User:Egm6321.f09.Team2/HW3

Homework Assignment #3 - due Wednesday, 10/7, 21:00 UTC

Problem 1
Find $$(m,n)$$ such that eqn. 1 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) is exact. A first integral is $$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Problem Statement : Given a L2_ODE_VC $$\sqrt{x}y''+2xy'+3y=0$$

Find (m,n) from the integrating factor (xm,yn) that makes the equation exact.

A first integral is $$\phi(x,y,p)=xp+\big(2x^{3/2}\big)y+k_1=k_2$$


 * $$\phi_p=x$$
 * $$\phi_x=p+3x^{1/2}$$
 * $$\phi_y=2x^{3/2}-1$$

$$\phi(x,y,p)=h(x,y)+\int_{}f(x,y,p)dp, f=\phi_p$$

$$\phi(x,y,p)=h(x,y)+\int_{}xdp={h(x,y)+xp}$$

$$g(x,y,p)=\phi_x+\phi_yp=h_x+\phi_x+(h_y+\phi_y)p$$

$$g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_x+x+(h_y+0)p$$


 * $$h_y=2x^{3/2}$$
 * $$h_x=3x^{1/2}-x$$

$$y^n=h_y\Rightarrow lny^n=ln(2x^{3/2})$$  $$n=ln(2x^{3/2})$$ $$x^m=h_x\Rightarrow lnx^m=ln(3x^{1/2}-x)$$  $$m=ln(3x^{1/2}-x)$$  You were asked to find the $$m,n$$ that make the ODE exact. Your expressions for m,n are functions of $$x$$ which is not constant. Start with the ODE and multiply by $$h(x,y)=x^my^n$$ and then enforce the exactness conditions to solve for $$m,n$$. --Egm6321.f09.TA 05:11, 15 October 2009 (UTC)

Problem 2
Solve eqn. 2 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) for $$y(x)$$.

Problem Statement: Given a first integral$$\phi$$ of a L2_ODE_VC, solve for $$y(x)$$.

$$\phi(x,y,p)=xp+(2x^{3\over 2}-1)y+k_1=k_2$$ (1)

where k1 and k2 are const, and $$p=y'$$

Eq. (1) is in the form $$M(x,y)+N(x,y)y'$$ where

$$M(x,y)=(2x^{3\over 2}-1)y$$

$$N(x,y)=x$$

so it satisfies the 1st condition of exactness.

Check if $$M_y=N_x$$ for the 2ndcondition of exactness

$$M_y=2x^{3\over2}-1$$

$$N_x=1$$

$$M_y\ne N_x$$ so we do not satisfy the 2nd condition of exactness.

We must apply the integrating factor method for a L1_ODE_VC.

$$xp+(2x^{3\over 2}-1)y=k_2-k_1$$, divide by x to obtain the form:

$$y'+a_0(x)y=b(x)$$ where:

$$a_o(x)={1\over x}(2x^{3\over 2}-1)=2\sqrt{x}-{1\over x}$$

$$b(x)={k_2-k_1\over x}$$

From our solution of a general non-homogeneous L1_ODE_VC p.8-1

$$h(x)=exp\int_{}^{x}a_0(s)ds$$

$$h(x)=exp\int_{}^{x}2\sqrt{x}-{1\over x}=exp\bigg({4\over 3}x^{3\over 2}-ln\left|x\right|\bigg)\bigg({{k_2-k_1}\over x}\bigg)=exp{4\over 3}x^{3\over 2}-x$$

 Something isn't right here. I think this is where you got off track.--Egm6321.f09.TA 23:49, 27 October 2009 (UTC)

From p.8-2 Eq. (4)

$$y(x)={1\over h(x)}\int_{}^{x}h(s)b(s)ds$$

Use the product rule of integration $$\int_{} ab=a\int_{} b-\int_{}a'\int_{}b$$

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-\int_{}^{x}h'(x)\int_{}^{x}b(x)\bigg]$$

In our example $$\int_{}^{x}h'(x)=h(x)$$ so,

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-h(x)\int_{}^{x}b(x)\bigg]$$



$$y(x)=0$$

 $$y(x)\neq 0$$. Check your steps. Egm6321.f09.TA 23:49, 27 October 2009 (UTC)

Problem 3
From ([[media:Egm6321.f09.p13-1.png|p.13-1]]), find the mathematical structure of $$\Phi$$ that yields the above class of ODE.

