User:Egm6321.f09.team4/HW2

Homework #2: Team 4 (Complete)  Please add 'EDIT' links to each problem so that I can leave feedback one problem at a time. Without these links I have to scroll through your entire page to find each line. This will also assist you in editing your problems. --Egm6321.f09.TA 03:29, 28 September 2009 (UTC)

The way you "add the 'EDIT' link" is to create sections and subsections using the equal signs "=", instead of using the html code H4; see the HW report of Team 1 or my Lecture Plan as examples. Egm6321.f09 13:14, 1 October 2009 (UTC) Problem #1:

(Lecture 7-1)

Given :
 * $$h_xN-h_yM+h(N_x-M_y)=0$$

 Start a line before this, with $$hM+hNy'=0$$ to show where this equation comes from. --Egm6321.f09.TA 03:45, 28 September 2009 (UTC) Find :
 * Solve for h(x,y) given that $$h_xN=0$$

Solution :


 * Substitute $$h_xN=0$$, then move one term to right-hand side of equation:


 * $$ \begin{align}

-h_{y}\left(x,y\right)M\left(x,y\right) + h\left(x,y\right)\left(N_{x}-M_{y}\right) &= 0 \\ h\left(x,y\right)\left(N_{x}-M_{y}\right) &= h_{y}M\left(x,y\right) \end{align} $$


 * Isolate terms involving $$h\left(x,y\right)$$, then simplify using definition of derivative of $$\text{ln}\left(u\right)$$:


 * $$ \begin{align}

\frac{h_{y}\left(x,y\right)}{h\left(x,y\right)} &= \frac{1}{M}\left(N_{x} - M_{y}\right) \\ \frac{\partial}{\partial y}\left\{\text{ln}\left[h\left(x,y\right)\right]\right\} &= \frac{1}{M}\left(N_{x}-M_{y}\right) \end{align} $$


 * Integrate both sides of the equation, then take the exponential function of both sides:


 * $$ \begin{align}

\text{ln}\left[h\left(x,y\right)\right] &= \int\frac{1}{M}\left(N_{x}-M_{y}\right)dy + C\left(x\right) \\ h\left(x,y\right) &= C\left(x\right) \text{exp}\left[ \int\frac{1}{M}\left(N_{x}-M_{y}\right)dy\right] \end{align}$$

Problem #2: (Lecture 8-2)

Given :


 * The non-homogeneous linear 1st Order ODE with Varying Coefficients $$y^{\prime}+\frac{1}{x}y=x^2$$

Find :


 * Verify that the integrating factor necessary to make the ODE exact is $$h(x)=x$$

Solution :

Multiplying through by h(x) yields:


 * $$xy^{\prime}+y=x^3$$

In order to be exact, both exactness conditions for 1st Order ODE's must be satisfied. To satisfy the first condition, the ODE must have the form:
 * $$F(x,y)=M(x,y)+N(x,y)y^{\prime}=0$$

Rearranging terms of our ODE after multiplication by the integration factor gives:


 * $$F(x,y)=\left (x \right )y^{\prime}+\left (y-x^3  \right )=0$$

Where $$N(x,y)=x$$ and $$M(x,y)=y-x^3$$ Therefore the first exactness condition is met. To verify the second exactness condition we must show:


 * $$M_y=N_x$$

We differentiate M and N with respect to y and x,


 * $$M_y(x,y)=1$$


 * $$N_x(x,y)=1$$

Therefore, both exactness conditions are satisfied and the 1st order ODE is exact.  You did not show that the solution is $$y(x)=1/4\cdot x^3 +\frac{1}{x}C$$. You showed that the integrating factor $$x$$ does indeed put the equation in an exact form, but you did not solve it. --Egm6321.f09.TA 03:45, 28 September 2009 (UTC) Problem #3:  (Lecture 9-2)

Given :


 * $$\frac{1}{2}x^2y^{\prime}+[x^4y+10]=0$$

Find :


 * Show that the above linear, 1st Order, ODE-VC is either exact, or exactly integrable by the integrating factors method.

