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


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

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

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

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

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


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

$$y^n=h_y\Rightarrow lny^n=ln(2x^{3/2})$$  $$n=ln(2x^{3/2})$$ $$x^m=h_x\Rightarrow lnx^m=ln(3x^{1/2}-x)$$  $$m=ln(3x^{1/2}-x)$$

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.[[media:Egm6321.f09.p8-1.png|8-1]]

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

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

From p.[[media:Egm6321.f09.p8-2.png|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,

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



$$y(x)=0$$

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,


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



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

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:$$f_1=\phi_y$$

$$F={d\phi\over dx}(x,y^{(0)},....y^{(n-1)}=\phi_x+\phi_y6{(0)}y^{(1)}$$

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

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

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

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

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

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

Problem 5
From ([[media:Egm6321.f09.p13-3.png|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 6
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 $$h(x,y)=x^my^n$$.

Problem 7
From ([[media:Egm6321.f09.p14-3.png|p.14-3]]), 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.

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

[[Media:Graph1.pdf]]

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

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

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, 

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

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

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

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.



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

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

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.

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

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

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

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

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.

Contributing Team Members
Joe Gaddone 16:46, 3 October 2009 (UTC)

Matthew Walker

Egm6321.f09.Team2.sungsik 21:22, 4 November 2009 (UTC)