User:Egm6321.f10.team2.zou/HW4

= Problem 1: Show Exactness of Second Order Linear Ordinary Differential Equations with Varying Coefficients (L2_ODE_VC) =

From the lecture slide Mtg 22-3

Given
For the following equation:
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$$ \displaystyle
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F\left(x,y,y',y\right)=\left(cos x \right)y+\left(x^{2}-sin x\right)y'+ 2xy=0

$$ (eq 1.1)
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(1) Show that Eq. 1.1 is exact

(2) Find the first integral,Φ, and

(3) Solve for y(x)

(1) Proof of Exactness
To be exact, the equation must satisfy two conditions.

1st Condition of Exactness
First, the equation must be presented in the following format
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$$ \displaystyle
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F\left(x,y,y',y\right)=g\left(x,y,y'\right)+f\left(x,y,y'\right)y

$$ (eq 1.2) By letting y' = p, the first condition of exactness is satisfied as shown in the following equation:
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$$ \displaystyle
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F\left(x,y,p,y\right)=\underbrace{\left(x^{2}-sin x \right)p+ 2xy}_{g\left(x,y,p\right)}+\underbrace{\left(cos x \right)}_{f\left(x,y,p\right)}y

$$

(eq 1.3)
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2nd Condition of Exactness
In order to meet the second condition of exactness, the following equations must be satisfied:


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$$ \displaystyle
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f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}

$$

(eq 1.4)
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$$ \displaystyle
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f_{xp}+pf_{yp}+2f_{y}=g_{pp}

$$

(eq 1.5) Each element of equations 1.4 and 1.5 are calculated to be the following:
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$$ \displaystyle f_{xx}=-cos\left(x\right)$$
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$$\displaystyle f_{xy}=f_{yy}=f_{y}=f_{xp}=f_{yp}=0$$

$$\displaystyle g_{xp}=2x-cos\left(x\right)$$

$$\displaystyle g_{y}=2x$$

$$\displaystyle g_{yp}=g_{pp}=0 $$


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Plugging each relevant element into equation 1.4 yields the following:
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$$ \displaystyle
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-cos\left(x\right)+2p\cancel{f_{xy}}^{0}+p^{2}\cancel{f_{yy}}^{0}=2x-cos\left(x\right)+p\cancel{g_{yp}}^{0}-2x $$

$$-cos\left(x\right)=-cos\left(x\right)$$ (eq 1.6)
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Plugging each relevant element into equation 1.5 yields the following:
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$$ \displaystyle
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\cancel{f_{xp}}^{0}+p\cancel{f_{yp}}^{0}+2\cancel{f_{y}}^{0}=\cancel{g_{pp}}^{0} $$

$$ \displaystyle 0=0$$ (eq 1.7) Equations 1.4 and 1.5 have been satisfied and therefore the equation is exact.
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(2) Solving for the First Integral,Φ
The first integral is defined by the following:


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$$ \displaystyle
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\phi=h\left(x,y\right)+\underbrace{\int \underbrace{f\left(x,y,p\right)}_{cos\left(x\right)}\,dp}_{pcos\left(x\right)}

$$

(eq 1.8)
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From equation 1.8 the following equations can be derived:
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$$ \displaystyle
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\phi_{x}=h_{x}+	\frac{\partial}{\partial x}\left[pcos\left(x\right)\right]=h_{x}-psin\left(x\right)

$$

(eq 1.9)
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$$ \displaystyle
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\phi_{y}=h_{y}+	\frac{\partial}{\partial y}\left[pcos\left(x\right)\right]=h_{y}

$$

(eq 1.10) Using the definition of g to solve for h, yields the following:
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$$ \displaystyle
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g\left(x,y,p\right)=\phi_{y}p+\phi_{x}=h_{y}p+\left[h_{x}-psin\left(x\right)\right]