$$F(x,y,y',y'')={d\phi\over dx}=\phi_x(x,y,p)+\phi_y(x,y,p)p+\phi_p(x,y,p)p'\ where\ p=y'$$

$$\phi=h(x,y)+\int_{}\phi_pdp=h(x,y)+\phi_pp$$

$$\phi_x=h_x+\phi_{px}p$$

$$\phi_y=h_y+\phi_{py}p$$

$$g=\phi_x+\phi_yp=h_x+\phi_{px}p+(h_y+\phi_{py}p)p$$

$$h_y=\phi_y-\phi_{px}$$

Take the integral of $$h_y$$

$$h(x,y)=(\phi_y-\phi_{px})y+k_1$$

Substitute back into the equation for $$\phi$$

$$\phi=(\phi_y-\phi_{px})y+k_1+\phi_pp$$

Rearrange the terms to obtain

 $$\phi(x,y,p)=\phi_pp+(\phi_y-\phi_{px})y+k$$ where,


 * $$P(x)=\phi_p$$
 * $$T(x)=(\phi_y-\phi_{px})y$$
 * $$k=k_1$$



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

Your solution is not easy to follow. The reader has to guess why you are taking the steps you chose. Egm6321.f09.TA 01:53, 28 October 2009 (UTC)

Problem 4
From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=1$$ (N1_ODE) $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$. Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$. Hint: Use $$f_1=\Phi_y$$. Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem Statement: Given a N1_ODE, for the case n=1 $$F(x,y,y')=0\Leftrightarrow {d\phi\over dx}(x,y)$$

Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$ Hint:$$f_1=\phi_y$$

$$F={d\phi\over dx}(x,y^{(0)},....y^{(n-1)}=\phi_x+\phi_y6{(0)}y^{(1)}$$  Why is there a 6 in here ? --Egm6321.f09.TA 14:03, 15 October 2009 (UTC)

$$F=\phi_x+\phi_yy'$$

$$f_i:={\partial F\over \partial y^i}$$

4.1
Find $$f_0$$ in terms of $$\phi$$.

$$f_0={\partial F\over \partial y}={\partial (\phi_x+\phi_yy')\over \partial y}=\phi_{xy}$$  $$f_0=\phi_{xy}$$

 $$f_0=\phi_{xy}+\Phi_{yy}y'$$ You are missing terms.

4.2
Find $$f_1$$ in terms of $$\phi_y$$

$$f_1={\partial F\over \partial y'}={\partial (\phi_x+\phi_yy')\over \partial y'}=\phi_y$$  $$f_1=\phi_y$$

4.3
Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$

$$\phi_{xy}-{d\phi_y\over df_x}=\phi_{xy}-\phi_{yx}=0$$  $$\phi_{xy}=\phi_{yx}$$

 You should get $$\phi_{xy}+\phi_{yy}y'=\Phi_{yx}+\phi_{yy}y'$$. --Egm6321.f09.TA 14:03, 15 October 2009 (UTC)

Problem 5
Problem statement : From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for a non-linear, second order, ordinary differential equation (N2_ODE), $$n=2$$:
 * 1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$
 * 2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$
 * 3) Show $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$
 * 4) Relate eqn. 5 to eqs. 4&5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]).

Before solving, there are a few definitions required. First,
 * $$F(x,y,y',y)=\frac{d}{dx}\phi(x,y,y')=\phi_x+\phi_{y}y'+\phi_{y'}y$$

Second,
 * $$f_i:=\frac{\partial F}{\partial y_i}$$

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

Therefore, we can write out $$f_0$$, $$f_1$$, and $$f_2$$ as,
 * $$f_0=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_yy'_y+\phi_{y'y}y+\phi_{y'}y_y\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=\phi_{xy'}+\phi_{yy'}y'+\phi_y+\phi_{y'y'}y+\phi_{y'}y_{y'}\!\,$$
 * $$f_2=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_yy'_{y}+\phi_{y'y}y+\phi_{y'}\!\,$$

Since $$\phi$$ is only a function of $$x$$, $$y$$, and $$y'$$, any partial with respect to $$y''$$ is zero. Additionally, partials of $$y$$ and its derivatives with respect to a different $$y$$ or its derivatives are zero. Therefore,
 * $$f_0=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_{y'y}y''\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=\phi_{xy'}+\phi_{yy'}y'+\phi_{y'y'}y''+\phi_y\!\,$$
 * $$f_2=\frac{\partial F}{\partial y''}=\phi_{y'}\!\,$$