Solution :

First, does this linear, first-order ODE satisfy the conditions for exactness?


 * If $$M$$ and $$N$$ are defined as follows:


 * $$ \begin{align}

M\left(x,y\right) &= x^{4}y + 10 \\ N\left(x,y\right) &= 0.5 x^{2} \end{align} $$


 * Then, can we find some function $$\Psi\left(x,y\right)$$ such that:


 * $$ \begin{align}

\frac{d\Psi\left(x,y\right)}{dx} &= M\left(x,y\right) \\ \frac{d\Psi\left(x,y\right)}{dy} &= N\left(x,y\right) \end{align} $$


 * For each equation, integrate both sides:


 * $$ \begin{align}

\Psi\left(x,y\right) &= \int{M\left(x,y\right)}dx + g_{1}\left(y\right) = \frac{x^5y}{5} + 10x + g_{1}\left(y\right) \\ \Psi\left(x,y\right) &= \int{N\left(x,y\right)}dy + g_{2}\left(x\right) = \frac{x^2y}{2} + g_{2}\left(x\right) \end{align} $$


 * Note that no choice of the arbitrary functions of integration $$g_{1}\left(y\right)$$ and $$g_{2}\left(x\right)$$ will make these two equations equal, due to the different powers of $$x$$ in the first term. So, this equation does not meet the conditions for exactness.

Since this ODE isn't exact, is it exactly integrable by the integrating factors method?


 * The equation under consideration can be rewritten in the following form, by multiplying both sides with $$2x^{-2}$$:


 * $$ \begin{align}

y' + p\left(x\right)&y = g\left(x\right) \\ \text{where} \quad &p\left(x\right) = 2x^{2} \\ &g\left(x\right) = -20x^{-2} \end{align} $$


 * Since this equation is a linear, first-order ODE with this particular structure, it is integrable via the integrating factors method. Let $$\mu\left(x\right)$$ be the integrating factor, then multiply both sides of the equation by it:


 * $$ \begin{align}

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


 * If we then assume that $$\mu\left(x\right)$$ satisfies $$\mu\left(x\right)p\left(x\right) = \mu'\left(x\right)$$, this equation becomes:


 * $$ \begin{align}

\mu\left(x\right)y' + \mu'\left(x\right)y &= \mu\left(x\right)g\left(x\right) \\ \frac{d}{dx}\left[ \mu\left(x\right)y \right] &= \mu\left(x\right)g\left(x\right) \quad \text{(product rule)} \end{align} $$


 * Integrating both sides of the equation, then solving for $$y$$, yields the following solution to the ODE:



y = \frac{\int{\mu\left(x\right)g\left(x\right)dx + c}}{\mu\left(x\right)} $$


 * To find $$\mu\left(x\right)$$, we then use the assumption described above:


 * $$\begin{align}

\mu\left(x\right)p\left(x\right) &= \mu'\left(x\right) \\ p\left(x\right) &= \frac{\mu'\left(x\right)}{\mu\left(x\right)} \\ p\left(x\right) &= \frac{d}{dx}\left[ \text{ln}\left(\mu\right) \right] \\ \int{p\left(x\right)}dx + k &= \text{ln}\left(\mu\right) \quad \text{(choose k=0)}\\ \int{2x^{2}}dx &= \text{ln}\left(\mu\right) \\ \mu\left(x\right) &= \text{exp}\left( \frac{2}{3}x^{3} \right) \end{align}$$


 * Finally, the solution to the equation is obtained by plugging in $$\mu\left(x\right)$$ and $$g\left(x\right)$$:



y = \frac{ -20\int{\text{exp}\left(\frac{2}{3}x^{3}\right)x^{-2}}dx + c }{\text{exp}\left(\frac{2}{3}x^{3}\right)} $$


 * While the solution to this integral is very difficult to obtain, it does exist. Therefore, this equation is exactly integrable via the integrating factors method.