$$

$$g\left(x,y,p\right)=\left[h_{y}-sin\left(x\right)\right]p+h_{x}$$

(eq 1.11) By equating equation 1.11 with the g(x,y,p) shown in equation 1.3, the partial derivatives of h can be solved as shown:
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$$ \displaystyle
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\left[h_{y}-sin\left(x\right)\right]p+h_{x}=\left[x^{2}-sin\left(x\right)\right]p+2xy $$ (eq 1.12)
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$$\displaystyle h_{y}=x^{2}$$ (eq 1.13)
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$$\displaystyle h_{x}=2xy$$
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(eq 1.14) Next, equations 1.13 and 1.14 can be integrated separately to solve for h(x,y) as follows:
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$$\displaystyle h=\int h_{x}\,dx +w\left(y\right)=x^{2}y+ w\left(y\right)$$
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(eq 1.15)
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$$\displaystyle h=\int h_{y}\,dy +v\left(x\right)=x^{2}y+ v\left(x\right)$$
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(eq 1.16)
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Equating equations 1.15 and 1.16 show that
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$$\displaystyle h=x^{2}y$$ (eq 1.17) Therefore, equation 1.8 can be rewritten as:
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$$ \displaystyle
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\phi=x^{2}y+{{y}'}cos\left(x\right)+k_{1}

$$

(eq 1.18)
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(3) Solving for y(x)
When combined with the definition of the first integral:
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$$ \displaystyle
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\phi=k_{2} $$

(eq 1.19)
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Equation 1.18 can be rewritten as:


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$$ \displaystyle
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\phi=x^{2}y+cos\left(x\right)y'=K $$

Where $$\displaystyle K=k_{2}-k_{1}$$ (eq 1.20) The equation is divided by cos(x) in order to find an integrating factor:
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$$ \displaystyle
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y'+\frac{x^{2}}{cos\left(x\right)}y=\frac{K}{cos\left(x\right)} $$

(eq 1.21)
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Multiply the entire equation by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields:
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$$ \displaystyle
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e^{\int \frac{x^{2}}{cos(x)}\,dx}y'+\frac{x^{2}}{cos\left(x\right)}e^{\int \frac{x^{2}}{cos(x)}\,dx}y=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.22) The left hand side of the equation can be condensed to the following:
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$$ \displaystyle
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(e^{\int \frac{x^{2}}{cos(x)}\,dx}y)'=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.23) Integrating both sides with respect to x gives:
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$$ \displaystyle
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e^{\int \frac{x^{2}}{cos(x)}\,dx}y=K\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.24) Dividing both sides by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields the solution for y:
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$$ \displaystyle
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y=Ke^{-\int \frac{x^{2}}{cos(x)}\,dx}\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.25)
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= Problem 2: Exactness of Bessel Equation =

From the lecture slide Mtg 24-1

Given
Bessel equation has the form as follows :


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$$ \displaystyle {{x}^{2}}{y}''+x{y}'+({{x}^{2}}-{{\nu }^{2}})y=0$$ (Eq 2.1)
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where $$\displaystyle \nu $$ is some positive real number.

Find

 * i)Verify the exactness of Bessel Equation Eq 2.1 using method of Mtg15-3


 * ii)Veryfy the exactness of Bessel Equation Eq 2.1 using method of Mtg22-4


 * iii)If Bessel equation Eq 2.1 can't be made exact, try integrating factor method with the factor of $$ \displaystyle {{x}^{m}}{{y}^{n}}$$ where $$ \displaystyle m$$ and $$ \displaystyle n$$ are all constants.

i) verifying exactness by method given by Mtg15-3
By Eq 2.1 we define,


 * $$\begin{align}

& f\equiv {{x}^{2}} \\ & g\equiv x{y}'+({{x}^{2}}-{{\nu }^{2}})y \\ \end{align}$$

thus Eq 2.1 can be written as,


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$$ \displaystyle f{y}''+g=0$$ (Eq 2.2)
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which satisfy the first condition of exactness in method of Mtg15-3.

From the above equation we derive,


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$$ \displaystyle \begin{align} & {{f}_{xx}}=2 \\ & {{f}_{xy}}=0 \\ & {{f}_{yy}}=0 \\ & {{f}_{xp}}=0 \\ & {{f}_{yp}}=0 \\ & {{f}_{y}}=0 \\ & {{g}_{xp}}=1 \\ & {{g}_{yp}}=0 \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{x}^{2}}-{{\nu }^{2}} \\ \end{align}$$ (Eqs 2.3)
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Substitute Eqs 2.3 into (1)15-3 and (2)15-3 for the second condition of exactness, we derive,


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$$ \displaystyle \begin{matrix} {{x}^{2}}-{{\nu }^{2}}+1=0 \\ 0=0 \\ \end{matrix}$$ (Eq 2.4)
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which can't be satisfied because $$ \displaystyle x$$ is a variable while $$ \displaystyle \nu $$ is a constant. Thus The original Bessel Equation is not exact.