Taking the total derivative of $$f_2$$ with respect to $$x$$,
 * $$\frac{df_2}{dx}=\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Notice that $$f_1-\frac{df_2}{dx}=\phi_y$$, therefore (1) is satisfied. Taking the total derivative of $$\phi_y$$ with respect to x we get,
 * $$\frac{d\phi_y}{dx}=\phi_{yx}+\phi_{yy}y'+\phi_{yy'}y''$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Comparing to $$f_0$$ we see that, $$f_0=\frac{d\phi_y}{dx}$$, therefore (2) is satisfied. Combining the derivative of (1) with (2) we get,
 * $$\frac{df_1}{dx}-\frac{d^2f_2}{dx^2}=\frac{d\phi_y}{dx}=f_0$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Rearranging leads to $$\frac{df_1}{dx}-\frac{d^2f_2}{dx^2}=f_0$$ therefore (3) is satisfied. Finally, we can relate (3) to equations 4 & 5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]). By redefining $$F\!\,$$ in terms of new functions $$f\!\,$$ and $$g\!\,$$ (used in eq. 4 & 5) yields,
 * $$F=fy''+g\!\,$$

Following all the same steps to get $$f_i\!\,$$, for $$i=0,1,2\!\,$$, in terms of these new variables yields,
 * $$f_0=\frac{\partial F}{\partial y}=f_yy''+g_y\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=f_{y'}y''+g_{y'}\!\,$$
 * $$f_2=\frac{\partial F}{\partial y''}=f\!\,$$

Taking the total derivative of $$f_1\!\,$$ with respect to $$x\!\,$$ once, and taking it twice for $$f_2\!\,$$ yields,
 * $$\frac{df_1}{dx}=\frac{df_{y'}y}{dx}+g_{y'x}+g_{y'y}y'+g_{y'y'}y\!\,$$
 * $$\frac{d^2f_2}{dx^2}=\frac{df_{y'}y}{dx}+f_{xx}+f_{xy}y'+f_{xy'}y+(f_{yx}+f_{yy}y'+f_{yy'}y)y'+f_yy\!\,$$

Now plugging $$f_0\!\,$$, $$\frac{df_1}{dx}\!\,$$, and $$\frac{d^2f_2}{dx^2}\!\,$$ into (3) yields,
 * $$2f_yy+g_y-g_{y'x}-g_{y'y}y'-g_{y'y'}y+f_{xx}+2f_{xy}y'+f_{xy'}y+f_{yy}y'^2+f_{yy'}yy'=0\!\,$$
 * $$=>(2f_y-g_{y'y'}+f_{xy'}+f_{yy'}y')y''+g_y-g_{y'x}-g_{y'y}y'+f_{xx}+2f_{xy}y'+f_{yy}y'^2=0\!\,$$

Since $$f\!\,$$ and $$g\!\,$$ are not functions of $$y\!\,$$, $$y\!\,$$ can be arbitrary without changing their values. Therefore, to ensure the equation equals zero, the equation can be split into two parts, each equaling zero independently allowing $$y''\!\,$$ to be arbitrary. <div style="width: 50%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Therefore,
 * $$2f_y-g_{y'y'}+f_{xy'}+f_{yy'}y'=0\!\,$$
 * $$g_y-g_{y'x}-g_{y'y}y'+f_{xx}+2f_{xy}y'+f_{yy}y'^2=0\!\,$$

These equations are equal to 4 & 5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]) if setting $$y'=p\!\,$$.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good work. Consider using the notation $$y'=p$$ and $$y''=q$$. Then you do not have to have primes in your subscripts and you can take partials with respect to $$p,q$$. --Egm6321.f09.TA 01:27, 16 October 2009 (UTC)

Problem 6
Problem statement : From ([[media:Egm6321.f09.p14-2.png|p.14-2]]), for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.) ([[media:Egm6321.f09.p10-3.png|p.10-3]]), Equations 4&5. 6.1b.) ([[media:Egm6321.f09.p14-1.png|p.14-1]]), Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

First, no matter which method is used, the form of the equation meets the first exactness criteria since, $$F$$ can be written as,
 * $$F(x,y,y',y)=f(x,y,y')y+g(x,y,y')=0\!\,$$

