User:Egm6321.f09.team7/HW3

Homework Three: Problem 1
Problem 1 Find $$(m,n)$$ such that eqn. 1 on 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.

Given: Equation to be solved: $$x^\frac12y'' + 2xy' + 3y = 0$$

Problem solution: First exactness condition is met with the form of the equation in that y'&apos; is unmodified Second exactness condition:

$$Fxx+2P*Fxy+P^2*Fyy=Gxp+P*Gyp-Gy$$

$$Fxp+P*Fyp+2Fy=Gpp$$

First step is to multiply through by the h term in the form $$h=x^m*y^n$$

giving us : $$ (x^{\frac12+m}y^n)y''+2x^{m+1}y^np+3x^my^{n+1}$$ then we can identify F and G for the equation in the form $$F*y''+G$$: F is given as: $$x^{\frac12+m}y^n$$ G is given as: $$2x^{m+1}y^np+3x^my^{n+1}$$

The next step will be to identify the necessary partial derivatives to fulfill the second exactness condition:

$$Fx=(m+0.5)x^{(m-0.5)}y^n$$ $$Fxp=0$$ $$Fxx=(m^2-0.25)x^{m-1.5}y^n$$ $$Fxy=(m+0.5)nx^{m-0.5}y^{n-1}$$ $$Fy=nx^{m+0.5}y^{n-1}$$ $$Fyy=n(n-1)x^{m+0.5}y^{n-2}$$ $$Fyp=0$$ $$Gp=2x^{m+1}y^n$$ $$Gx=2p(m+1)x^my^n+3mx^{m-1}y^{n+1}$$ $$Gy=2pnx^{m+1}y^{n-1}+3(n+1)x^my^n$$ $$Gpp=0$$ $$Gxp=2(m+1)x^my^n$$ $$Gyp=2nx^{m+1}y^{n-1}$$

Finally we see the above partials should be inserted into the second exactness condition:

condition 2 part 2: $$0+p(0)+2nx^{m+0.5}y^{n-1}=0\rightarrow n=0$$

this condition can be used to simplify the arguments for the first part of the second exactness condition

condition 2 part 1: $$(m^2-0.25)x^{m-1.5}=2(m+1)x^m-3x^m\rightarrow m=0.5$$ note:m=0.5 found through trial and error and recognizing that this forces the LHS of the above equation to zero meaning that $$2(m+1)x^m=3x^m$$which is true for 0.5  You do not have to use trial and error. Start with the shorter of the two equations for the second exactness condition and you will find that $$n=0$$. Then use this result in the other equation and it will naturally fall out that $$m=1/2$$. --Egm6321.f09.TA 04:35, 15 October 2009 (UTC)

thus using the found m and n to modify the original equation and make it exact we can find the exact form of this equation is: $$(x^my^n)*[x^\frac12y'' + 2xy' + 3y] = 0$$ Meeting both conditions of exactness

 Express the final form with the values you found for m,n: i.e., $$xy''+2x\sqrt{x}y'+3\sqrt{x}y=0$$.--Egm6321.f09.TA 04:35, 15 October 2009 (UTC)

Homework Three: Problem 2
Problem Statement and Given:  Solve eqn. 2 on for $$y(x)$$. $$ xy' + (2x^{\frac{3}{2}}-1)y = k_2 - k_1 $$

Solution:

Divide by x to put the ODE into a general form: $$ y' + p(x)y = q(x) $$ Where: $$ p(x)=\frac{(2x^{\frac{3}{2}}-1)}{x}=2x^{\frac{1}{2}}-x^{-1}$$ $$ q(x)=\frac{k_2 - k_1}{x}$$ The ODE can be solved by finding an integrating factor $$u(x)$$ such that: $$\frac{d(u(x)y)}{dx}=u(x)q(x)$$ Dividing by $$u(x)y$$ gives: $$\frac{1}{y}\frac{dy}{dx}+\frac{1}{u(x)}\frac{du(x)}{dx}=\frac{q(x)}{y}$$ $$p(x)=\frac{1}{u(x)}\frac{du(x)}{dx}$$ so: $$\frac{1}{y}\frac{dy}{dx}+p(x)=\frac{q(x)}{y}$$ Integrate both sides to obtain: $$\int p(x)dx=\int \frac{du(x)}{u(x)}=ln(u(x))+c$$ This gives: $$u(x)=exp(\int p(x)dx)$$ Integrating $$\frac{d(u(x)y)}{dx}=u(x)q(x)$$ gives: $$u(x)y=\int u(x)q(x)dx+c $$ Solving for $$y$$ and substituting in $$u(x)=exp(\int p(x)dx)$$ gives: $$y=\frac{\int exp(\int dxp(x))q(x)+c}{exp(\int p(x)dx)}$$ $$\int p(x)dx=\int 2x^{\frac{1}{2}}-x^{-1}=\frac{4}{3}x^{\frac{3}{2}}-ln(x)+C$$ The final solution becomes: $$y=\frac{\int C_2x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})\frac{k_2 - k_1}{x}+C_1}{ C_2x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})}$$