ii) verifying exactness by method given by Mtg24-4
Define according to the Bessel Equation,


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$$ \displaystyle F\equiv {{x}^{2}}{y}''+x{y}'+({{x}^{2}}-{{\nu }^{2}})y$$ so as for the form the Eq 2.1 satisfy the first condition in method of Mtg22-2.
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We derive from the previous definition that,


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$$ \displaystyle \begin{align} & \frac{\partial F}{\partial y}={{x}^{2}}-{{\nu }^{2}} \\ & \frac{\partial F}{\partial {y}'}=x \\ & \frac{d}{dx}\left( \frac{\partial F}{\partial {y}'} \right)=1 \\ & \frac{\partial F}{\partial {y}''}={{x}^{2}} \\ & \frac{d{{x}^{2}}}\left( \frac{\partial F}{\partial {y}''} \right)=2 \\ \end{align}$$ (Eqs 2.5)
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We learn from Mtg22-4 that the second condition of this method with respect to 2nd order ODE is,


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$$ \displaystyle \frac{\partial F}{\partial y}+\frac{d}{dx}\left( \frac{\partial F}{\partial {y}'} \right)+\frac{d{{x}^{2}}}\left( \frac{\partial F}{\partial {y}''} \right)=0$$ (Eq 2.6)
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Substituting Eqs 2.5 into Eq 2.6 we have,


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$$ \displaystyle {{x}^{2}}-{{\nu }^{2}}+1=0$$ (Eq 2.7)
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This equation won't be satisfied for the same reason as Eq 2.4. Thus second method also shows that Bessel Equation is not exact.

iii) the trial of IFM method
Define integrating factor $$ \displaystyle h$$ as ,


 * $$\displaystyle h\equiv {{x}^{m}}{{y}^{n}}$$

where $$ \displaystyle m$$ and $$ \displaystyle n$$ are some constants.

Then we multiple $$ \displaystyle h$$ to both sides of Eq 2.1 changing it as,


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$$ \displaystyle {{F}_{1}}={{x}^{m+2}}{{y}^{n}}{y}''+{{x}^{m+1}}{{y}^{n}}{y}'+({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}=0$$ (Eq 2.8)
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This of course satisfy the first condition of exactness of method of Mtg22-4.

Then we have,


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$$ \displaystyle \begin{align} & \frac{\partial {{F}_{1}}}{\partial y}=n{{x}^{m+2}}{{y}^{n-1}}{y}''+n{{x}^{m+1}}{{y}^{n-1}}{y}'+(n+1){{x}^{m+2}}{{y}^{n}}-(n+1){{\nu }^{2}}{{x}^{m}}{{y}^{n}} \\ & \frac{\partial {{F}_{1}}}{\partial {y}'}={{x}^{m+1}}{{y}^{n}} \\ & \frac{d}{dx}\left( \frac{\partial {{F}_{1}}}{\partial {y}'} \right)=(m+1){{x}^{m}}{{y}^{n}} \\ & \frac{\partial {{F}_{1}}}{\partial {y}''}={{x}^{m+2}}{{y}^{n}} \\ & \frac{d{{x}^{2}}}\left( \frac{\partial {{F}_{1}}}{\partial {y}''} \right)=(m+2)(m+1){{x}^{m}}{{y}^{n}} \\ \end{align}$$ (Eqs 2.9)
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Then substitute Eq 2.9 into Eq 2.6, we get,


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$$ \displaystyle n{{x}^{m+2}}{{y}^{n-1}}{y}''+n{{x}^{m+1}}{{y}^{n-1}}{y}'+{{x}^{m}}{{y}^{n}}\left[ (n+1)({{x}^{2}}-{{\nu }^{2}})+{{(m+1)}^{2}} \right]=0$$ (Eq 2.10)
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Since this equation is supposed to be free from $$ \displaystyle {{y}}$$, thus we must eliminate terms with $$ \displaystyle {{y}}$$, so we derive,


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$$ \displaystyle n=0$$ (Eq 2.11)
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and then,


