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

Problem 1

Find $(m,n)$ such that eqn. 1 on (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.

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

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

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

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$\phi _{p}=x$

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

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

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

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

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

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

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

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

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

$n=ln(2x^{3/2})$

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

$m=ln(3x^{1/2}-x)$

Problem 2

Solve eqn. 2 on (p.13-1) for $y(x)$.

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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 $p=y'$

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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 1^st condition of exactness.

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

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

$N_{x}=1$

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

We must apply the integrating factor method for a L1_ODE_VC.

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

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

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

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,

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

$y(x)=0$

Problem 3

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

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

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

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

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

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

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

Take the integral of $h_{y}$

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

Substitute back into the equation for $\phi$

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

Rearrange the terms to obtain

$\phi (x,y,p)=\phi _{p}p+(\phi _{y}-\phi _{px})y+k$ 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 (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 $f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \Phi _{xy}=\Phi _{yx}$.

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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:$f_{1}=\phi _{y}$

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

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

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

4.1

Find $f_{0}$ in terms of $\phi$.

$f_{0}={\partial F \over \partial y}={\partial (\phi _{x}+\phi _{y}y') \over \partial y}=\phi _{xy}$

$f_{0}=\phi _{xy}$

4.2

Find $f_{1}$ in terms of $\phi _{y}$

$f_{1}={\partial F \over \partial y'}={\partial (\phi _{x}+\phi _{y}y') \over \partial y'}=\phi _{y}$

$f_{1}=\phi _{y}$

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 (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) $f_{0}-{\frac {df_{1}}{dx}}+{\frac {d^{2}f_{2}}{dx^{2}}}=0$
5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem 6

From (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.) (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 $h(x,y)=x^{m}y^{n}$.

Problem 7

From (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 (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
$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_{t}\right)$
$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}\right)$

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

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, $x^{2}y''-2xy'+2y=0$ 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.

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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.

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Problem 12

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

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

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

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

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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.

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Contributing Team Members

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

Matthew Walker

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