ g_{pp} &= \Phi_{xpp} + 2\Phi_{yp} + \Phi_{ypp}p \end{align} $$


 * Expand equation (2a):


 * $$ \begin{align}

f_{xx} + 2pf_{xy} + p^{2}f_{yy} &= g_{xp} + pg_{yp} -g_{y} \\ \Phi_{pxx} + 2p\Phi_{pxy} + p^{2}\Phi_{pyy} &= \Phi_{xxp} + \Phi_{px} + \Phi_{yxp}p + p\Phi_{xyp}+p\Phi_{yy}+p^{2}\Phi_{yyp}-\Phi_{xy}-\Phi_{yy}p \\ \\ \Phi_{xxp} + 2p\Phi_{xyp} + p^{2}\Phi_{yyp} &= \Phi_{xxp} + 2p\Phi_{xyp} + p^{2}\Phi_{yyp} \end{align} $$


 * Expand equation (2b):


 * $$ \begin{align}

f_{xp} + pf_{yp} + 2f_{y} &= g_{pp} \\ \\ \Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{py} &= \Phi_{xpp} + 2\Phi_{py} + p\Phi_{ypp} \end{align} $$


 * Next, we'll expand the new exactness condition. First, note the following:


 * $$ \begin{align}

f_{0} &= \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' = \frac{d}{dx}\left(\Phi_y\right) \quad \text{(from part 2)} \\ \frac{df_{1}}{dx} &= \frac{d}{dx}\left[ \Phi_{xp} + \Phi_{yp}p + \Phi_{y} + \Phi_{pp}p' \right]\\ &= \left[\Phi_{xpx}+\Phi_{xpy}p+\Phi_{xpp}p'\right] + \left[p'\Phi_{yp}+\Phi_{ypx}p+\Phi_{ypy}p^{2}+\Phi_{ypp}pp'\right] + \left[\Phi_{ppx}p'+\Phi_{ppy}pp'+\Phi_{ypp}\left(p'\right)^{2}\right] +f_{0} \end{align} $$


 * Also, remember from the derivation of Part 3:


 * $$ \begin{align}

\frac{d^{2}f_{2}}{dx^{2}} = \frac{df_{1}}{dx} - f_{0} \end{align} $$


 * This allows us to rewrite the new exactness condition as follows:


 * $$ \begin{align}

&\frac{df_{1}}{dx} - f_{0} = \frac{df_{1}}{dx} - f_{0} \\ &\Phi_{xpx}+\Phi_{xpy}p+\Phi_{xpp}p' + p'\Phi_{yp}+\Phi_{ypx}p+\Phi_{ypy}p^{2}+\Phi_{ypp}pp' + \Phi_{ppx}p'+\Phi_{ppy}pp'+\Phi_{ypp}\left(p'\right)^{2}    = \cdots \quad \text{(both sides the same)}\\ \end{align} $$


 * We can then group the terms of this equation by powers of $$p'$$:


 * $$ \begin{align}

&\left[\Phi_{xpx}+\Phi_{xpy}p+\Phi_{ypx}p+\Phi_{ypy}p^{2}\right] + \left[\Phi_{xpp}+\Phi_{yp}+\Phi_{ypp}p+\Phi_{ppx}+\Phi_{ppy}p\right]p' + \left[\Phi_{ppp}\right]\left(p'\right)^{2} = \cdots \quad \text{(both sides the same)}\\ &\left[\Phi_{xxp}+2p\Phi_{xyp}+p^{2}\Phi_{yyp}\right] + \left[2\Phi_{xpp}+2p\Phi_{ypp}+\Phi_{yp}\right]p' + \left[\Phi_{ppp}\right]\left(p'\right)^{2} = \cdots \quad \text{(both sides the same)} \end{align} $$


 * Then, subtract $$\left[\Phi_{xpp}+p\Phi_{ypp}-\Phi_{yp}\right]p' + \Phi_{ppp}p^{2}$$ from both sides:


 * $$ \begin{align}

&\left[\Phi_{xxp}+2p\Phi_{xyp}+p^{2}\Phi_{yyp}\right] + \left[\Phi_{xpp}+p\Phi_{ypp}+2\Phi_{yp}\right]p' = \cdots \quad \text{(both sides the same)} \end{align} $$


 * Finally, we can split this single equality into two parts by equating the coefficients of each $$p'$$ term.


