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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 $\mathrm{\Phi }(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2$ where $k_1,k_2$ are constants.

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Problem Statement: Given a L2_ODE_VC $\sqrt{x}{y}^{″}+2x{y}^{\prime }+3y=0$

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

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

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${\varphi }_p=x$

${\varphi }_x=p+3x^{1/2}$

${\varphi }_y=2x^{3/2}-1$

$\varphi (x,y,p)=h(x,y)+\underset{}{\int }f(x,y,p)dp,f={\varphi }_p$

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

$g(x,y,p)={\varphi }_x+{\varphi }_{y}p=h_x+{\varphi }_x+(h_y+{\varphi }_y)p$

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

$h_y=2x^{3/2}$

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

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

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

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

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

Problem 2

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

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

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

where k₁ and k₂ are const, and $p=y^{\prime }$

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

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

$N(x,y)=x$

so it satisfies the 1^st condition of exactness.

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

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

$N_x=1$

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

We must apply the integrating factor method for a L1_ODE_VC.

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

$y^{\prime }+a_0(x)y=b(x)$ where:

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

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

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

$h(x)=exp\underset{}{\overset{x}{\int }}{a}_0(s)ds$

$h(x)=exp\underset{}{\overset{x}{\int }}2\sqrt{x}-\frac{1}{x}=exp(\frac{4}{3}{x}^{\frac{3}{2}}-ln\left|x\right|)\left(\frac{k_2-k_1}{x}\right)=exp\frac{4}{3}{x}^{\frac{3}{2}}-x$

From p.8-2 Eq. (4)

$y(x)=\frac{1}{h(x)}\underset{}{\overset{x}{\int }}h(s)b(s)ds$

Use the product rule of integration $\underset{}{\int }ab=a\underset{}{\int }b-\underset{}{\int }{a}^{\prime }\underset{}{\int }b$

$y(x)=\frac{1}{h(x)}[h(x)\underset{}{\overset{x}{\int }}b(x)-\underset{}{\overset{x}{\int }}{h}^{\prime }(x)\underset{}{\overset{x}{\int }}b(x)]$

In our example $\underset{}{\overset{x}{\int }}{h}^{\prime }(x)=h(x)$ so,

$y(x)=\frac{1}{h(x)}[h(x)\underset{}{\overset{x}{\int }}b(x)-h(x)\underset{}{\overset{x}{\int }}b(x)]$

$y(x)=0$

Problem 3

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

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$F(x,y,y^{\prime },y^{″})=\frac{d\varphi }{dx}={\varphi }_x(x,y,p)+{\varphi }_y(x,y,p)p+{\varphi }_p(x,y,p)p^{\prime }wherep=y^{\prime }$

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

${\varphi }_x=h_x+{\varphi }_{px}p$

${\varphi }_y=h_y+{\varphi }_{py}p$

$g={\varphi }_x+{\varphi }_{y}p=h_x+{\varphi }_{px}p+(h_y+{\varphi }_{py}p)p$

$h_y={\varphi }_y-{\varphi }_{px}$

Take the integral of $h_y$

$h(x,y)=({\varphi }_y-{\varphi }_{px})y+k_1$

Substitute back into the equation for $\varphi$

$\varphi =({\varphi }_y-{\varphi }_{px})y+k_1+{\varphi }_{p}p$

Rearrange the terms to obtain

$\varphi (x,y,p)={\varphi }_{p}p+({\varphi }_y-{\varphi }_{px})y+k$ where,

$P(x)={\varphi }_p$

$T(x)=({\varphi }_y-{\varphi }_{px})y$

$k=k_1$

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

Problem 4

From (p.13-3), for the case $n=1$ (N1_ODE) $F(x,y,y^{\prime })=0=\frac{d\mathrm{\Phi }}{dx}(x,y)$. Show that $f_0-\frac{d{f}_1}{dx}=0\iff {\mathrm{\Phi }}_{xy}={\mathrm{\Phi }}_{yx}$. Hint: Use $f_1={\mathrm{\Phi }}_y$.
Specifically:
4.1) Find $f_0$ in terms of $\mathrm{\Phi }$
4.2) Find $f_1$ in terms of $\mathrm{\Phi }$($f_1={\mathrm{\Phi }}_y$)
4.3) Show that $f_0-\frac{d{f}_1}{dx}=0\iff {\mathrm{\Phi }}_{xy}={\mathrm{\Phi }}_{yx}$.

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Problem Statement: Given a N1_ODE, for the case n=1 $F(x,y,y^{\prime })=0\iff \frac{d\varphi }{dx}(x,y)$

Show that $f_0-\frac{d{f}_1}{d{f}_x}=0\iff {\varphi }_{xy}={\varphi }_{yx},$ Hint:$f_1={\varphi }_y$

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

$F={\varphi }_x+{\varphi }_y{y}^{\prime }$

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

4.1

Find $f_0$ in terms of $\varphi$.

$f_0=\frac{\partial F}{\partial y}=\frac{\partial ({\varphi }_x+{\varphi }_y{y}^{\prime })}{\partial y}={\varphi }_{xy}$

$f_0={\varphi }_{xy}$

4.2

Find $f_1$ in terms of ${\varphi }_y$

$f_1=\frac{\partial F}{\partial y^{\prime }}=\frac{\partial ({\varphi }_x+{\varphi }_y{y}^{\prime })}{\partial y^{\prime }}={\varphi }_y$

$f_1={\varphi }_y$

4.3

Show that $f_0-\frac{d{f}_1}{d{f}_x}=0\iff {\varphi }_{xy}={\varphi }_{yx},$

${\varphi }_{xy}-\frac{d{\varphi }_y}{d{f}_x}={\varphi }_{xy}-{\varphi }_{yx}=0$

${\varphi }_{xy}={\varphi }_{yx}$

Problem 5

From (p.13-3), for the case $n=2$ (N2_ODE) show:
5.1) Show $f_1=\frac{d{f}_2}{dx}+{\mathrm{\Phi }}_y$
5.2) Show $\frac{d}{dx}({\mathrm{\Phi }}_y)=f_0$
5.3) $f_0-\frac{d{f}_1}{dx}+\frac{d^2{f}_2}{d{x}^2}=0$
5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem 6

From (p.14-2), for the Legendre differential equation $F=(1-x^2)y^{″}-2x{y}^{\prime }+n(n+1)y=0$,
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 $\mathrm{\forall }u,v$ functions of $x$, $L(u+v)=L(u)+L(v)$. and
7.2 $\mathrm{\forall }\lambda \in ℝ,L(\lambda u)=\lambda L(u)\mathrm{\forall }$ functions of $x$.
are equivalent to equation 3 on p.3-3.

Problem 8

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

Problem 9

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

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$y_{xxx}=(d/dx)\left[\frac{d/dx}{(d/dx)y}\right]=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y$

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

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

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

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

$y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))⟨(dt/dx)(d/dt)y⟩$

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

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

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

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

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

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

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

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

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, $x^2{y}^{″}-2x{y}^{\prime }+2y=0$ using the method of trial solution $y=e^{rx}$ directly for the boundary conditions $\{\begin{array}{cc}y(1)=& 3\\ y(2)=& 4\end{array}$
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 $Z(x)=\frac{c}{u_1^2}\exp (-{\int }^x{a}_1(s)ds)$ 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:

$y=a{x}^b$

$y=e^{rx}$

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)