Next we test the second set of exactness relations. The first two for part 1 are,
 * $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y\!\,$$
 * $$f_{xp}+pf_{yp}+2f_y=g_{pp}\!\,$$

where,
 * $$f=1-x^2\!\,$$
 * $$g=-2xy'+n(n+1)y\!\,$$:$$\!\,$$

Therefore we can calculate all the partials required to test exactness,
 * $$f_{xx}=-2, f_{xy}=f_{yy}=f_{xp}=f_{yp}=f_y=0\!\,$$
 * $$g_{xp}=-2, g_y=n(n+1), g_{yp}=g_{pp}=0\!\,$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Plugging into the exactness relations we see,
 * $$n(n+1)=0\!\,$$ (satisfied for $$n=-1$$ and $$n=0$$)

and,
 * $$0=0\!\,$$ (satisfied identically)

Last, we test the second exactness relation for part 2. That is,
 * $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0\!\,$$

where $$f_i=\frac{\partial F}{\partial y^i}\!\,$$, $$i=0,1,2\!\,$$. Writing out each $$f_i\!\,$$,
 * $$f_0=n(n+1)\!\,$$
 * $$f_1=-2x=>\frac{df_1}{dx}=-2\!\,$$
 * $$f_2=1-x^2=>\frac{df_2}{dx}=-2x=>\frac{d^2f_2}{dx^2}=-2\!\,$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Plugging into the relationship yields,
 * $$n(n+1)+2-2=0\!\,$$ (satisfied for $$n=-1$$ and $$n=0$$)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work. What about the case when $$n\neq 0,1$$? Can you make the equation exact using an integration factor?--Egm6321.f09.TA 01:29, 16 October 2009 (UTC)

Problem 7
Problem statement: From ([[media:Egm6321.f09.p14-3.png|p.14-3]]). Show that equations 1 and 2, namely 7.1 $$\forall u(x)\!\,$$ and $$v(x)\!\,$$, $$L(u+v)=L(u)+L(v)\!\,$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\!\,$$ with $$u\!\,$$ a function of $$x\!\,$$. are equivalent to equation 3 on [[media:Egm6321.f09.p3-3.png|p.3-3]].

An operator is linear if it follows the relationship $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)\!\,$$. This is equivalent to saying that the operator must satisfy simultaneously. <div style="width: 40%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> Given $$L(\alpha u+\beta v)\!\,$$, it can be split into two parts using (1),
 * 1) $$L(u+v)=L(u)+L(v)\!\,$$
 * 2) $$L(\alpha u)=\alpha L(u)\!\,$$
 * $$L(\alpha u+\beta v)=L(\alpha u)+L(\beta v)\!\,$$

Then each part can be used with (2)
 * $$L(\alpha u)=\alpha L(u)\!\,$$
 * $$L(\beta v)=\beta L(v)\!\,$$

Therefore, the two relations (1) and (2) are equivalent to the composite relation if taken together only. < <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> clear explanation. --Egm6321.f09.TA 04:02, 16 October 2009 (UTC)

Problem 8
From ([[media:Egm6321.f09.p15-2.png|p.15-2]]), plot the shape function $$N_{j+1}^{2}(x)$$.

[[Media:Graph1.pdf]]

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

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

Problem 9
Problem Statement: From (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$

$$y_{xxx}=(d/dx)\bigg[{d/dx\over (d/dx)y}\bigg]=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

What does this notation mean? Egm6321.f09.TA 03:20, 28 October 2009 (UTC)

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$ 'Chain Rule'

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(t_{tt}))$$

$$y_{xxx}=(e^{-t})(d/dt)(-e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

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

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

Factor out $$e^{-3t}$$ and re-arrange terms in ordre of derivative, <div style="width: 35%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;">

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

$$y_{xxxx}=(d/dx)(d/dx)(d/dx)(d/dx)y$$

$$y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))\langle(dt/dx)(d/dt)y\rangle$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(-2e^{-2t}(y_t)+e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

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

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

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

Factor out $$e^{-4t}$$ and re-arrange terms in order of derivative.

<div style="width: 45%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;">

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

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

It is not clear at all here which terms your differential operators are acting on. At a minimum, you need some parenthesis or brackets to clarify the grouping of terms. Egm6321.f09.TA 03:20, 28 October 2009 (UTC)

Problem 10
Problem Statement: Solve equation 1 on ([[media:Egm6321.f09.p16-1.png|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 [[media:Egm6321.f09.p16-3.png|p.16-3]]. Use MATLAB to plot the solutions.