 * $$ \begin{align}

\text{(1)} \quad &\Phi_{xxp} + 2p\Phi_{xyp} +\Phi_{yyp}p^{2} = \Phi_{xxp} + 2p\Phi_{xyp} +\Phi_{yyp}p^{2}\\ \text{(2)} \quad &\Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{yp} = \Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{yp} \end{align} $$


 * These two equations are equal to the expansions of equations (2a) and (2b) that we performed earlier. Therefore, we've shown that the general exactness condition specialized to second-order equations is equivalent to the specific 2nd order conditions derived earlier in the course.

 Good work. Your solution could be more concise but it is complete. --Egm6321.f09.TA 01:36, 16 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.palubin.d 03:03, 7 October 2009 (UTC)
 * Egm6321.f09.team4.sls 19:48, 7 October 2009 (UTC)

Problem #6:
(Lecture 14-2)


 * Given :The form of the Legendre linear 2nd order ODE with varying coefficients:

$$F=(1-x^2)y^{\prime\prime}-2xy^{\prime}+n(n+1)y=0$$


 * Find :

1)Verify exactness using:

a)Equations 4 and 5 on p. 10-3

b)Equation 5 on p. 14-1

2)If not exact, see whether it can be made exact by using the integrating factor method with $$h(x,y)=x^my^n$$

Solution:

1a) Equations 4 and 5 on p. 10-3 are as follows:


 * $$ \begin{align}

\text{(a)} \quad & f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} \\ \text{(b)} \quad & f_{xp}+pf_{yp}+2f_{y}=g_{pp} \end{align} $$

Where $$f=\Phi_p=(1-x^2)$$ and $$g=\Phi_x+\Phi_y p=n(n+1)y-2xp$$

Solving for the partial derivatives, we have:

$$f_x=-2x$$

$$f_y=0$$

$$f_{xx}=-2$$

$$f_{xy}=0$$

$$f_{yy}=0$$

$$f_{xp}=0$$

$$f_{yp}=0$$

$$g_x=-2p$$

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

$$g_p=-2x$$

$$g_{xp}=-2$$

$$g_{yp}=0$$

$$g_{pp}=0$$

Substituting into the original expressions we get:

$$f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}\Rightarrow -2+0+0\neq -2+0+n(n+1)$$

and

$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}\Rightarrow 0+0+0=0$$

The first condition is not met, therefore the Legendre ODE is not exact.

1b) Equation 5 on p.14-1 is:

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

And by definition, $$f_i=\frac{\partial F}{\partial y^{(i)}}$$

From this we see that:

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

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

$$f_2=\frac{dF}{dy''}=(1-x^2)$$

Substituting into the exactness condition we get:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0\Rightarrow n(n+1)-\frac{d}{dx}( -2x )+\frac{d^2}{dx^2}(1-x^2)=n(n+1)+2-2=n(n+1)\neq 0$$

Therefore the Legendre ODE is not exact.

2) To make the Legendre ODE exact, we attempt to use an integrating factor of the form $$x^m y^n$$, which yields:

$$F=x^My^N(1-x^2)y''-2x^My^Nxy'+x^My^Nn(n+1)y$$

This simplifies to:

$$(x^My^N-x^{M+2}y^N)y''-2x^{M+1}y^Ny'+x^My^{N+1}n(n+1)=0$$

Working in a similar fashion to part 1b, we get:

$$f_0=\frac{dF}{dy}=(N+1)x^My^Nn(n+1)$$

$$f_1=\frac{dF}{dy'}=-2x^{M+1}y^N$$

$$f_2=\frac{dF}{dy''}=x^My^N-x^{M+2}y^N$$

Substituting in to the original exactness condition we get:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=(N+1)n(n+1)x^My^N-(-2(M+1)x^My^N)+\frac{d}{dx}(Mx^{M-1}y^N-(M+2)x^{M+1}y^N)$$