 You were asked to show exactness, not solve for the solution. Calculating $$h(x,y)$$ (or in your case)$$\mu(x,y)$$ and verifying that$$\overline{N}_x=\overline{M}_y$$ would have sufficed. --Egm6321.f09.TA 03:45, 28 September 2009 (UTC)

Problem #4:  (Lecutre 9-3) Lecture 9-3 Given :


 * $$\frac{1}{3}x^3(y^4)y^{\prime}+(5x^3+2)(\frac{1}{5}y^5)=0$$

Find :


 * Show that the above equation is an exact nonlinear 1st Order ODE or integrable by the integrating factors method.

Solution :

To simplify, first divide through by $$y^4$$


 * $$\frac{1}{3}x^3y^{\prime}+(5x^3+2)(\frac{1}{5}y)=0$$

Notice that


 * $$\underset{N}{\underbrace{\frac{1}{3}x^3}}y^{\prime}+\underset{M}{\underbrace{{}(5x^3+2)(\frac{1}{5}y)}}=0$$

For exactness,


 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

For our case,


 * $$\frac{\partial }{\partial y}((5x^3+2)(\frac{1}{5}y))=\frac{\partial }{\partial x}(\frac{1}{3}x^3)$$


 * $$\frac{1}{5}(5x^3+2)\neq x^2$$

This equation is therefore not exact. To see if it is integrable, we manipulate the original equation


 * $$\frac{1}{3}x^3y^{\prime}+(5x^3+2)(\frac{1}{5}y)=0$$


 * $$y'+\left (\frac{x^{3}+\frac{2}{5}}{\frac{1}{3}x^3} \right )y=0$$

This equation now has the form of,


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

Multiple by each side of the above by integrating factor, $$h(x)$$


 * $$h(x)y'+h(x)p(x)y=0$$

We shall chose choose $$h(x)$$ such that


 * $$h(x)p(x)=h'(x)$$


 * $$p(x)=\frac{h'(x)}{h(x)}$$

Integrate,


 * $$\int p(x)dx=ln(h(x))$$


 * $$\int \left [\frac{x^{3}+\frac{2}{5}}{\frac{1}{3}x^3} \right ]dx=ln(h(x))$$


 * $$\int \left [3+\frac{6}{5}x^{-3} \right ]dx=ln(h(x))$$

Therefore,


 * $$h(x)=C_{1}e^{3x-\frac{6}{10}x^{-2}}$$

Let $$C_{1}=1$$ because it is arbitrary

Recall that this was selected so that


 * $$h(x)y'+h(x)p(x)y=0$$


 * $$\frac{\partial }{\partial x} \left [h(x)y(x) \right ]=0$$


 * $$\left [h(x)y(x) \right ]=C_{2}$$


 * $$y(x)=\frac{C_{2}}{h(x)}$$

And so,


 * $$y(x)=\frac{C_{2}}{e^{3x-\frac{6}{10}x^{-2}}}$$

This equation is integrable only for intervals where $$x\neq 0$$  Calculate $$h(x,y)$$ and then verify $$\overline{N}_x=\overline{M}_y$$. You were not asked to determine the solution, merely verify an exact form. --Egm6321.f09.TA 03:45, 28 September 2009 (UTC) Problem #5: </H4> (Lecture 10-3)

Given :


 * $$(xy)y{}''+x(y{}'^{2})+yy{}'=0$$

Find :


 * Determine if the 2nd condition for exactness is satisfied for the above equation.