We solve by trial solution. First attempt $$y(x)=e^{rx}\!\,$$. The derivatives are,
 * $$y'=re^{rx}\!\,$$
 * $$y''=r^2e^{rx}\!\,$$

Plugging into the differential equation we obtain,
 * $$x^2r^2e^{rx}-2xre^{rx}+2e^{rx}=0\!\,$$

Dividing by $$e^{rx}\!\,$$ yields,
 * $$x^2r^2-2xr+2=0\!\,$$

which cannot be solved for uniquely. Therefore we change the trial solution to, $$y(x)=x^r\!\,$$. The derivatives are,
 * $$y'=rx^{r-1}\!\,$$
 * $$y''=r(r-1)x^{r-2}\!\,$$

And plugging into the differential equation yields,
 * $$r(r-1)x^r-2rx^r+2x^r=0\!\,$$

Dividing by $$x^r\!\,$$ since $$x=0\!\,$$ is a trivial solution yields,
 * $$r^2-3r+2=(r-2)(r-1)=0\!\,$$

The roots are then $$r=1\!\,$$ and $$r=2\!\,$$, exactly the same answer as equation 10 on [[media:Egm6321.f09.p16-3.png|p.16-3]]. The form of the answer is then,
 * $$y(x)=C_1x+C_2x^2\!\,$$

<div style="width: 30%; margin-left: 0; margin-right: 0; padding: 8px; border: 2px solid #0000FF; text-align: center;"> The boundary conditions can be used to obtain $$C_1=4\!\,$$ and :$$C_2=-1\!\,$$. Therefore the solution is,
 * $$y(x)=4x-x^2\!\,$$



<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

very good. Egm6321.f09.TA 03:45, 28 October 2009 (UTC)

Problem 11
Problem Statement: From (| p.17-4) obtain equation 2 from p.17-3 $$

Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$ using the integrator factor method.

Solution:

$$u_{1}(x) Z'+\left[a_{1}u_{1}(x)+2u_{1}' (x) \right]Z=0 $$

$$h(x)Z'+h(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 $$

$$\frac{d}{dx}\left[h(x)Z(x) \right]=h(x)Z'(x)+h'(x)Z$$

Let $$h'(x)= h(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$\frac{h'(x)}{h(x)}= \left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

Integrating both sides

$$ \int{\frac{h'(x)}{h(x)}} = \int\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$ ln|h| = \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]$$

$$ h = \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}$$

$$ [h(x)Z(x)]' = 0$$ ,where hZ(x) is

$$ hZ(x) = \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)$$

$$ [hZ(x)]' = [\exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)]'$$

$$ \int[hZ(x)]' = \int[\exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)]'=\int0$$

$$ \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)= C $$

Where C is constant

Therefore,

$$ Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

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

Problem 12
Problem Statement: From (| p.18-1), develop reduction of order method using the following algebraic options

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

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

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

Solution:

a. $$ y(x) = U(x) \pm u_1(x) $$  <eq(1)>

$$y'(x) = U' \pm u_1'$$        <eq(2)>

$$y(x) = U \pm u_1''$$        <eq(3)>

Substituding Eq(1),(2) and(3) into following Eq

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

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

Therefore

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

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.

b.

$$y(x)=\frac{U(x)}{u_1 (x)}$$   <eq(1)>

$$y'(x) = \frac{U'u_{1}-Uu_1'}{u_{1}^2}$$   <eq(2)>

$$y''(x) = \frac{(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}{u_{1}^4}$$    <eq(3)>

Substituding eq(1),(2) and (3) into following Equation

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

$$ y''+a_1y'+a_0y = \frac{(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}{u_{1}^4}

+a_1{\frac{U'u_{1}-Uu_1'}{u_{1}^2}}

+a_0{\frac{U(x)}{u_1 (x)}}$$

Multify equation by $$ u_{1}^4 $$

Rearrage

$$ y''+a_1y'+a_0y = {(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}

+a_1{U'u_{1}-Uu_1'}{u_{1}^2}{u_{1}^2}

+a_0{U(x)}{u_1 (x)}{u_{1}^3}=0$$

The above y(x) here is not valid because the equation above is not missing U(x). The Reduction of Order Method is possible only when dependent variable is missing.

c.

$$y(x)=\frac{u_1 (x)}{U(x)}$$    <eq(1)>

$$y'(x) = \frac{u_{1}'U-u_1U'}{U^2}$$    <eq(2)>

$$y''(x) = \frac{(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)'}{U^4}$$   <eq(3)>

Substituding eq(1),(2) and (3) into following Equation

y'' + a_1y' + a_0y = 0

$$ y''+a_1y'+a_0y = \frac{(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)'}{U^4}

+a_1\frac{u_{1}'U-u_1U'}{U^2}

+a_0\frac{u_1 (x)}{U(x)}$$

Multify equation by $$ U^4 $$

Rearrage

$$ y''+a_1y'+a_0y = [(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)']

+a_1[u_{1}'U-u_1U']U^2

+a_0[u_1]U^3 = 0 $$

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.