Solving for the last derivative gives:

$$(N+1)n(n+1)x^My^N+2(M+1)x^My^N+M(M-1)x^{M-2}y^N-(M+2)(M+1)x^My^N=0$$

We can simplify the problem by pulling out the term $$x^My^N$$ giving:

$$(N+1)n(n+1)+2(M+1)+M(M-1)x^{-2}-(M+2)(M+1)=0$$

We can see that in order to eliminate n(n+1) from this equation, N must equal -1. M can be solved for by trial and error. Attempting the solution where M=0 yields:

$$2(M+1)+M(M-1)x^{-2}-(M+2)(M+1)\Rightarrow 2(1)+0-(2)(1)=0$$

Therefore, the Legendre ODE can be made exact using the integrating factor $$h(y)=y^{-1}$$

Contributors:
 * Egm6321.f09.team4.palubin.d 16:24, 7 October 2009 (UTC)

Problem #7:
(Lecture 14-3)


 * Given :That an operator $$L(\cdot )$$ is linear if and only if:

1)For all u(x),v(x), $$L(u+v)=L(u)+L(v)$$

and

2)For all $$\lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)$$


 * Find : Show that the above conditions for linearity are equivalent to equation 3 on p.3-3:

$$\forall\alpha,\beta \in\mathbb{R}$$, $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$ where u and v are any functions of x

Solution:

Define A and B such that: $$A=\alpha u$$ and $$B=\beta v$$

A and B are therefore still both functions of x, and applying the first given relation (equation 1 on p.14-3) we can write:

$$L(A+B)=L(A)+L(B)$$ or $$L(\alpha u + \beta v)=L(\alpha u) + L(\beta v)$$

Applying the second relation (equation 2 on p.14-3) and considering each term individually, we get:

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

Therefore the combination of relations 1 and 2 on p.14-3 are equivalent to equation 3 on p.3-3,

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

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

Contributors:
 * Egm6321.f09.team4.palubin.d 23:21, 6 October 2009 (UTC)

Problem #8:
Lecture 15-2


 * Given :
 * Shape functions for $$N_{j}^1(x),N_j^2(x),N_{j+1}^1(x)$$


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


 * Solution :


 * [[File:Egm6321_hw3_team4.PNG]]



correct. Egm6321.f09.TA 03:07, 28 October 2009 (UTC)

Contributors: :Egm6321.f09.team4.sls 17:32, 7 October 2009 (UTC) :Egm6321.f09.team4.palubin.d 17:32, 7 October 2009 (UTC)

Problem #9:
(Lecture 16-2)


 * Given : $$y(x)=y(x(t))$$ and $$x(t)=e^t$$


 * Find :Show that:

$$y_{xxx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

and

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

Solution:

We start from equations 5 on p.16-1 and 7 on p.16-2:

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

$$y_{xx}=e^{-2t}(y_{tt}-y_t)$$

Solving for $$y_{xxx}$$ yields:

$$y_{xxx}=\left ( y_{xx} \right )_x=\frac{d}{dt}(e^{-2t}(y_{tt}-y_t))\frac{dt}{dx}$$ where $$\frac{dt}{dx}=e^{-t}$$

Applying the chain rule, we get:

$$\left [ \left ( -2e^{-2t} \right )\cdot \left ( y_{tt}-y_t \right )+(e^{-2t})\cdot (y_{ttt}-y_{tt}) \right ]e^{-t}=e^{-3t}(-2y_{tt}+2y_t+y_{ttt}-y_{tt})$$

Which equals the solution

$$y_{xxx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

Solving for $$y_{xxxx}$$ is similar. We start from:

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

Substituting in $$\frac{dt}{dx}=e^{-t}$$ and applying the chain rule we get:

$$\left [ (-3e^{-3t})\cdot (y_{ttt}-3y_{tt}+2y_t)+(e^{-3t})\cdot (y_{tttt}-3y_{ttt}+2y_{tt}) \right ]e^{-t}$$

which equals:

$$e^{-4t}(-3y_{ttt}+9y_{tt}-6y_t+y_{tttt}-3y_{ttt}+2y_{tt})$$

This simplifies down to the final solution:

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



good - very legible. Egm6321.f09.TA 03:25, 28 October 2009 (UTC)

Contributors:

Egm6321.f09.team4.palubin.d 22:10, 6 October 2009 (UTC)

Problem #10:
(Lecture 16-4)


 * Given : The equation $$x^2 y^{\prime\prime} -2xy^{\prime}+2y=0$$


 * Find :
 * (a)Try to solve for this equation directly using the method of trial solutions with $$y=e^{rx}$$. For boundary conditions, set y(1)=3 and y(2)=4.
 * (b)Compare with the solution on p.16-3: $$y(x)=C_1 x^{r_1}+C_2 x^{r_2}$$
 * (c)Use Matlab to plot the solution.


