User:Egm6321.f10.team4/HW 4

= Problem 1 -  Computing $$\phi$$ and y from an exact L2_ODE =

From

Given

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

$$\displaystyle F={cos(x)y''+(x^2-sin(x))p+2xy}=0 $$ $$
 * $$\displaystyle (Eq. 1-1)
 * }
 * }

Find
a) Is F exact?

b) Find $$\phi$$

c) Solve for $$y$$

Solution
Part a

In order for F to be exact, the following two equations need to be satisfied.


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$$\displaystyle {g_{pp}}=f_{xp}+pf_{yp}+2f_{y} $$ $$
 * $$\displaystyle (Eq. 1-2)
 * }
 * }


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$$\displaystyle {f_{xx}}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}+g_{y} $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }

Before computing the differentials let's first identify $$f$$ and $$g$$


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$$\displaystyle f(x,y,p)=cos(x) $$ $$
 * $$\displaystyle (Eq. 1-4)
 * }
 * }


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$$\displaystyle g(x,y,p)=(x^2-sin(x))p+2xy $$ $$
 * $$\displaystyle (Eq. 1-5)
 * }
 * }

The differentials are as follows


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$$\displaystyle g_y=2x $$
 * }
 * }


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


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$$\displaystyle g_x=2y+(2x-cox(x))p $$
 * }
 * }


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


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$$\displaystyle f_x=-sin(x) $$
 * }
 * }


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$$\displaystyle f_{xp}=0 $$
 * }
 * }


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$$\displaystyle f_{xy}=0 $$
 * }
 * }


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$$\displaystyle f_{xx}=-cos(x) $$
 * }
 * }


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$$\displaystyle f_y=0 $$
 * }
 * }


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

If we take the above differentials and replace them in Eq (1-2)


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$$\displaystyle 0=0 $$
 * }
 * }

thus the first condition of exactness is met. Next, let's take the corresponding differentials and replace them into Eq. (1-3)
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$$\displaystyle -cos(x)=2x-cos(x)-2x $$
 * }
 * }

therefore


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$$ 0=0 $$
 * }
 * }

Since both Eq (1-2) and Eq. (1-3) are satisfied, F= EXACT

Part b

We will first identify the different components of a L2_ODE in symbolic form


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$$\displaystyle {\phi_{p}}=cos(x) $$ $$
 * $$\displaystyle (Eq. 1-6)
 * }
 * }

integrating Eq. (1-6)


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$$\displaystyle \phi=\int{\phi_{p}}dp=P(x)p+C(x,y) $$ $$
 * $$\displaystyle (Eq. 1-7)
 * }
 * }

Differentiating $$\phi$$ with respect to $$x$$
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$$\displaystyle \phi_{x}=P'(x)p+\frac{\delta(C)}{\delta(x)} $$ $$
 * $$\displaystyle (Eq. 1-8)
 * }
 * }

and with respect to $$y$$


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$$\displaystyle \phi_{y}=\frac{\delta(C)}{\delta(y)} $$ $$
 * $$\displaystyle (Eq. 1-9)
 * }
 * }

Now we need to connect the symbols with Eq. (1-1)


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$$\displaystyle P(x)=cos(x)=\phi_p $$ $$
 * $$\displaystyle (Eq. 1-10)
 * }
 * }


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$$\displaystyle Q(x)=P'(x)-\frac{\delta(C)}{\delta(y)}=x^2-sin(x) $$ $$
 * $$\displaystyle (Eq. 1-11)
 * }
 * }

and the last term


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$$\displaystyle \frac{\delta(C)}{\delta(x)}=R(x)y=(2x)y $$ $$
 * $$\displaystyle (Eq. 1-12)
 * }
 * }

We will now proceed to calculate $$C$$


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$$\displaystyle C(x,y)=y\int{2\xi}d\xi+K(y)=yx^2+k(y) $$ $$
 * $$\displaystyle (Eq. 1-13)
 * }
 * }


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$$\displaystyle \phi=cos(x)p+x^2y+k(y) $$ $$
 * $$\displaystyle (Eq. 1-14)
 * }
 * }

But in ordet to get a complete solution we need to identify $$k(y)$$


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$$\displaystyle P_x(x)+C_y(x,y)=Q(x) $$
 * }
 * }

Rearranging


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$$\displaystyle C_y(x,y)=Q(x)-P_x(x) $$ which is only a function of $$x$$, that leads us to the conclusion that $$k=$$constant. Which will yield the final form of $$\phi$$
 * }
 * }

$$\displaystyle \phi(x,y,p)=cos(x)p+x^2y+k $$ $$
 * $$\displaystyle (Eq. 1-15)
 * }
 * }