Good work - nice solution. Egm6321.f09.TA 01:33, 28 October 2009 (UTC)

Homework Three: Problem 3
Problem Statement and Given:  From, find the mathematical structure of $$\Phi$$ that yields the above class of ODE.

Solution:

'''See attached file contains the solution! '''



You are using the result for the derivation. This is circular. Egm6321.f09.TA 02:13, 28 October 2009 (UTC)

Homework Three: Problem 4
Problem 4 From, 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 solution: For the universal solution given in class the following statements apply $$F=\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

since this problem is for a first order equation n=1 and therefore:

$$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}\rightarrow F= \Phi_{x}+\Phi_{y}y'$$

So the first function to find is $$f_0$$

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy}$$  This should be a partial differential with respect to $$y$$. --Egm6321.f09.TA 14:41, 15 October 2009 (UTC) $$f_0=\Phi_{xy}+\Phi_{yy}P$$

$$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}$$

Since Phi is only a function of x and y

$$f_1=\Phi_{y}$$

now to take the full derivative of f_1 W.R.T. x

$$\frac{df_1}{dx}\rightarrow \frac{d\Phi_{y}}{dx}$$ Phi being a function of x and y

$$=\frac{d\Phi_{y}}{dx}\frac{dx}{dx}+\frac{d\Phi_{y}}{dy}\frac{y}{dx}\rightarrow \Phi_{yx}+\Phi_{yy}P$$  This first term should be $$\frac{\partial \Phi_y}{\partial x}\frac{dx}{dx}$$. The chain rule uses partials with respect to the first variable of differentiation, and then total derivatives for the implicit dependencies. Same thing with $$\frac{\partial\phi_y}{\partial x}\frac{dy}{dx}$$ --Egm6321.f09.TA 14:41, 15 October 2009 (UTC) $$ f_0-\frac{df_1}{dx}=\Phi_{xy}+\Phi_{yy}P-\Phi_{yx}+\Phi_{yy}P=0$$

Finally cancelling terms leaves us with:

$$\Phi_{yx}-\Phi_{xy}=0$$ or $$\Phi_{yx}=\Phi_{xy}$$

Homework Three: Problem 5
Problem 5 From (p.13-3), for the case n = 2 (N2_ODE) show:

5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$

5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$

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

5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem solution: For the universal solution given in class the following statements apply $$F=\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

since this problem is for a first order equation n=1 and therefore:

$$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+\Phi_{y^{(1)}}y^{(2)}\rightarrow F= \Phi_{x}+\Phi_{y}y'+\Phi_{y'}y''$$

So the first step is to find the values for the functions $$f_0,f_1,f_2$$

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y'')}{dy}$$

$$f_0=\Phi_{xy}+\Phi_{yy}P+\Phi_{py}y''$$

$$f_1=\frac{dF}{dp}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y'')}{dp}$$

$$f_1=\Phi_{xp}+\Phi_{yp}P+\Phi_{y}+\Phi_{pp}y''$$

$$f_2=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y)}{dy''}$$

$$f_2=\Phi_{p}$$

Starting with the first problem statement:

5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$

$$f_1 and f_2$$ are shown above so now we will solve $$\frac{df_2}{dx}+\Phi_y$$

the full derivative is: $$=\frac{d\Phi_{p}}{dx}\frac{dx}{dx}+\frac{d\Phi_{p}}{dy}\frac{y}{dx}+\frac{d\Phi_{p}}{dp}\frac{dp}{dx}\rightarrow \Phi_{px}+\Phi_{py}P+\Phi_{pp}y''$$

adding this to the problem statement and description of the first f function gives the answer:

$$\Phi_{px}+\Phi_{py}P+\Phi_{pp}y+\Phi_{y}=\Phi_{px}+\Phi_{py}P+\Phi_{pp}y+\Phi_{y}$$

