User:Egm6321.f10.team03/Hwk4

=Problem 1: Solution of a L2-ODE-VC=

Given
The following Linear second order ODE with Varying Coefficients is given:
 * {| style="width:100%" border="0"

$$\,F=2xy+\left( {{x}^{2}}-\sin (x) \right){y}'+\cos (x){{y}'}'=0$$
 * $$\,(Eq. 1.1) $$
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 * }

Find
It has to be shown that Eq. 1.1 is exact (or either can be made exact), the function $$\Phi $$ has to be found and the solution for y needs to be calculated.

Exactness condition 1
The first exactness conditon is the one where the form of the given function F in Eq. 1.1 has to match the pattern $$g(x,y,p)+f(x,y,p)y''=0$$. Looking at Eq. 1.1 it can be seen that:
 * {| style="width:100%" border="0"

$$\underbrace{2xy+\left( {{x}^{2}}-\sin (x) \right){y}'}_{g(x,y,p)}+\underbrace{\cos \left( x \right)}_{f(x,y,p)}{{y}'}'=0$$ and thus the first condition of exactness is met.
 * $$\,(Eq. 1.2) $$
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 * }

Exactness condition 2.1
As known, the second exactness condition for an L2-ODE has two sub-conditions, the first one says that:
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$${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}\overset{!}{\mathop{=}}\,{{g}_{xp}}+p{{g}_{yp}}-{{g}_{y}}$$ where in this particular problem the single values are:
 * $$\,(Eq. 1.3) $$
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 * }
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$$\begin{align} & {{f}_{x}}=-\sin \left( x \right) \\ & {{f}_{xx}}=-\cos \left( x \right) \\ & {{f}_{xy}}={{f}_{xp}}={{f}_{yy}}={{f}_{y}}={{f}_{yp}}=0 \\ \end{align}$$

and
 * $$\,(Eq.set 1.4) $$
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 * }
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$$\begin{align} & {{g}_{x}}=2y+p\left( 2x-\cos \left( x \right) \right) \\ & {{g}_{y}}=2x \\ & {{g}_{xp}}=2x-\cos \left( x \right) \\ & {{g}_{yp}}={{g}_{pp}}=0 \\ \end{align}$$

Inserting the values of Eq.set 1.4 and 1.5 into Eq. 1.3 the first sub-condition of the second exactness condition becomes:
 * $$\,(Eq.set 1.5) $$
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 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=2x-\cos \left( x \right)-2x$$ which is equal to
 * $$\,(Eq. 1.6) $$
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 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=-\cos \left( x \right)$$ and thus the first sub-condition is also met.
 * $$\,(Eq. 1.7) $$
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 * }

Exactness condition 2.2
The second sub-condition of the second exactness condition for L2-ODE's, says that:
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$${{f}_{xp}}+p{{f}_{yp}}+2{{f}_{y}}\overset{!}{\mathop{=}}\,{{g}_{pp}}$$ inserting the appropriate values from Eq.set 1.4 and 1.5 in to Eq. 1.8 yields to:
 * $$\,(Eq. 1.8) $$
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 * }
 * {| style="width:100%" border="0"

$$\,0=0$$ and which clearly states that the left and right side of Eq. 1.8 are equal. '''Thus also the second sub-condition is met and therefore the given function Equation in Eq. 1.1 is exact!'''
 * $$\,(Eq. 1.9) $$
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 * }

Find $$\Phi $$
From Eq. (3) of the lecture notes in pp. 15-3, we know that function $$\Phi $$ is composed of the following sum:
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$$\Phi =h(x,y)+\int{f\left( x,y,p \right)}\,dp$$ The second term of Eq. 1.0 can be easily calculated to be:
 * $$\,(Eq. 1.10) $$
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 * }
 * {| style="width:100%" border="0"

$$\int{f\left( x,y,p \right)}\,dp=\cos \left( x \right)p$$ And thus the function $$\Phi$$ can be expressed as:
 * $$\,(Eq. 1.11) $$
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 * }
 * {| style="width:100%" border="0"

$$\Phi =h\left( x,y \right)+\cos \left( x \right)p$$ Now, from Eq. (3) of the lecture notes in pp. 15-2, it's known that
 * $$\,(Eq. 1.12) $$
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 * }
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$$\,g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}$$ which in our specific case is equal to
 * $$\,(Eq. 1.13) $$
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 * }
 * {| style="width:100%" border="0"