Solution :


 * First, note that this equation can be rewritten in the following form:


 * $$\begin{align}

f\left(x,y,p\right)y'' + g\left(x,y,p\right) &= 0 \\ \\ \text{where} \quad f\left(x,y,p\right) &= xy \\ g\left(x,y,p\right) &= xp^{2} + yp \\ p &= y' \end{align}$$


 * This means that the first condition of exactness is satsified. For the second condition, two equations involving the partial derivatives of $$f$$ and $$g$$ must be satisfied:


 * $$\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}$$


 * Values for each of the required partial derivatives are obtained from the definitions of $$f\left(x,y,p\right)$$ and $$g\left(x,y,p\right)$$:


 * $$\begin{align}

&f_{xx}=f_{yy}=f_{xp}=f_{yp}=0, \quad f_{xy}=1, \quad f_{y} = x \\ &g_{xp}=2p,\quad g_{yp}=1,\quad g_{pp}=2x,\quad g_{y}=p \end{align}$$


 * These values are then substituted into the two equations from the second condition:


 * $$\begin{align}

\text{(a)} \quad &\left(0\right) + 2p\left(1\right) +p^{2}\left(0\right) = \left(2p\right) + p\left(1\right) - \left(p\right) \\ \text{(b)} \quad &\left(0\right) + p\left(0\right) + 2\left(x\right) = \left(2x\right) \\ \\

\text{(a)} \quad & 2p = 2p +p - p = 2p \\ \text{(b)} \quad & 2x = 2x \end{align}$$


 * Conditions (a) and (b) are both satisfied. Therefore, this ODE is exact.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Instead of working two equations simultaneously, show one equation from the second exactness condition and then show the other. Working a.) b.) .... a.) b.)  is confusing to the reader. --Egm6321.f09.TA 03:56, 28 September 2009 (UTC) <H4>Problem #6: </H4> (Lecture 11-2)

Given : Equation (5) of 10-2,


 * $$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

Find :


 * Derive the above equation by differentiating Equation (3) of 10-1, which is,


 * $$g(x,y,p)=\Phi _{x}+\Phi _{y}p$$


 * with respect to $$p=y{}'$$.

Solution :

First find the partial derivative of g(x,y,p) with respect to p:


 * $$g_p=\frac{\partial}{\partial p}\left ( \Phi_x+\Phi_y p \right )$$


 * $$g_p=\Phi_{xp}+p\Phi_{yp}+\Phi_y$$     by the Product rule

The second partial derivative of g with respect to p is similar:


 * $$g_{pp}=\frac{\partial }{\partial p}\left ( \Phi_{xp}+p\Phi_{yp}+\Phi_y \right )=\Phi_{xpp}+\Phi_{yp}+p\Phi_{ypp}+\Phi_{yp}$$    by the Product rule

Because of the nature of the exactness condition, the order of differentiation does not matter, therefore:
 * $$\Phi_{xpp}=\Phi_{pxp}=\frac{\partial^2}{\partial p \partial x}\left ( \frac{\partial \Phi}{\partial p} \right )=\frac{\partial^2}{\partial p \partial x}\left ( f \right)$$
 * where $$f=\frac{\partial \Phi}{\partial p}=\Phi_{p}$$

Making the appropriate substitutions for every term with a partial derivative of $$\Phi$$ with respect to p.


 * $$g_{pp}=\Phi_{xpp}+\Phi_{yp}+p\Phi_{ypp}+\Phi_{yp}$$


 * $$g_{pp}=(\Phi_{p})_{xp}+(\Phi_{p})_{y}+p(\Phi_{p})_{yp}+(\Phi_{p})_{y}$$


 * $$g_{pp}=f_{xp}+f_{y}+pf_{yp}+f_{y}$$

The final result is:


 * $$g_{pp}=f_{xp}+pf_{yp}+2f_y$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> nice work. --Egm6321.f09.TA 03:56, 28 September 2009 (UTC)

<H4>Problem #7:</H4> (Lecture 12-1)

Given :


 * $$\Phi _{xy}=\Phi _{yx}$$

Find :