5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$

$$f_o$$ es show above so now we will solve $$\frac{d\phi_y}{dx}$$

the full derivative is: $$ = \frac{d\Phi_y}{dx}\frac{dx}{dx}+\frac{d\Phi_{y}}{dy}\frac{y}{dx}+\frac{d\Phi_{y}}{dp}\frac{dp}{dx}$$

$$= \Phi_{yx}+\Phi_{yy}P+\Phi_{yp}y''$$

This is exactly the same as the statement for $$f_0$$ and satisfies the problem statement

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

As was shown in the first problem $$f_1=\frac{df_2}{dx}+\Phi_y$$ so when subtracted out these leave only the $$\Phi_y$$ term which reduces the problem statement to:

$$f_0-\frac{\Phi_y}{dx}=0$$

since $$f_0$$ is known from above the only task is to find $$\frac{\Phi_y}{dx}=0$$:

$$\frac{\Phi_y}{dx}+\frac{\Phi_y}{dy}\frac{dy}{dx}+\frac{\Phi_y}{dp}\frac{dp}{dx} \rightarrow \Phi_{yx}+\Phi_{yy}P+\Phi_{yp}y''$$

This is the same function as $$f_0$$ and since they are subtracted = 0 (answer could have also been inferred from a combination of the first and second parts which are also related to the two exactness condition mentioned in 5.4)

 Your results for this problem are all correct - nice write up. You want to restate the expression that you have proven (and possibly box it) so that it is evident you have arrived at the expected result. --Egm6321.f09.TA 02:00, 16 October 2009 (UTC)

Homework Three: Problem 6
Problem Statement and Given:  From, 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.), Equations 4&5. 6.1b.) , Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Solution 6.1a:

The general form is: $$F=f(x,y,p)y''+g(x,y,p)$$ Where: $$p=y' $$ $$f(x,y,p)=1-x^2$$ $$g(x,y,p)=-2xp+n(n-1)y$$

, Equations 4&5. i)$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ ii)$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

The partial derivatives are: $$f_{xx}=-2$$ $$f_{xy}=0$$ $$f_{yy}=0$$ $$f_{xp}=0$$ $$f_{yp}=0$$ $$f_{y}=0$$ $$g_{xp}=-2$$ $$g_{yp}=0$$ $$g_{y}=n(n-1)$$ $$g_{pp}=0$$

Substituting in the partial derivatives: i) $$-2+2p0+p^20=-2+p0-n(n-1)$$ i) $$-2\neq-2-n(n-1)$$ and ii) $$0+p0+20=0$$ ii) $$0=0$$ Since i) does not equal, this ODE is not exact.

Solution 6.1b:

, Equation 5. $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

Where: $$f_0=\frac{dF}{dy}=n(n+1)$$ $$f_1=\frac{dF}{dy'}=-2x$$ $$\frac{f_1}{dx}=-2$$ $$f_2=\frac{dF}{dy''}=1-x^2$$ $$\frac{d^2f_2}{dx^2}=-2$$

Substituting into $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ gives: $$n(n+1)+2-2\neq0$$ so the ODE is not exact.

Solution 6.1c:

With an integrating factor $$h(x,y)=x^my^n$$, the ODE becomes: $$F=(x^my^n)(1-x^2)y''-(x^my^n)2xy'+(x^my^n)n(n+1)y=0$$ Where: $$p=y' $$ $$f(x,y,p)=x^my^n-x^{m+2}y^n$$ $$g(x,y,p)=-2x^{m+1}y^np+n(n-1)x^my^{n+1}$$

, Equations 4&5. i)$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ ii)$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

The partial derivatives are: $$f_{xx}=m(m-1)x^{m-2}y^n-(m+2)(m+1)x^my^n$$ $$f_{xy}=mnx^{m-1}y^{n-1}-n(m+2)x^{m+1}y^{n-1}$$ $$f_{yy}=n(n-1)x^my^{n-2}-n(n-1)x^{m+2}y^{n-2}$$ $$f_{xp}=0$$ $$f_{yp}=0$$ $$f_{y}=nx^my^{n-1}-nx^{m+2}y^{n-1}$$ $$g_{xp}=-2(m+1)x^my^n$$ $$g_{yp}=-2nx^{m+1}y^{n-1}$$ $$g_{y}=-2nx^{m+1}y^{n-1}p+a(a-1)x^m(n+1)y^n$$ $$g_{pp}=0$$