$$g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}=2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p$$ If we take the derivatives of $$\Phi$$ in Eq. 1.12 with respect to x and with respect to y, we get:
 * $$\,(Eq. 1.14) $$
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 * }
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$${{\Phi }_{x}}={{h}_{x}}-p\cdot \sin \left( x \right)$$
 * $$\,(Eq. 1.15) $$
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 * }
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$$\,{{\Phi }_{y}}={{h}_{y}}$$ Inserting Eq. 1.15 and Eq. 1.16 into Eq. 1.13 enables to write Eq. 1.14 as
 * $$\,(Eq. 1.16) $$
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 * }
 * {| style="width:100%" border="0"

$$2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p={{h}_{x}}-p\cdot \sin \left( x \right)+{{h}_{y}}p={{h}_{x}}+\left( {{h}_{y}}-\sin \left( x \right) \right)p$$ and by comparison of the left and right side of the Eq. 1.17, it can be seen that:
 * <p style="text-align:right;">$$\,(Eq. 1.17) $$
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 * }
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$$\,{{h}_{x}}=2xy$$
 * <p style="text-align:right;">$$\,(Eq. 1.18) $$
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 * }
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$$\,{{h}_{y}}={{x}^{2}}$$ The function h(x,y) can be obtained by integrating $${{h}_{x}}$$ with respect to x, that for our particular case yields to:
 * <p style="text-align:right;">$$\,(Eq. 1.19) $$
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 * }
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$$\int{{{h}_{x}}\,dx}=h\left( x,y \right)={{x}^{2}}y+{{k}_{1}}(y)$$ If we differentiate Eq. 1.20 with respect to y and compare the result with Eq. 1.19, then the value for $${{k}_{1}}(y)$$ can be found:
 * <p style="text-align:right;">$$\,(Eq. 1.20) $$
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 * }
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$$\frac{dh\left( x,y \right)}{dy}={{x}^{2}}+{{k}_{1}}^{\prime }(y)$$
 * <p style="text-align:right;">$$\,(Eq. 1.21) $$
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 * }
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$${{h}_{y}}={{x}^{2}}+{{k}_{1}}^{\prime }(y)\overset{!}{\mathop{=}}\,{{x}^{2}}$$ thus,
 * <p style="text-align:right;">$$\,(Eq. 1.22) $$
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 * }
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$${{k}_{1}}^{\prime }\left( y \right)=0\Rightarrow {{k}_{1}}\left( y \right)=C$$ With the knowledge of Eq. 1.23 we can rewrite Eq. 1.20 as follows for the function h(x,y):
 * <p style="text-align:right;">$$\,(Eq. 1.23) $$
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 * }
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$$\,h(x,y)={{x}^{2}}y+C$$ and thus the function $$\Phi$$ can be written as being:
 * <p style="text-align:right;">$$\,(Eq. 1.24) $$
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 * }
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$$\Phi \left( x,y,p \right)={{x}^{2}}y+\cos \left( x \right)p+C={{x}^{2}}y+\cos \left( x \right){y}'+C=0$$
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 * <p style="text-align:right;">$$\,(Eq. 1.25) $$
 * style= |
 * }

Solve for y
At this point solving for y is not very difficult anymore, since we only need to integrate p which is the derivative of y. For that, Eq. 1.25 is rewritten as:
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$$p\left( x \right)=\frac{-({{x}^{2}}y+C)}{\cos \left( x \right)}=\frac{-{{x}^{2}}y}{\cos \left( x \right)}-\frac{C}{\cos \left( x \right)}={y}'$$ Integrating Eq. 1.26 with respect to x gives the solution for y:
 * <p style="text-align:right;">$$\,(Eq. 1.26) $$
 * style= |
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team03.sigillo 01:01, 19 October 2010 (UTC)


 * Reviewed by Egm6321.f10.team03 03:02, 19 October 2010 (UTC) Schulze


 * Reviewed by Egm6321.f10.team03 13:19, 19 October 2010 (UTC) Hong SJ


 * Solved, Proof read and edited by Adam Franklin 02:09, 20 October 2010 (UTC)