 * Use the above relationship to derive the following 1st relationship in exactness (condition 2) for N2, ODEs;


 * $$f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

Solution :

Working from the relations: $$\Phi_{xy}=\Phi_{yx}$$,$$\Phi_{xp}=\Phi_{px}$$,$$\Phi_{py}=\Phi_{yp}$$,

and the expressions: $$f=\Phi_p$$ and $$g=\Phi_x+\Phi_yp$$

We obtain:

$$\Phi_{xy}=\Phi_{yx}\Rightarrow \left ( \Phi_x \right )_y=\left ( \Phi_y \right )_x$$ where $$\Phi_x=g-p\Phi_y$$ and $$\Phi_y=\frac{1}{p}\left ( g-\Phi_x \right )$$

Plugging in and taking the derivative, we obtain: $$\left ( \Phi_x \right )_y=\left ( \Phi_y \right )_x\Rightarrow \left ( g-p\Phi_y \right )_y=\left ( \frac{1}{p}\left [ g-\Phi_x \right ] \right )_x\Rightarrow g_y-p\Phi_{yy}=\frac{1}{p}g_x-\frac{1}{p}\Phi_{xx}$$

From another relation, we can obtain: $$\Phi_{yp}=\Phi_{py}\Rightarrow \left ( \Phi_y \right )_p=\left ( \Phi_p \right )_y\Rightarrow\left (\frac{1}{p}\left [ g-\Phi_x \right ] \right )_p=f_y\Rightarrow -\frac{1}{p^2}\left ( g-\Phi_x \right )+\frac{1}{p}\left ( g_p-\Phi_{xp} \right )=f_y$$

From here we see that $$\Phi_{xp}=\Phi_{px}\Rightarrow\ \left ( \Phi_p \right )_x=f_x$$ and rearranging terms gives:

$$\Phi_x=g-p\left ( g_p-f_x \right )+p^2f_y$$

From the last relation we have: $$\Phi_{xp}=\Phi_{px}\Rightarrow \left ( \Phi_x \right )_p=\left ( \Phi_p \right )_x\Rightarrow \left ( g-p\Phi_y \right )_p=f_x\Rightarrow g_p-\Phi_y-p\Phi_{yp}=f_x$$

Again we see that $$\Phi_{yp}=\Phi_{py}\Rightarrow\ \left ( \Phi_p \right )_y=f_y$$, and rearranging terms gives:

$$\Phi_y=g_p-pf_y-f_x$$

$$\Phi_x$$ and $$\Phi_y$$ are then substituted back into the relation we obtained from $$\Phi _{xy}=\Phi _{yx}$$ as follows:

$$g_y-p\left ( g_p-pf_y-f_x \right )_y=\frac{1}{p}g_x-\frac{1}{p}\left ( g-p(g_p-f_x)+p^2f_y \right )_x$$

Taking the remaining derivatives and simplifying yields:

$$g_y-p\left ( g_{py}-pf_{yy}-f_{xy} \right )=-\frac{1}{p}\left ( g_x-g_x-p(g_{px}-f_{xx})+p^2f_{yx}) \right )$$

$$g_y-pg_{py}+p^2f_{yy}+pf_{xy}=g_{px}-f_{xx}-pf_{yx}$$

$$g_y-pg_{py}-g_{px}=-2pf_{xy}-p^2f_{yy}-f_{xx}$$

Multiplying through by negative one gives the final solution:

$$f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> You made a great deal of extra work for yourself. Start with $$\Phi_{xy}=\Phi_{yx}$$ and proceed from there, using expressions for $$\Phi_x$$ and $$\Phi_y$$. You do not need to have these $$\frac{1}{p}$$ terms in your solution. --Egm6321.f09.TA 03:56, 28 September 2009 (UTC)

<H4>Problem #8:</H4> (Lecture 12-2)

Given :


