University of Florida/Egm6321/f09.Team2/HW3

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

Find ${\displaystyle (m,n)}$ such that eqn. 1 on (p.13-1) is exact. A first integral is ${\displaystyle \Phi (x,y,p)=xp+(2x^{\frac {3}{2}}-1)y+k_{1}=k_{2}}$ where ${\displaystyle k_{1},k_{2}}$ are constants.

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

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

A first integral is ${\displaystyle \phi (x,y,p)=xp+{\big (}2x^{3/2}{\big )}y+k_{1}=k_{2}}$

${\displaystyle \phi _{p}=x}$

${\displaystyle \phi _{x}=p+3x^{1/2}}$

${\displaystyle \phi _{y}=2x^{3/2}-1}$

${\displaystyle \phi (x,y,p)=h(x,y)+\int _{}f(x,y,p)dp,f=\phi _{p}}$

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

${\displaystyle g(x,y,p)=\phi _{x}+\phi _{y}p=h_{x}+\phi _{x}+(h_{y}+\phi _{y})p}$

${\displaystyle g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_{x}+x+(h_{y}+0)p}$

${\displaystyle h_{y}=2x^{3/2}}$

${\displaystyle h_{x}=3x^{1/2}-x}$

${\displaystyle y^{n}=h_{y}\Rightarrow lny^{n}=ln(2x^{3/2})}$

${\displaystyle x^{m}=h_{x}\Rightarrow lnx^{m}=ln(3x^{1/2}-x)}$

Solve eqn. 2 on (p.13-1) for ${\displaystyle y(x)}$.

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

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

where k₁ and k₂ are const, and ${\displaystyle p=y'}$

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

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

${\displaystyle N(x,y)=x}$

so it satisfies the 1^st condition of exactness.

Check if ${\displaystyle M_{y}=N_{x}}$ for the 2^ndcondition of exactness

${\displaystyle M_{y}=2x^{3 \over 2}-1}$

${\displaystyle N_{x}=1}$

${\displaystyle M_{y}\neq N_{x}}$ so we do not satisfy the 2^nd condition of exactness.

We must apply the integrating factor method for a L1_ODE_VC.

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

${\displaystyle y'+a_{0}(x)y=b(x)}$ where:

${\displaystyle a_{o}(x)={1 \over x}(2x^{3 \over 2}-1)=2{\sqrt {x}}-{1 \over x}}$

${\displaystyle b(x)={k_{2}-k_{1} \over x}}$

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

${\displaystyle h(x)=exp\int _{}^{x}a_{0}(s)ds}$

${\displaystyle h(x)=exp\int _{}^{x}2{\sqrt {x}}-{1 \over x}=exp{\bigg (}{4 \over 3}x^{3 \over 2}-ln\left|x\right|{\bigg )}{\bigg (}{{k_{2}-k_{1}} \over x}{\bigg )}=exp{4 \over 3}x^{3 \over 2}-x}$

From p.8-2 Eq. (4)

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

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

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

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

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

From (p.13-1), find the mathematical structure of ${\displaystyle \Phi }$ that yields the above class of ODE.

${\displaystyle F(x,y,y',y'')={d\phi \over dx}=\phi _{x}(x,y,p)+\phi _{y}(x,y,p)p+\phi _{p}(x,y,p)p'\ where\ p=y'}$