= Problem 2 - Verify Exactness of Bessel Equation =

Given

 * Bessel Equation is given by
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$$
 * $${{x}^{2}}{{y}^{''}}+x{{y}^{'}}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * }
 * }


 * Where, $$\displaystyle\upsilon \in \mathbb{R}$$

Find

 * $$\displaystyle\left( 1 \right)$$ Verify Exactness of $$\displaystyle Eq.2.1$$ using 2 methods


 * <H4>Method 1</H4>


 * Second conditions of Exactness:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 * <p style="text-align:right;">$$\displaystyle (Eq.2.2)
 * }
 * }


 * and


 * {| style="width:100%" border="0" align="left"

$$ $$
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 * <p style="text-align:right;">$$\displaystyle (Eq.2.3)
 * }
 * }


 * <H4>Method 2</H4>


 * Another form of second Exactness condition for a $$\displaystyle N2-ODE$$:


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f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0 $$ $$.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.2.4)
 * }
 * }
 * $$\displaystyle\left( 2 \right)$$ If $$\displaystyle Eq.2.1$$ not Exact, see whether it can be made Exact using Integrating Factor Method with:


 * {| style="width:100%" border="0"

$$
 * $$h\left( x,y \right)={{x}^{m}}{{y}^{n}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * }
 * }


 * That is,find $$\displaystyle (m,n) $$ such that Eq 2.1 is exact.

Solution

 * <H4>PART 1: Verify Exactness of Bessel Equation</H4>


 * The first condition for exactness of $$\displaystyle N2-ODE$$ is that it is in the following form:
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F(x,y,p) = f(x,y,p)y'' + g(x,y,p)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.6)
 * }
 * }


 * where


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p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.7)
 * }
 * }


 * Now comparing Eq.2.1 with the form of Eq.2.6,we can see that:


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$$
 * $$\displaystyle f(x,y,p)={{x}^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.8)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.8)
 * }
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 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle g(x,y,p)=x{p}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.9)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.9)
 * }
 * }


 * Hence the first condition for exactness is satisfied.


 * Now verify second condition of Exactness:


 * (a) Using Method 1- Eq.2.2 and Eq.2.3


 * Taking derivative of Eq.2.8 yields:


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_x = 2x} & {f_{xx}  = 2} & {f_{xy}  = 0} & {f_{xp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {f_{yp}  = 0}\\ \end{array} $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * Similarly taking derivative of Eq.2.9 yields:


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {g_x = p+2xy} & {g_{xp} = 1} & {{g}_{y}}={{x}^{2}}-{{\upsilon }^{2}} & {g_{yp}  = 0 }  \\ {g_p = x} & {g_{pp}  = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting the above values to Eq.2.3 (first relation) and Eq.2.4 (second relation):


 * {| style="width:100%" border="0" align="left"


 * $$2+2p\times 0+{{p}^{2}}\times 0=1+p\times 0-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$
 * $$\Rightarrow 2=1-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.10)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.10)
 * }
 * }


 * So first relation for second exactness condition is not satisfied.


 * {| style="width:100%" border="0" align="left"

$$
 * $$0+p\times 0+2\times 0=0\Rightarrow 0=0$$, second relation is satisfied
 * <p style="text-align:right;">$$\displaystyle (Eq.2.11)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.11)
 * }
 * }


 * Since the first relation is not satisfied, the Eq.2.1 is not Exact


 * (b) Using Method 2- Eq.2.4


 * Differentiating $$\displaystyle Eq.2.1$$, partially with respect to $$\displaystyle y,{{y}^{'}},{{y}^{''}}$$,


 * {| style="width:100%" border="0" align="left"


 * $${{f}_{0}}=\frac{\partial F}{\partial y}={{x}^{2}}-{{\upsilon }^{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $${{f}_{1}}=\frac{\partial F}{\partial {{y}^{'}}}=x$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $${{f}_{2}}=\frac{\partial F}{\partial {{y}^{''}}}={{x}^{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * Now, $$\frac{\partial {{f}_{1}}}{\partial x}=1$$, and $$\frac{\partial {{x}^{2}}}=2$$
 * }
 * }
 * }


 * Substittuting above values in $$\displaystyle Eq.2.4$$ yields,


 * {| style="width:100%" border="0" align="left"