 * $$f=8x^{5}p$$


 * $$g=2x^{2}p+20x^{4}p^{2}+4xy$$

Find :


 * Show that equations (4) and (5) on 10-2 for exactness are satisfied for the above equations. Equations (4) and (5) are


 * $$ \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} $$

Solution :


 * Values for each of the required partial derivatives are obtained from the definitions of $$f$$ and $$g$$:


 * $$ \begin{align}

&f_{xx} = 160x^{3}p, \quad f_{xp}=40x^{4} \\ &f_{xy} = f_{yy}=f_{yp}=f_{y}=0 \\ &g_{xp} = 4x + 160x^{3}p, \quad g_{pp}=40x^{4} \\ &g_{yp} = 0, \quad g_{y}=4x \end{align}$$


 * These values are then substituted into the two equations from the second condition:


 * $$ \begin{align}

\text{(a)} \quad & 160x^{3}p + 2p\left(0\right) + p^{2}\left(0\right) = \left(4x+160x^{3}p\right) + p\left(0\right) - 4x \\ \text{(b)} \quad & 40x^{4} + p\left(0\right) + 2\left(0\right) = 40x^{4} \\ \\ \text{(a)} \quad & 160x^{3}p - 160x^{3}p = 4x - 4x \\ \text{(b)} \quad & 40x^{4} = 40x^{4} \\ \\ \text{(a)} \quad & 0 = 0 \\ \text{(b)} \quad & 0 = 0 \end{align} $$


 * Both equations (a) and (b) are satisfied.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Again, work on one equation at a time. Mathematical explanations never evolve two equations simultaneously. --Egm6321.f09.TA 03:56, 28 September 2009 (UTC) <H4>Problem #9: </H4> (Lecture 12-3)

Given :


 * $$\sqrt{x}y{}''+2xy{}'+3y=0$$

Find :


 * Verify exactness of the above ODE.

Solution :


 * The first condition of exactness is that the equation follows the following expression:


 * $$\begin{align}

f\left(x,y,p\right)y'' + g\left(x,y,p\right) &= 0 \\ \\ \text{where} \quad f\left(x,y,p\right) &= x^{\frac{1}{2}} \\ g\left(x,y,p\right) &= 2xp + 3y \\ p &= y' \end{align}$$


 * The first condition of exactness is satsified. For the second condition, two equations must be satisfied:


 * $$\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}$$


 * Values for each of the required partial derivatives are obtained from the definitions of $$f\left(x,y,p\right)$$ and $$g\left(x,y,p\right)$$:


 * $$\begin{align}

&f_{xx}=-\frac{1}{4}x^{-\frac{3}{2}} &f_{xy}=f_{yy}=f_{xp}=f_{yp}=f_{y}=0 \\ &g_{xp}=2,\quad g_{yp}=g_{pp}=0,\quad g_{y}=3 \end{align}$$


 * These values are then substituted into the two equations from the second condition:


 * $$\begin{align}

\text{(a)} \quad &\left(-\frac{1}{4}x^{-\frac{3}{2}}\right) + 2p\left(0\right) +p^{2}\left(0\right) = \left(2\right) + p\left(0\right) - \left(3\right) \\ \text{(b)} \quad &\left(0\right) + p\left(0\right) + 2\left(0\right) = \left(0\right) \\ \\

\text{(a)} \quad & -\frac{1}{4}x^{-\frac{3}{2}} = -1 \\ \text{(b)} \quad & 0 = 0 \end{align}$$


 * Only condition (b) is satisfied. Therefore, this ODE is not exact.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good. --Egm6321.f09.TA 03:56, 28 September 2009 (UTC)

<H4>Contributing Team Members:</H4> Egm6321.f09.Team4.meyerat 03:33, 23 September 2009 (UTC)

Egm6321.f09.team4.sls 16:31, 23 September 2009 (UTC)

Egm6321.f09.team4.palubin.d 22:56, 22 September 2009 (UTC)