Substituting in the partial derivatives: i)$$m(m-1)x^{m-2}y^n-(m+2)(m+1)x^my^n+2pmnx^{m-1}y^{n-1}-n(m+2)x^{m+1}y^{n-1}+p^2n(n-1)x^my^{n-2}-n(n-1)x^{m+2}y^{n-2}=-2(m+1)x^my^n+p-2nx^{m+1}y^{n-1}+2nx^{m+1}y^{n-1}p+a(a+1)x^m(n+1)y^n$$ ii)$$0+p0+2nx^my^{n-1}-nx^{m+2}y^{n-1}=0$$

ii)$$2nx^my^{n-1}-nx^{m+2}y^{n-1}=0$$

From equation ii, n=0. This gives i as: i)$$m(m-1)x^{m-2}-(m+2)(m+1)x^m=-2(m+1)x^m+a(a+1)x^m$$ i)$$m(m-1)x^{-2}-(m+2)(m+1)=-2(m+1)+a(a+1)$$

Homework Three: Problem 7
Problem Statement and Given:  From, Show that equations 1 and 2, namely 7.1 $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$. are equivalent to equation 3 on p.3-3.

Solution:

Alternate definition of Linearity:  L(.) is linear if and only if: 1) $$L(u+v)  =  L(u) + L(v) $$ 2)  $$L(\lambda u) = \lambda L(u)$$

Let, $$\bar u = u \lambda_1 $$ $$\bar v = v \lambda_2 $$

Use property 1 for the above definitions: $$L(\bar u + \bar v) =  L(\bar u) + L(\bar v) $$

Rewrite in terms of u: $$L(u \lambda_1 + v \lambda_2) =  L(u \lambda_1) + L(v \lambda_2) $$

Use the second property twice to get: $$L(u \lambda_1) =  \lambda_1 L(u)$$ $$L(v \lambda_2) =  \lambda_2 L(v)$$

Substitute the above expressions: $$L(u \lambda_1 + v \lambda_2) =   \lambda_1 L(u) + \lambda_2 L(v) $$

The result is the former definition of linearity in 3-3 but different nomenclature.

 Good explanation. Why is your font so small? --Egm6321.f09.TA 04:08, 16 October 2009 (UTC)

Homework Three: Problem 8
Problem Statement and Given:  From, plot the shape function $$N_{j+1}^{2}(x)$$.

Solution:



correct. Egm6321.f09.TA 03:10, 28 October 2009 (UTC)

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

 Part 1 $$y_{xxx} = \frac{d}{dx}(y_{xx}) = \frac{dt}{dx} \frac{d}{dt}(y_{xx})$$

$$y_{xx} = \frac{d}{dx}(\frac{d}{dx}(y))= \frac{dt}{dx} \frac{d}{dt}(\frac{dt}{dx} \frac{d}{dt}(y))$$

$$ y_{xxx} = e^{-t} \frac{d}{dt} (e^{-2t}(\ddot y - \dot y))$$

Define the following because Wiki Latex won’t allow a triple dot accent: $$\frac{d}{dt}(\ddot y ) = \check{Y}$$

$$ y_{xxx} = e^{-t} (e^{-2t}(\check{Y}  - \ddot y  - 2 e^{-2t}(\ddot y  - \dot y)))$$

$$ y_{xxx} = e^{-3t} (\check{Y} - 3 \ddot y +2  \dot y))$$

The above equation matches!

 Part 2 $$ y_{xxxx} =\frac{d}{dx}(y_{xxx})= e^{-t} \frac{d}{dt}( e^{-3t} (\check{Y} - 3 \ddot y  +2  \dot y))$$

Define the following because Wiki Latex won’t allow a quadruple dot accent: $$ \frac{d}{dt}(\check{Y}) = \dot \check{Y}$$

$$ y_{xxxx} = e^{-t} [ -3 e^{-3t} (\check{Y} - \ddot y - 3 \ddot y  -2  \dot y) + e^{-3t} (\dot \check{Y}- 3 \check{Y}  +2  \ddot y)]$$

$$ y_{xxxx} = e^{-4t} [\dot \check{Y} - 6 \check{Y} + 11 \ddot y - 6 \dot y]$$

The above equation matches!



Very good. Boxing your answers would make the results more visible. Nice concise approach. Egm6321.f09.TA 03:30, 28 October 2009 (UTC)

Homework Three: Problem 10
Problem Statment and Given:  From (| p.16-4) Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ or $$ y=x^{r}$$ 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.

Solution:  First Try: $$ y=e^{rx}$$

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

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

Plug into the ODE and you will notice that $$ e^{rx}$$ can factored out:

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

The quadratic equation can be used to get $$r_1(x)$$ and $$r_2(x)$$ ; however, this contradicts the implicit assumption made in calculating $$y'$$ and $$y''$$. We assumed r was constant therefore obtaining r(x) is in direct violation of our assumption.