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$$ \displaystyle ({{x}^{2}}-{{\nu }^{2}})+{{(m+1)}^{2}}=0$$ (Eq 2.12)
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which can't be satisfied because $$ \displaystyle x$$ is a variable but $$ \displaystyle m$$ and $$ \displaystyle \nu$$ are all constants. This means the recipe of Mtg19 doesn't make Bessel Equation exact.

iiii) another perspective of explanation
To illustrate our ideas in another way, we can firstly assume that Eq 2.1 is exact and see whether we can find any $$ \displaystyle \Phi $$ such that,


 * $$\displaystyle \frac{d\Phi }{dx}=F$$

Since we know that ,


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$$ \displaystyle \begin{align} & f={{x}^{2}}={{\Phi }_{p}} \\ & g=x{y}'+({{x}^{2}}-{{\nu }^{2}})y={{\Phi }_{x}}+{{\Phi }_{y}}{y}' \\ \end{align}$$ (Eq 2.13)
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where $$ \displaystyle p$$ denotes $$ \displaystyle {{y}'}$$.

we can write $$ \displaystyle \Phi $$ as,


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$$ \displaystyle \Phi =H(x,y)+{{x}^{2}}p$$ (Eq 2.14)
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and from that we get,


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$$ \displaystyle \begin{align} & {{\Phi }_{x}}={{H}_{x}}+2xp \\ & {{\Phi }_{y}}={{H}_{y}} \\ \end{align}$$ (Eq 2.15)
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Substituting Eq 2.15 back into Eq 2.13 we get,


 * $$\displaystyle {{H}_{x}}+{{H}_{y}}p+xp=({{x}^{2}}-{{\nu }^{2}})y$$

or


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$$ \displaystyle {{H}_{x}}=-{{H}_{y}}p-xp+({{x}^{2}}-{{\nu }^{2}})y$$ (Eq 2.16)
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while we always have,


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$$ \displaystyle dH={{H}_{x}}dx+{{H}_{y}}dy$$ (Eq 2.17)
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If we substitute Eq 2.16 into Eq 2.17, we derive,


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$$ \displaystyle dH=({{x}^{2}}-{{\nu }^{2}})ydx-xdy$$ (Eq 2.18)
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We can see that the term $$ \displaystyle {{H}_{y}}$$ is cancelled and there is no $$ \displaystyle {{H}_{x}}$$ left either which means the function $$\displaystyle H$$ always have the form of Eq 2.18 regardless of what you assume $$ \displaystyle {{H}_{x}}$$ or $$ \displaystyle {{H}_{y}}$$ to be. This also answers for the "2 choices" tricks of Mtg18 and the last problem of HW3("Finishing the story").

Besides that, we know that the prerequisite of the existense of $$ \displaystyle H$$ is that the RHS of Eq 2.18 is exact, meaning,


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$$ \displaystyle \begin{align} & \frac{\partial ({{x}^{2}}-{{\nu }^{2}})y}{\partial y}=\frac{\partial (-x)}{\partial x} \\ & \Rightarrow {{x}^{2}}-{{\nu }^{2}}=-1 \\ \end{align}$$ (Eq 2.19)
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Unfortunately Eq 2.19 is not satisfied thus this $$\displaystyle H$$ never exist and so is $$ \displaystyle \Phi $$. Therefore equations from Eq 2.13 to Eq 2.19(except Eq 2.17) they all become meaningless and we conclude that Bessel Equation is non-exact. In fact those equations are all based on the assumption that "Eq 2.1 is exact". If we know Eq 2.1 is non-exact in advance, then the effort to find $$ \displaystyle H$$ is predestined to fail.

One may argue that the integrating factor may turn Eq 2.1 Bessel Equation into exact one.

From Eq 2.8 we difine,


 * $$\displaystyle \begin{align}

& {{f}_{1}}\equiv {{x}^{m+2}}{{y}^{n}} \\ & {{g}_{1}}\equiv {{x}^{m+1}}{{y}^{n}}{y}'+{{x}^{m+2}}{{y}^{n+1}}-{{\nu }^{2}}{{x}^{m+1}}{{y}^{n+1}} \\ \end{align}$$

in which,


 * $$\begin{align}

& {{f}_{1}}={{\Phi }_{P}} \\ & {{g}_{1}}={{\Phi }_{x}}+{{\Phi }_{y}}{y}' \\ \end{align}$$

So we can write $$ \displaystyle \Phi $$ in such form:


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$$ \displaystyle {{\Phi }_{}}={{H}_{1}}+{{x}^{m+2}}{{y}^{n}}p$$ (Eq 2.20)
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or


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$$ \displaystyle d\Phi =d{{H}_{1}}+d({{x}^{m+2}}{{y}^{n}}p)$$ (Eq 2.21)
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Note that,


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$$ \displaystyle \frac{d\Phi }{dx}={{F}_{1}}={{x}^{m+2}}{{y}^{n}}{y}''+{{x}^{m+1}}{{y}^{n}}{y}'+({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}$$ (Eq 2.22)
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Then we derive $$ \displaystyle d{{H}_{1}}$$ without calculating its partial derivatives.


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$$ \displaystyle d{{H}_{1}}=({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}dx+\left[ -(m+1){{x}^{m+1}}{{y}^{n}}-np{{x}^{m+2}}{{y}^{n-1}} \right]dy$$ (Eq 2.23)
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Since $$ \displaystyle {{H}_{1}}$$ is supposed to be independent of $$ \displaystyle p$$ but integrating according to $$ \displaystyle dy$$ won't make $$ \displaystyle p$$ disappear. This means one may expect the term including $$ \displaystyle p$$ to be eliminated, thus $$ \displaystyle n=0$$.

If $$ \displaystyle {{H}_{1}}$$ really exist, the RHS of Eq 2.23 must be exact, meaning,


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$$ \displaystyle \frac{\partial \left[ ({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{0+1}} \right]}{\partial y}=\frac{\partial \left[ -(m+1){{x}^{m+1}}{{y}^{0}}-0*p{{x}^{m+2}}{{y}^{0-1}} \right]}{\partial x}$$ (Eq 2.24)
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Then we have,
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$$ \displaystyle {{(m+1)}^{2}}+({{x}^{2}}-{{\nu }^{2}})=0$$ (Eq 2.25)
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which can't be satisfied. We still can't find legitimate $$ \displaystyle m$$ and $$ \displaystyle n$$ to make $$ \displaystyle {{H}_{1}}$$ exist and the same with $$ \displaystyle \Phi$$ which means integrating factor can't save Bessel Equation from being non-exact.

=Problem 3: FindX(x) in terms of sin,cos,sinh,cosh=

From the lecture slide Mtg 25-2

Given


Equation of motion for Euler-Bernoulli Beam given as :


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$$  \displaystyle -EI\frac{\partial^4 u}{\partial x^4}+f=m\frac{\partial^2 u}{\partial t^2} $$     (Eq 3.1)
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where

f=disrubuted load

u=transverse displacement

For f=0 Which equals the free vibration condition :


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$$  \displaystyle -EI\frac{\partial^4 u}{\partial x^4}=m\frac{\partial^2 u}{\partial t^2} $$     (Eq 3.2)
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Euler thought that this equation has a solution in the form of exponential.

Method of variable separation

Since $$\displaystyle u=u(x,t) $$ we have to use separation of variables method separation of variables to separate u into two functions both depends on its own variable.Then we can conclude that:


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$$  \displaystyle u(x,t)=X(x).T(t) $$     (Eq 3.3)
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$$  \displaystyle -EIX^{(4)}T=mX\ddot{T} $$     (Eq 3.4)
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Assuming that


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$$  \displaystyle T(t)=e^{iwt} $$     (Eq 3.5)
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$$  \displaystyle \ddot{T}=-w^{2}T $$     (Eq 3.6)
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$$  \displaystyle \frac{EI}{mw^{2}}X^{(4)}=X $$     (Eq 3.7)
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$$  \displaystyle K^{4}=\frac{EI}{mw^{2}} $$     (Eq 3.8)
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$$  \displaystyle K^{4}X^{(4)}=X $$     (Eq 3.9)
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Method Of Trial Solution:Assuming that the solution in the form of:


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$$  \displaystyle X(x)=exp(rx) $$     (Eq 3.10)
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r=unknown coefficient

r's are can be found by solving ths characteristic equation. Since the characteristic equation has a power of 4 we have four roots.