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

Problem 13
Problem Statement: From (| p.18-1), Find $$u_{1}(x)$$ and $$ u_{2}(x)$$ of equation 1 on p.18-1 using 2 trial solutions:

$$ y=ax^b$$

$$ y=e^{rx}$$

Compare the two solutions using boundary conditions $$y(0)=1$$ and $$ y(1)=2$$ and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Solution:

1. When $$ y=ax^b $$

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

$$ y=ax^{b} $$ <eq(2)>

$$ y'=bax^{b-1} $$  <eq(3)>

$$ y''=ab(b-1)x^{(b-2)} $$  <eq(4)>

Substituting eq(2),(3) and (4) into Eq(1)

$$ (1-x^2)ab(b-1)x^{(b-2)} - 2xabx^{(b-1)} + 2ax^b = 0 $$ $$ ab(b-1)x^{(b-2)}-ab(b-1)x^b - 2abx^b + 2ax^b = 0 $$

$$ ab(b-1)x^{(b-2)}-\left[ab(b-1) + 2ab - 2a \right]x^b = 0 $$

In the first Term of LHS $$ x^{(b-2)}$$,

$$ ab(b-1) = 0 $$

a and b both cannot be zero, Therefore b=1

In the second Term of LHS, $$ x^b$$

$$ ab(b-1) + 2ab - 2a = 0$$

$$ a(b^2+b-2) = 0$$

a can not be zero ,hence $$ b =1,-2$$

$$ u_1(x) = ax$$

$$ u_2(x) = \frac{a}{x^2} $$

The trail solution is

$$ y = c_1u_1(x) + c_2u_2(x) $$

$$ y = C_1ax + C_2\frac{a}{x^2} $$

Using the boundary conditions,

$$ y(0) = 1 $$

$$ y(1) = 2 $$

$$ 1 = C_1a(0)+C_2\frac{a}{0} $$

$$ 2 = C_1a(1)+C_2\frac{a}{1}$$

This Equation cannot be established.

Therefore

$$ y=ax^b$$ cannot be a solution.

2. When $$ y=e^{rx}$$ $$ (1-x^{2})y''-2xy'+2y=0 $$ <eq(1)>

$$ y=e^{rx} $$  <eq(2)>

$$ y'= re^{rx} $$  <eq(3)>

$$ y''= r^{2}e^{rx} $$   <eq(4)>

Substituting Eq(2),(3) and(4) into Eq(1)

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

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Missing a term from the line above. Egm6321.f09.TA 04:52, 28 October 2009 (UTC)

Rearrange

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

$$ r = \pm \sqrt2i$$

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How can you simply this? Why did you set $$x=0$$? Egm6321.f09.TA 04:52, 28 October 2009 (UTC)

The trial solution becomes,

$$\displaystyle y = e^{\sqrt2ix}$$

$$\displaystyle y = C_1 cos(\sqrt2x) + C_2 sin(\sqrt2x) $$

Boundary Condition y(0)=1 and y(1)=2

$$\displaystyle 1 = C_1 cos(0) + C_2 sin(0)$$

$$ C_{1}= 1$$

$$\displaystyle 2 = cos(\sqrt2) + C_2 sin(\sqrt2)$$

$$ C_2= \frac{2-cos(\sqrt2)}{sin(\sqrt2)}$$

The solution is

$$ y= cos(\sqrt2x) + \frac{2-cos(\sqrt2)}{sin(\sqrt2)}sin(\sqrt2x)$$

[[Media:Graph13.pdf|Graph]]

Contributing Team Members
Joe Gaddone 16:46, 3 October 2009 (UTC) Kumanchik 20:37, 7 October 2009 (UTC) Matthew Walker Egm6321.f09.Team2.sungsik 21:16, 7 October 2009 (UTC)