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

The ODE's solution is not of the form $$y = e^{rx}$$ !

Instead, the solution takes the form: $$y = x^{r}$$  $$y'=rx^{r-1}$$

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

Plug into the ODE and you will notice that $$ x^{r}$$ can factored out: $$x^r(r(r-1)-2r+2)=0$$

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

$$ (r-2)(r-1)=0$$

$$r_1=2$$ and $$r_2=1$$


 * {| style="width:100%" border="0" align="left"

y(x) = c_1 x^2 + c_2 x$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

The above solution to the ODE identical to that in equation 10 on p.16-3!

Use the BVC to get the constants:

$$3 = c_1 + c_2$$

$$4 = 4 c_1 + 2 c_2$$

$$ c_1 = -1$$ and $$c_2 = 4$$

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

The solution is plotted:



very good. Egm6321.f09.TA 03:56, 28 October 2009 (UTC)

Homework Three: Problem 11
Problem Statement and Given: 

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.

Problem solution:

Starting from equation 1 from p.17-3.

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

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

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

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

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

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

Integrating this will give:

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

$$\mu(x)Z'+\mu'(x)Z=0$$

$$\int_{}^{}\mu(x)*Z(x)=\int_{}^{}0$$

$$\mu(x)Z(x)=K$$

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

$$Z(x)= C exp \left[-\int_{}^{x}a_{1}(s)ds-ln\left|u_{1} \right|^{2} \right]$$

$$Z(x)= \frac{C}{u_{1}^{2}} exp \left[-\int_{}^{x}a_{1}(s)ds \right]$$



very good. Egm6321.f09.TA 04:10, 28 October 2009 (UTC)

Homework Three: Problem 12
Problem Statement and Given: 

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

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

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

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

Problem solution:

Part a:

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

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

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

$$ (U (x)\pm u_1  (x))+a_1(U' (x)\pm u_1 ' (x))+a_0(U(x)\pm u_1 (x))=0$$

$$ (U (x)+a_1 U'(x) + a_0 U(x))\pm ((u_1(x)+a_1 u_1'(x)+a_0 u_1(x))=0$$

The reduction of order method can not be implemented since U is not missing.

Part b:

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

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

$$y (x)= U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U''(x)}{u_{1}(x)}$$

$$\left[U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U(x)}{u_{1}(x)} \right]+a_{1}\left[-\frac{U(x)}{u_{1}^{2}(x)}u_{1}'(x)+\frac{U'(x)}{u_{1}(x)}\right]+a_{0}\left[\frac{U(x)}{u_{1}(x)} \right]=0$$

The reduction of order method can not be implemented since U is not missing.

Part c:

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

$$y' (x)= -\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}$$

$$y (x)= u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1''(x)}{U(x)}$$

$$\left[u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1(x)}{U(x)} \right]+a_{1}\left[-\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}\right]+a_{0}\left[\frac{ u_1 (x)}{ U (x) } \right]=0$$

The reduction of order method can not be implemented since U is not missing.



very good. Egm6321.f09.TA 04:18, 28 October 2009 (UTC)

Homework Three: Problem 13
Problem Statement and Given: 

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.

Problem solution:

Part 1:

Starting with the equation:

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

Using the trial solution given:

$$ y=ax^{b}$$

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

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

The equation becomes:

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

Now divide by: $$ ax^{b}$$

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

It is obvious that at x=0 the equation would be undefined unless b^2-b=0 which would make b=1 and at x=1 it is obvious too that b=1.

$$ y(x)=ax$$

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y=ax^{b}$$ cannot be a solution to this problem.

Part 2:

Starting with the equation:

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

Using the trial solution given:

$$ y= exp(rx)$$

$$ y'=r exp(rx)$$

$$ y''=r^{2}exp(rx) $$

The equation becomes:

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

Now divide by: $$ exp(rx)$$

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

It is obvious that at x=0 the value of r=(-2)^(1/2) and at x=1 the value of r= 1.

$$ y(x)= exp(x)+ exp(xi\sqrt[]{2})$$

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y= exp(rx)$$ cannot be a solution to this problem since the result should only be real numbers. 

Where is your plot? Egm6321.f09.TA 06:00, 28 October 2009 (UTC)

Contributing Authors
Egm6321.f09.team7.mm 03:16, 6 October 2009 (UTC) Egm6321.f09.team7.benedict 01:10, 7 October 2009 (UTC) Egm6321.f09.team7.hua 04:05, 7 October 2009 (UTC) User:Emg6321.f09.blanco 11:17, 7 October 2009 (UTC)