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$$  \displaystyle K^{4}r^{4}=1 $$     (Eq 3.11)
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$$  \displaystyle r^{4}=\frac{1}{K^{4}} $$     (Eq 3.12)
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$$  \displaystyle r_{1}=\frac{1}{K} $$     (Eq 3.13)
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$$  \displaystyle r_{2}=-\frac{1}{K} $$     (Eq 3.14)
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$$  \displaystyle r_{3}=\frac{i}{K} $$     (Eq 3.15)
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$$  \displaystyle r_{4}=-\frac{i}{K} $$     (Eq 3.16)
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Find
Show X(x) in terms of sin,cos,sinh,cosh

Solution
If we set up solution


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$$  \displaystyle X(x)=c_{1}exp(\frac{x}{k})+c_{2}exp(-\frac{x}{k})+c_{3}exp(i\frac{x}{k})+c_{4}exp(-i\frac{x}{k}) $$     (Eq 3.17)
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$$  \displaystyle e^{i\phi }=cos\phi +isin\phi $$     (Eq 3.18)
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$$  \displaystyle e^{-i\phi }=cos\phi-isin\phi $$     (Eq 3.19)
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Now we can manupulate equation by part by part.


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$$  \displaystyle c_{3}e^{i\frac{x}{k}}=c_{3}\left [ cos\frac{x}{k}+isin\frac{x}{k} \right ] $$     (Eq 3.20)
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$$  \displaystyle c_{4}e^{-i\frac{x}{k}}=c_{4}\left [ cos\frac{x}{k}-isin\frac{x}{k} \right ] $$     (Eq 3.21)
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$$  \displaystyle c_{3}e^{i\frac{x}{k}}+c_{4}e^{-i\frac{x}{k}}=(c_{3}+c_{4})cos\frac{x}{k}+i(c_{3}-c_{4})sin\frac{x}{k} $$     (Eq 3.22)
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$$  \displaystyle c_{1}exp(\frac{x}{k})=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(\frac{x}{k}) $$     (Eq 3.23)
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$$  \displaystyle c_{2}exp(-\frac{x}{k})=\frac{1}{2}c_{2}exp(-\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k}) $$     (Eq 3.24)
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$$  \displaystyle c_{1}exp(\frac{x}{k})+c_{2}exp(-\frac{x}{k})=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k}) $$     (Eq 3.25)
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Let's add and subtract $$\displaystyle\frac{1}{2}c_{1}exp(-\frac{x}{k})$$ and $$ \displaystyle \frac{1}{2}c_{2}exp(\frac{x}{k}) $$

Hyperbolic forms of sin and cos are given as:


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$$  \displaystyle sinhx=\frac{1}{2}(e^{x}-e^{-x}) $$     (Eq 3.26)
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$$  \displaystyle coshx=\frac{1}{2}(e^{x}+e^{-x}) $$     (Eq 3.27)
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$$  \displaystyle c_{1}cosh\frac{x}{k}=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(-\frac{x}{k}) $$     (Eq 3.28)
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$$  \displaystyle c_{1}sinh\frac{x}{k}=\frac{1}{2}c_{1}exp(\frac{x}{k})-\frac{1}{2}c_{1}exp(-\frac{x}{k}) $$     (Eq 3.29)
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Then we can conclude same things for c2 and if we put all terms in the Eq.(3.17)


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$$  \displaystyle X(x)=(c_{1}+c_{2})cosh\frac{x}{k}+(c_{1}-c_{2})sinh\frac{x}{k}+(c_{3}+c_{4})cos\frac{x}{k}+i(c_{3}-c_{4})sin\frac{x}{k}
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$$     (Eq 3.30)
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$$  \displaystyle X(x)=A_{1}cosh\frac{x}{k}+A_{2}sinh\frac{x}{k}+A_{3}cos\frac{x}{k}+A_{4}sin\frac{x}{k}
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$$     (Eq 3.31)
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Where


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$$  \displaystyle A_{1}=(c_{1}+c_{2}) ,
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A_{2}=(c_{1}-c_{2}) ,

A_{3}=(c_{3}+c_{4}) ,

A_{4}=i(c_{3}-c_{4})

$$
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 * }

=Contributions Of Team Members=


 * Problem 1:    Solved by: Patterson, Rasul

Author: Patterson

Proofread by: Rasul, Dube


 * Problem 2:    Solved by: Zou, Oztekin

Author: Zou, Yaakov, Ismail

Proofread by: Yaakov, Ismail


 * Problem 3:    Solved by: Oztekin

Author: Oztekin, Dube

Proofread by: Dube

= References =