 * Solution (a) :


 * $$y=e^{rx}$$
 * $$y^{\prime}=re^{rx}$$
 * $$y^{\prime\prime}=r^2e^{rx}$$


 * Substitute back in,


 * $$x^2(r^2e^{rx})-2xre^{rx}+2e^{rx}=0$$
 * $$e^{rx}(r^2x^2-2xr+2)=0$$
 * $$r^2x^2-2xr+2=0$$


 * Find roots,



This equation is quadratic in 2 variables. You cannot use the quadratic equation here. Egm6321.f09.TA 03:50, 28 October 2009 (UTC)


 * $$r_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
 * $$r_{1,2}=\frac{2\pm \sqrt{2^2-4(1)(2)}}{2(1)}$$
 * $$r_{1,2}=1\pm i$$


 * Solution has the form,


 * $$y=e^{1\pm i}$$
 * $$y=e^{x}e^{\pm i}$$
 * $$y=e^{x}(C_{1}cos(x)+C_{2}sin(x))$$


 * Apply boundary conditions,


 * $$3=e^{1}(C_{1}cos(1)+C_{2}sin(1))$$


 * and,


 * $$4=e^{2}(C_{1}cos(2)+C_{2}sin(2))$$


 * we find that,


 * $$C_{1}=\frac{3}{e^{1}cos(1)}-C_{2}tan(1)$$


 * and,


 * $$C_{2}=\frac{1}{tan(1)-tan(2)}\left ( \frac{3}{e^{1}cos(1)}-\frac{4}{e^{2}cos(2)} \right )$$


 * these are simply constants that correspond to $$C_{1}=0.6513$$ and $$C_{2}=0.8934$$


 * Solution:
 * $$y=e^{x}(C_{1}cos(x)+C_{2}sin(x))$$
 * $$C_{1}=\frac{3}{e^{1}cos(1)}-C_{2}tan(1)$$
 * $$C_{2}=\frac{1}{tan(1)-tan(2)}\left ( \frac{3}{e^{1}cos(1)}-\frac{4}{e^{2}cos(2)} \right )$$


 * Solution (b) :
 * We see that the solution has the same form as Equation (10) of 16-3; however, in this instance the polynomial solution is expressed in terms of an exponential function where $$x$$ has been substituted with $$e^{x}$$.


 * Solution (c) :
 * The plot of the solution is as follows:




 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:23, 7 October 2009 (UTC)

Problem #11:
(Lecture 17-3)


 * Given :$$u_1(x)z^{\prime}+(a_1(x)u_1(x)+2u_1^{\prime}(x))z=0$$


 * Find :Solve for z using the integrating factor method


 * Solution :


 * $$u_1(x)z^{\prime}+(a_1(x)u_1(x)+2u_1^{\prime}(x))z=0$$
 * $$z^{\prime}+\frac{(a_1(x)u_1(x)+2u_1^{\prime}(x))}{u_1(x)}z=0$$
 * $$z^{\prime}+\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )z=0$$


 * Choose an integrating factor $$h(x)$$ and multiply through,


 * $$h(x)z^{\prime}+\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )h(x)z=0$$


 * And force $$h(x)$$ such that,


 * $$h^{\prime}(x)=\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )h(x)$$


 * $$\frac{h^{\prime}(x)}{h(x)}=\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )$$


 * $$ln\left (h(x) \right )=\int \left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)}  \right )dx$$


 * $$ln\left (h(x) \right )=2ln(u_1(x))+\int a_1(x)dx+C$$


 * $$h(x)=e^{(2ln(u_1(x))+\int a_1(x)dx)+C}$$


 * $$h(x)=C_1(u_1(x))^{2}e^{\int a_1(x)dx}$$


 * Going back, recall we chose $$h(x)$$ such that,


 * $$h(x)z^{\prime}+h^{\prime}(x)z=0$$


 * Therefore,


 * $$\frac{\partial }{\partial x}\left (h(x)z) \right )=0$$


 * $$h(x)z=C_2$$


 * Plug in our expression for $$h(x)$$,


 * $$z=\frac{C_2}{C_1\left(u_1(x) \right )^{2}e^{\int a_1(x)dx}}$$


 * $$z=C_3(u_1(x))^{-2}e^{-\int a_1(x)dx}$$


 * (Answer)



very good. Egm6321.f09.TA 04:06, 28 October 2009 (UTC)