 * $${{x}^{2}}-{{\upsilon }^{2}}-1+2\ne 0\Rightarrow $$not satisfied and hence the Eq.2.1 is not Exact.
 * }
 * }
 * }


 * <H4>PART 2: Make Bessel Equation exact using Integrating Factor Method</H4>


 * Multiply $$\displaystyle Eq.2.1$$ by Integrating Factor ($$\displaystyle Eq.2.5$$)$$\Rightarrow $$


 * {| style="width:100%" border="0" align="left"

$$
 * $$\underbrace_{f}{{y}^{''}}+\underbrace{{{x}^{m+1}}{{y}^{n}}{{y}^{'}}+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}}_{g}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.12)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.12)
 * }
 * }


 * New $$\displaystyle f$$ and $$\displaystyle g$$ are:


 * {| style="width:100%" border="0" align="left"

$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}{{y}^{n}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.13)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.13)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}{{y}^{n}}p+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.14)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.14)
 * }
 * }


 * {| style="width:100%" border="0" align="left"



Analyzing the second relation of second exactness condition ($$\displaystyle Eq.2.3 $$):$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}$$, it can be noted that most of the terms drop out since $$\displaystyle f$$ is independent of $$\displaystyle p$$. Consequently all derivatives of $$\displaystyle f$$ with respect to $$\displaystyle p$$ become zero. Moreover $$\displaystyle {{g}_{pp}}=0$$ and hence $$\displaystyle Eq.2.3 $$ is simplified to,
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle 2{{f}_{y}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.15)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.15)
 * }
 * }


 * Taking derivative of $$\displaystyle (Eq.2.13)$$ with repect to $$\displaystyle y$$ and substituting to$$\displaystyle (Eq.2.15)$$ yields,


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle n{{x}^{m+2}}{{y}^{n-1}}=0$$
 * }
 * }
 * }


 * The above equation is true only when,


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$$\displaystyle n=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.16)
 * }
 * }


 * As $$\displaystyle n=0 $$, the function $$\displaystyle f$$ and $$\displaystyle g$$ can be rewritten as:


 * {| style="width:100%" border="0" align="left"

$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.17)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.17)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}p+{{x}^{m+2}}y-{{x}^{m}}y{{\upsilon }^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.18)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.18)
 * }
 * }


 * Taking derivative of $$\displaystyle Eq.2.17$$ yields,


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {{f}_{x}}=\left( m+2 \right){{x}^{m+1}}, & {{f}_{xx}}=\left( m+2 \right)\left( m+1 \right){{x}^{m}}, & {f_{xy} = 0}, & {f_{xp}  = 0 }  \\ {f_y = 0}, & {f_{yy}  = 0}, & {f_{yp}  = 0}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Similarly taking derivative of $$\displaystyle Eq.2.18$$ yields,


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {{g}_{x}}=\left( m+1 \right){{x}^{m}}p+\left( m+2 \right){{x}^{m+1}}y-m{{x}^{m-1}}y{{\upsilon }^{2}}\\ {{g}_{xp}}=\left( m+1 \right){{x}^{m}}\\ {{g}_{y}}={{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}}\\ {g_{yp} = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above differentials in $$\displaystyle Eq.2.2$$ gives,


 * {| style="width:100%" border="0" align="left"


 * $$\left( m+2 \right)\left( m+1 \right){{x}^{m}}+0+0=\left( m+1 \right){{x}^{m}}+0-\left( {{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }


 * Simplifying,


 * {| style="width:100%" border="0" align="left"


 * $$\left( m+2 \right)\left( m+1 \right)=m+1+{{\upsilon }^{2}}-{{x}^{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\Rightarrow {{m}^{2}}+2m+\left( {{x}^{2}}-{{\upsilon }^{2}}+1 \right)=0$$
 * }
 * }
 * }


 * $$\displaystyle m$$ is obtained by solving the above quadratic equation:


 * {| style="width:100%" border="0"

$$m=-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.19)
 * }
 * }


 * So the integration factor is obtained by substituting $$\displaystyle Eq. 2.16 $$ and $$\displaystyle Eq. 2.19 $$ in $$\displaystyle Eq. 2.5 $$:


 * {| style="width:100%" border="0"

$$h={{x}^{-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.20)
 * }
 * }


 * Thus Bessel Equation can be made exact using the above integrating factor.

