User:EGM6321.F10.TEAM1.HW4

Find
1) Show Eq1.1 is exact. 2) Find $$ \phi\ $$ 3) Solve for $$ y(x)\ $$

Solution
1a) Solve the first condition of exactness, which is given by: For:

and the fist condition is satisfied.

1b) Solve the second condition of exactness, which is given by: and: Substituting 1.3 and 1.4 into 1.06 and 1.7, we get: and

and

and the second condition is satisfied.

Therefore equation 1.1 is exact.

2)

3)

When $$\begin{align} cosx\neq0 \end{align} $$ , we can divide both sides of (1.16) with $$\begin{align} cosx \end{align} $$

Author EGM6321,F10.TEAM1.WILKS EGM6321,F10.TEAM1.Taeho Kim

Find
1) Verify exactness of eq2.1 using the following two methods 1a) 1a)

1b)

2) If equation is not exact, make it exact by using the integrating factor method with:

Solution
1a) First we will use our original method, which we should note is just a particular case of a more general method to determine whether or not a even equation is exact.

1st criteria for exactness is the equation must be able to be put into the following form: $$F(x,y,{y}',{y})=f(x,y,p){y}+g(x,y,p)$$ Where we have: $$p={y}'$$ If we compare the above equation to the equation given in the problem statement we find: $$f={{x}^{2}}$$ $$g=xp+({{x}^{2}}-{{\nu }^{2}})y$$ Therefore the first condition for exactness is satisfied. The second criteria for exactness is the following: $${{g}_{pp}}={{f}_{xp}}+p{{f}_{yp}}+{{p}^{2}}{{f}_{y}}$$ Which we can easily see is: $$0=0+0+0$$ $${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}={{g}_{xp}}+p{{g}_{yp}}-{{g}_{y}}$$ Which after evaluation is: $$2+0+0=1+0-{{x}^{2}}+{{\nu }^{2}}$$ $${{x}^{2}}={{\nu }^{2}}-1$$ We see that this equation has failed to satisfy the second condition for exactness and therefore will be attempted to be made exact through the use of an integrating factor in part 2 of this problem. 1b) We know the given equation can be written in the following form: $$F(x,y,p,q)=f(x,y,p)q+g(x,y,p)={{\phi }_{p}}q+{{\phi }_{x}}+p{{\phi }_{y}}$$ The more general criteria for exactness is the following: $${{f}_{0}}-\frac{d{{f}_{1}}}{dx}+\frac{d{{x}^{2}}}=0$$ Where $${{f}_{0}}=\frac{\partial F(x,y,p,q)}{\partial y}={{f}_{y}}q+{{g}_{y}}$$ $${{f}_{1}}=\frac{\partial F(x,y,p,q)}{\partial p}={{f}_{p}}q+{{g}_{p}}$$ $${{f}_{2}}=\frac{\partial F(x,y,p,q)}{\partial q}=f$$ $${{f}_{0}}={{f}_{y}}q+{{g}_{y}}$$ $$\frac{d{{f}_{1}}}{dx}={{f}_{px}}q+{{f}_{py}}pq+{{f}_{pp}}{{q}^{2}}+{{f}_{p}}{q}'+{{g}_{px}}+p{{g}_{py}}+q{{g}_{pp}}$$ $$\frac{d{{f}_{2}}}{dx}={{f}_{x}}+{{f}_{y}}p+{{f}_{p}}q$$ $$\frac{d{{x}^{2}}}={{f}_{xx}}+{{f}_{yx}}p+{{f}_{xy}}p+{{f}_{yy}}{{p}^{2}}+{{f}_{yp}}pq+{{f}_{xp}}q+{{f}_{y}}q+{{f}_{px}}q+{{f}_{py}}pq+{{f}_{pp}}{{q}^{2}}+{{f}_{p}}{q}'$$ Now evaluating the various terms and plugging into the above equation gives us: $${{f}_{0}}=0+{{x}^{2}}-{{\nu }^{2}}$$ $$\frac{d{{f}_{1}}}{dx}=0+0+0+0+1+0+0$$ $$\frac{d{{x}^{2}}}=2+0+0+0+0+0+0+0+0+0+0$$ Plug above results into the following expression to test for exactness: $${{f}_{0}}-\frac{d{{f}_{1}}}{dx}+\frac{d{{x}^{2}}}=0$$ And we see it simplifies to the same expression as before: $${{x}^{2}}={{\nu }^{2}}-1$$ Therefore, the given equation fails the criteria for exactness and is not exact. 2) We assume an exponential form of the integrating factor and multiply the given equation in the problem statement. $${{x}^{m}}{{y}^{n}}[{{x}^{2}}{y}''+x{y}'+({{x}^{2}}-{{v}^{2}})y]=0$$ Now we organize the equation into an exact form. $$f={{x}^{m+2}}{{y}^{n}}$$ $$g={{x}^{m+1}}{{y}^{n}}p+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{v}^{2}}$$ Stating the first of the two exactness conditions for second order non-linear ODE's. $${{g}_{pp}}={{f}_{xp}}+{{f}_{yp}}p+2{{f}_{y}}$$ We find when plugging in the terms. $$0=0+0+2n{{y}^{n-1}}$$ $$\therefore n=0$$ $$\therefore f={{x}^{m+2}}$$ $$\therefore g={{x}^{m+1}}p+{{x}^{m+2}}y-{{x}^{m}}y{{v}^{2}}$$ Stating the second of the exactness conditions for second order non-linear ODE's. $${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}={{g}_{px}}+p{{g}_{py}}-{{g}_{y}}$$ Now plug in for the terms in the equation above and simplify. $$(m+2)(m+1){{x}^{m}}+0+0=(m+1){{x}^{m}}+0+{{x}^{m+2}}-{{x}^{m}}{{v}^{2}}$$ $$(m+2)(m+1){{x}^{m}}=(m+1){{x}^{m}}+{{x}^{m+2}}-{{x}^{m}}{{v}^{2}}$$ $$({{m}^{2}}+3m+2){{x}^{m}}-(m+1){{x}^{m}}+{{x}^{m}}{{v}^{2}}={{x}^{m+2}}$$ $${{m}^{2}}{{x}^{m}}+2m{{x}^{m}}+{{x}^{m}}+{{x}^{m}}{{v}^{2}}={{x}^{m+2}}$$ $$({{m}^{2}}+2m+1){{x}^{m}}+{{x}^{m}}{{v}^{2}}={{x}^{m+2}}$$ $$[{{(m+1)}^{2}}+{{v}^{2}}]{{x}^{m}}={{x}^{m+2}}$$ Solve for m so that both sides of the equation are of equal powers. $$\therefore m=v-1$$ Author 70.185.108.64 16:00, 20 October 2010 (UTC)EGM6321,F10.TEAM1.Lang

Find
$$ X(x)\ $$ in terms of $$ sin\ $$, $$ cos\ $$ , $$ sinh\ $$ and $$ cosh\ $$ ,

Solution
This means that a linear combination of each of these solutions will be the final solution. Then substitutions in terms of cos, sin, cosh, and sinh are made as: which will transform equation 3.4 into redefine the constants for neatness. i is just a constant and can be absorbed as such.

Author Egm6321.f10.team1.allison 18:31, 20 October 2010 (UTC)

Signatures
Solved problem 1 -- 192.91.147.35 21:22, 12 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 2 -- 70.185.108.64 16:01, 20 October 2010 (UTC)EGM6321,F10.TEAM1.LANG

Solved problem 3 -- Egm6321.f10.team1.allison 18:32, 20 October 2010 (UTC)