 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:24, 7 October 2009 (UTC)

Problem #12:
(Lecture 18-1)


 * Given :


 * $$y^{\prime\prime}+a_1(x)y^{\prime}+a_0(x)y=0$$ and


 * a) $$y(x)= U(x)\pm u_1(x)$$


 * b) $$y(x)= \frac{U(x)}{u_1(x)}$$


 * c) $$y(x)= \frac{u_1(x)}{U(x)}$$


 * Find : Develop a reduction of order method 2 using algebraic operations (a), (b), and (c) above.


 * Solution :


 * Part (a)
 * $$y(x)= U(x)\pm u_1(x)$$
 * $$y^{\prime}(x)= U^{\prime}(x)\pm u_1^{\prime}(x)$$
 * $$y^{\prime\prime}(x)= U^{\prime\prime}(x)\pm u_1^{\prime\prime}(x)$$


 * Plug in:


 * $$U^{\prime\prime}(x)\pm u_1^{\prime\prime}(x)+a_1(x)\left (U^{\prime}(x)\pm u_1^{\prime}(x) \right )+a_0\left (U(x)\pm u_1(x)  \right )=0$$


 * Factoring out the $$u_1(x)$$ homogenous solution, we are left with:


 * $$U^{\prime\prime}(x)+a_1(x)U^{\prime}(x)+a_0U(x)=0$$


 * We have not reduced the order of the equation; we did not achieve an ODE with a "missing" dependent variable. Therefore, operation (a) cannot be used in reduction of order processes.


 * Part (b)
 * Drop "(x)" for convenience only
 * $$y= \frac{U}{u_1}$$
 * $$y^{\prime}= \frac{-U}{u_1^{2}}u_1^{\prime}+\frac{U^{\prime}}{u_1}$$
 * $$y^{\prime\prime}= \frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-\frac{U'u_1^{\prime}}{u_1^2}-\frac{u_1'^2}{u_1^2}+\frac{U''}{u_1}$$


 * Plug back into ODE:
 * $$y^{\prime\prime}+a_1y^{\prime}+a_0y=0$$


 * $$\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-\frac{U'u_1^{\prime}}{u_1^2}-\frac{u_1'^2}{u_1^2}+\frac{U''}{u_1} \right )+a_1\left (\frac{-Uu_1'}{u_1^{2}}+\frac{U^{\prime}}{u_1}  \right )+a_0\frac{U}{u_1}=0$$


 * Group terms,


 * $$\frac{U''}{u_1}+\left (a_1\frac{U^{\prime}}{u_1}-\frac{U'u_1^{\prime}}{u_1^2} \right )+\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-a_1\frac{Uu_1'}{u_1^{2}}+a_0\frac{U}{u_1}  \right )-\frac{u_1'^2}{u_1^2}=0$$


 * The only way to have a reduction in order is if


 * $$\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-a_1\frac{Uu_1'}{u_1^{2}}+a_0\frac{U}{u_1} \right )=0$$


 * Working on this


 * $$U\left (\frac{2u_1'^{2}}{u_1^{3}}-\frac{u_1^{\prime\prime}}{u_1^2}-a_1\frac{u_1'}{u_1^{2}}+\frac{a_0}{u_1} \right )=0$$


 * $$a_0u_1^2+2u_1'^{2}-a_1u_1u_1'-u_1''u_1=0$$


 * $$u_1''+a_1u_1'-a_0u_1-\frac{2u_1'^{2}}{u_1}=0$$


 * However, because $$u_1$$ is a solution to the original such that


 * $$u_1''+a_1u_1'+a_0u_1=0 $$


 * We know that,


 * $$u_1''+a_1u_1'-a_0u_1-\frac{2u_1'^{2}}{u_1}\neq0$$


 * Answer.