${\displaystyle \phi =h(x,y)+\int _{}\phi _{p}dp=h(x,y)+\phi _{p}p}$

${\displaystyle \phi _{x}=h_{x}+\phi _{px}p}$

${\displaystyle \phi _{y}=h_{y}+\phi _{py}p}$

${\displaystyle g=\phi _{x}+\phi _{y}p=h_{x}+\phi _{px}p+(h_{y}+\phi _{py}p)p}$

${\displaystyle h_{y}=\phi _{y}-\phi _{px}}$

Take the integral of ${\displaystyle h_{y}}$

${\displaystyle h(x,y)=(\phi _{y}-\phi _{px})y+k_{1}}$

Substitute back into the equation for ${\displaystyle \phi }$

${\displaystyle \phi =(\phi _{y}-\phi _{px})y+k_{1}+\phi _{p}p}$

Rearrange the terms to obtain

${\displaystyle P(x)=\phi _{p}}$

${\displaystyle T(x)=(\phi _{y}-\phi _{px})y}$

${\displaystyle k=k_{1}}$

From (p.13-3), for the case ${\displaystyle n=1}$ (N1_ODE) ${\displaystyle F(x,y,y')=0={\frac {d\Phi }{dx}}(x,y)}$. Show that ${\displaystyle f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \Phi _{xy}=\Phi _{yx}}$. Hint: Use ${\displaystyle f_{1}=\Phi _{y}}$.
Specifically:
4.1) Find ${\displaystyle f_{0}}$ in terms of ${\displaystyle \Phi }$
4.2) Find ${\displaystyle f_{1}}$ in terms of ${\displaystyle \Phi }$(${\displaystyle f_{1}=\Phi _{y}}$)
4.3) Show that ${\displaystyle f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \Phi _{xy}=\Phi _{yx}}$.

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

Show that ${\displaystyle f_{0}-{df_{1} \over df_{x}}=0\Leftrightarrow \phi _{xy}=\phi _{yx},}$ Hint:${\displaystyle f_{1}=\phi _{y}}$

${\displaystyle F={d\phi \over dx}(x,y^{(0)},....y^{(n-1)}=\phi _{x}+\phi _{y}6{(0)}y^{(1)}}$

${\displaystyle F=\phi _{x}+\phi _{y}y'}$

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

From (p.13-3), for the case ${\displaystyle n=2}$ (N2_ODE) show:
5.1) Show ${\displaystyle f_{1}={\frac {df_{2}}{dx}}+\Phi _{y}}$
5.2) Show ${\displaystyle {\frac {d}{dx}}(\Phi _{y})=f_{0}}$
5.3) ${\displaystyle f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}=0}$
5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

From (p.14-2), for the Legendre differential equation ${\displaystyle F=(1-x^{2})y''-2xy'+n(n+1)y=0}$,
6.1 Verify exactness of this equation using two methods:
6.1a.) (p.10-3), Equations 4&5.
6.1b.) (p.14-1), Equation 5.
6.2 If it is not exact, see whether it can be made exact using the integrating factor with ${\displaystyle h(x,y)=x^{m}y^{n}}$.

From (p.14-3), Show that equations 1 and 2, namely
7.1 ${\displaystyle \forall u,v}$ functions of ${\displaystyle x}$, ${\displaystyle L(u+v)=L(u)+L(v)}$. and
7.2 ${\displaystyle \forall \lambda \in \mathbb {R} ,L(\lambda u)=\lambda L(u)\forall }$ functions of ${\displaystyle x}$.
are equivalent to equation 3 on p.3-3.

From (p.15-2), plot the shape function ${\displaystyle N_{j+1}^{2}(x)}$.

Media:Graph1.pdf

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

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

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

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

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

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

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

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

Factor out ${\displaystyle e^{-3t}}$ and re-arrange terms in ordre of derivative,

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

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

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

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

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

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

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

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

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

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

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

Problem Statement: From (p.16-4 ) Solve equation 1 on p.16-1, ${\displaystyle x^{2}y''-2xy'+2y=0}$ using the method of trial solution ${\displaystyle y=e^{rx}}$ directly for the boundary conditions ${\displaystyle \left\{{\begin{array}{rl}y(1)=&3\\y(2)=&4\\\end{array}}\right.}$
Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Problem Statement: From (p.17-4 ) obtain equation 2 from p.17-3 ${\displaystyle Z(x)={\frac {c}{u_{1}^{2}}}\exp \left(-\int ^{x}a_{1}(s)ds\right)}$ using the integrator factor method.

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

${\displaystyle y(x)=U(x)\pm u_{1}(x)}$

${\displaystyle y(x)={\frac {U(x)}{u_{1}(x)}}}$

${\displaystyle y(x)={\frac {u_{1}(x)}{U(x)}}}$

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

${\displaystyle y=ax^{b}}$

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

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

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

Matthew Walker

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