Contributing Members

 * Solved and Posted by Egm6321.f10.team3.Sudheesh 01:51, 19 October 2010 (UTC)


 * Solved and Proof read by Adam Franklin 02:50, 20 October 2010 (UTC)

=Problem 3: Solution of Free Vibration Equation for Euler bernoulli Beam using Method of Trial Solution=

Given
Free vibration equation for Euler Bernoulli beam is given by,


 * {| style="width:100%" border="0"


 * $$\displaystyle -EI\frac{{{\partial }^{4}}u}{\partial {{x}^{4}}}=m\frac{{{\partial }^{2}}u}{\partial {{t}^{2}}}$$
 * style= |
 * }
 * }


 * {| style="width:100%" border="0"


 * Assuming, $$u(x,t)=\chi \left( x \right)T\left( t \right)$$, and using the separation of variable technique,we obtain the following relation:
 * style= |
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{k}^{4}}{{\chi }^{\left( 4 \right)}}=\chi $$
 * style= |
 * }
 * }


 * {| style="width:100%" border="0"


 * Where, $${k}^{4}=\frac{EI}{m{{\omega }^{2}}}$$
 * style= |
 * }
 * }

Now using Method of trial solution(see King p 513, A5.3), that is assuming $$X\left( x \right)={{e}^{rx}}$$,we obtain undetermined coefficient(roots) $$\displaystyle r$$ as given in [[media:2010_10_14_14_56_10.djvu|page 25-2]] of the lecture notes


 * {| style="width:100%" border="0"

$$ {r}_{1,2} = \pm  \frac{1}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (3.1)
 * }


 * {| style="width:100%" border="0"

$$ {r}_{3,4} = \pm  \frac{i}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (3.2)
 * }

where $$ i $$ is the imaginary number and $$ k =  \sqrt[4]{ \frac{EI}{m { \omega}^{2}}} $$ given in pg 25-1 of the lecture notes

Find
Find
 * {| style="width:100%" border="0"

\chi (x) = {e}^{rx} $$ $$ in terms of sin, cos, sinh and cosh
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3.3)
 * style= |
 * }

Solution
First we solve $$\displaystyle (3.3) $$ for the root in $$\displaystyle (3.2) $$ using the following relationships


 * {| style="width:100%" border="0"

{e}^{i \theta} = \cos( \theta) + i\sin( \theta) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (3.4)
 * style= |
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle {{e}^{-i\theta }}=\cos (\theta )-i\sin (\theta )$$
 * <p style="text-align:right;">$$\displaystyle (3.5)
 * <p style="text-align:right;">$$\displaystyle (3.5)
 * style= |
 * }

The solution to $$\displaystyle (3.3) $$ becomes

Next we solve $$\displaystyle (3.3) $$ for the root in $$\displaystyle (3.1) $$ using the following relationships


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle {{e}^{\theta }}=\cosh (\theta )+\sinh (\theta )$$
 * <p style="text-align:right;">$$\displaystyle (3.7)
 * <p style="text-align:right;">$$\displaystyle (3.7)
 * style= |
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle {{e}^{-\theta }}=\cosh (\theta )-\sinh (\theta )$$
 * <p style="text-align:right;">$$\displaystyle (3.8)
 * <p style="text-align:right;">$$\displaystyle (3.8)
 * style= |
 * }

The solution to $$\displaystyle (3.3) $$ becomes

Using superposition, the solution to $$\displaystyle (3.3) $$ is

Contributing Members

 * Solved and proofread by James Roark 11:42, 19 October 2010 (UTC)
 * Posted by Chris Cook --98.71.213.114 22:18, 18 October 2010 (UTC)
 * Solved and proofread by Adam Franklin 02:55, 20 October 2010 (UTC)
 * Solved,proofread Egm6321.f10.team03.sigillo 03:57, 20 October 2010 (UTC)
 * Solved, edited and proofread by Egm6321.f10.team3.Sudheesh 14:56, 20 October 2010 (UTC)

=Contributing Team Members=
 * Egm6321.f10.team3.Sudheesh 14:56, 20 October 2010 (UTC)- Solved and posted problem 2 and proofread problem 3