 * Part (c)


 * Similar to Part (b), but here $$u_1$$ is replaced by $$U$$ and vice versa:


 * $$y= \frac{u_1}{U}$$
 * $$y^{\prime}= \frac{-u_1}{U^{2}}U^{\prime}+\frac{u_1^{\prime}}{U}$$
 * $$y^{\prime\prime}= \frac{2u_1U'^{2}}{U^{3}}-\frac{u_1U^{\prime\prime}}{U^2}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}+\frac{u_1''}{U}$$


 * Plug back into ODE:
 * $$y^{\prime\prime}+a_1y^{\prime}+a_0y=0$$


 * $$\left (\frac{2u_1U'^{2}}{U^{3}}-\frac{u_1U^{\prime\prime}}{U^2}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}+\frac{u_1''}{U} \right )+a_1\left (\frac{-u_1}{U^{2}}U^{\prime}+\frac{u_1^{\prime}}{U}  \right )+a_0\frac{u_1}{U}=0$$


 * Group terms,


 * $$-\frac{u_1U^{\prime\prime}}{U^2}+\left (\frac{2u_1U'^{2}}{U^{3}}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}-a_1\frac{u_1U'}{U^{2}} \right )+a_1\frac{u_1^{\prime}}{U}+a_0\frac{u_1}{U}+\frac{u_1''}{U}=0$$


 * Simplify,


 * $$-u_1U^{\prime\prime}+\left (\frac{2u_1U'^{2}}{U}-u_1'U^{\prime}-U'^2-a_1u_1U' \right )+U\left (a_1u_1^{\prime}+a_0u_1+u_1''  \right )=0$$


 * $$U^{\prime\prime}+\left (\frac{-2U'^{2}}{U}+\frac{u_1'U^{\prime}}{u_1}+\frac{U'^2}{u_1}+a_1U' \right )-\frac{U}{u_1}\left (a_1u_1^{\prime}+a_0u_1+u_1''  \right )=0$$


 * And although the last term goes to zero, it does not eliminate the $$U$$ term in the second term. Therefore, we cannot perform a reduction of order for Part (c).


 * Answer



very good. Egm6321.f09.TA 04:16, 28 October 2009 (UTC)


 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:26, 7 October 2009 (UTC)

Problem #13:
(Lecture 18-2)


 * Given :


 * Equation 1: $$(1-x^2)y''-2xy'+2y=0$$
 * a) $$y=ax^b$$
 * b) $$y=e^{rx}$$


 * Find : $$u_1$$ and $$u_2$$ of Equation (1) using trial solutions (a) and (b).


 * Solution :


 * Trial Solution (a)


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


 * Plug into Equation 1,


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


 * This is only true if b=1. Therefore,


 * $$y=ax$$


 * The value of a will be determined later.


 * $$u_1=ax$$


 * To find $$u_2$$, we then perform the following operation


 * $$y=Uu_1$$
 * $$y'=U'u_1+Uu_1'$$
 * $$y=Uu_1+2U'u_1'+u_1''U$$


 * Plug into Equation (1) and get,


 * $$(1-x^2)\left (Uu_1+2U'u_1'+u_1U \right )-2x(U'u_1+Uu_1')+2(Uu_1)=0$$


 * $$(1-x^2)Uu_1+U'\left ((1-x^2)2u_1'-2xu_1 \right )+U(-2xu_1'+2u_1+(1-x^2)u_1)=0$$


 * $$(1-x^2)Uu_1+U'\left ((1-x^2)2u_1'-2xu_1 \right )+\underset{0}{\underbrace{{}U(-2xu_1'+2u_1+(1-x^2)u_1)}}=0$$


 * $$U''+U'\left (\frac{2u_1'}{u_1}-\frac{2x}{(1-x^2)} \right )=0$$


 * With a change of variables $$U'=Z$$, we can show that,


 * $$U=\int Z=C_3u_1+C_2u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx \right ) dx$$


 * $$U=C_3u_1+C_2\underset{u_2}{\underbrace{u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx \right ) dx}}$$


 * Recall that $$y=Uu_1$$ and $$y=ku_1+k_2u_2$$. Therefore,


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


 * Which is evaluated as:


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


 * and


 * $$y(x)=C_3x+C_2u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx\right ) dx$$


 * ANSWER.



What about $$y=e^{rx}$$ ? Egm6321.f09.TA 05:34, 28 October 2009 (UTC)


 * Contributors:
 * Egm6321.f09.Team4.meyerat 13:04, 7 October 2009 (UTC)