Part c

Let's assume $$k=0$$ the


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$$\displaystyle \phi=cos(x)\frac{dy}{dx}+x^2y=0 $$ $$  Rearranging Eq. (1-16)
 * $$\displaystyle (Eq. 1-16)
 * }
 * }


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$$\displaystyle cos(x)y'=-x^2y $$
 * }
 * }


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$$\displaystyle \frac{y'}{y}=-\frac{x^2}{cos(x)} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1-17)
 * }
 * }

Integrating Eq. (1-17)


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$$\displaystyle ln(y)=\int{\frac{-x^2}{cos(x)}}dx+C_1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1-18)
 * }
 * }

Solving for $$y$$


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$$\displaystyle y(x)=C_2e^{-\int{\frac{-x^2}{cos(x)}}dx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1-19)
 * }
 * }

Egm6321.f10.team4.petralanda.n 23:09, 16 October 2010 (UTC)egm6321.f10.team4.petralanda.n Egm6321.f10.team4.ejm 17:20, 20 October 2010 (UTC) - Editor

= Problem 2 : Bessel Equation(L2_ODE_VC) = From Meeting 13, p 24-1

Find
Part 1) Is the given Bessel Equation exact? where,
 * First condition of Exactness:


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$$ \begin{align} F(x,y,p) = f(x,y,p)p' + g(x,y,p)=0 \end{align} $$     (Eq 2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * Second conditions of Exactness:


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$$ \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$     (Eq 2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \end{align} $$     (Eq 2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 2) If the Bessel Equation is not exact, find an integrating factor $$\displaystyle h(x,y)$$, if one exists, to make it exact.

Solution
PART 1: Exactness of Bessel Equation

 Applying the first exactness condition 

The first condition of exactness for N2-ODE is showing whether the given equation has the following form,


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$$ \begin{align} F(x,y,p) = f(x,y,p)p' + g(x,y,p)=0 \end{align} $$ (Eq 2-2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * where, $$\displaystyle p:=y' $$ and $$\displaystyle p':=y'' $$

above bessel equation can be transformed as follow,


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$$ \begin{align} f(x,y,p)={{x}^{2}} \end{align} $$ (Eq 2-5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$\begin{align} g(x,y,p)=x{p}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y \end{align}$$ (Eq 2-6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Hence, the first condition of exactness is satisfied.

 Applying the second exactness condition 

Taking derivatives of f(x) yields,


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\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
 * }.
 * }.

Then, taking derivatives of g(x) yields,


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\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 derivatives into the second exactness conditions(the first and the second equation) yields,


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$$ \begin{align} \left( 2 \right)+ \left(2p \times 0 \right) + \left(p^2 \times 0 \right) = \left(1\right) + \left(p \times 0 \right)- \left( x^2 - {{\upsilon }^{2}} \right) \end{align} $$     (Eq 2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} 2 \neq 1-{x}^{2}+{\upsilon }^{2} \end{align} $$     (Eq 2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * the first equation of the second exactness condition isn't satisfied.


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$$ \begin{align} \left( 0 \right) + \left( p \times 0 \right) + \left( 2 \times 0 \right) = 0 \end{align} $$     (Eq 2.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} 0=0 \end{align} $$     (Eq 2.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * the second equation of the second exactness condition is satisfied.

Even if the second equation of the second exactness is satisfied, the first equation is not satisfied.

Thus, the Bessel equation(Eq.2.1) is not Exact(N2_ODE).

Alternatively, the exactness (or lack thereof) can be determined using the single relation


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

$$ \begin{align} f_0 +  \frac {df_1}{dx}  +  \frac {d^2 f_2}{dx^2}  =  0 \end{align} $$     (Eq 2.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting the values $$\displaystyle p = y' $$  and  $$\displaystyle  q = y'' $$, our original equation becomes


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$$ \begin{align} F(x,y,p,q) =  x^2 q + xp + (x^2 - \upsilon^2)y  =  0 \end{align} $$     (Eq 2.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where


 * $$\displaystyle

\begin{array}{*{20}l} f(x,y,p) = x^2, & & & g(x,y,p) = xp + (x^2 - \upsilon^2)y\\ \end{array} $$

Taking the appropriate derivatives


 * $$\displaystyle

\begin{array}{*{20}l} {f_y = 0} & {f_p = 0}\\ {g_y = (x^2 - \upsilon^2)} & {g_p = x}\\ \end{array} $$

and using these derivatives to determine expressions for $$\displaystyle f_0 $$, $$\displaystyle f_1 $$ , and $$\displaystyle f_2 $$


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$$ \begin{align} f_0 =  \frac {dF}{dy^{(0)}}  =  f_y q + g_y  =  0 + (x^2 - \upsilon^2) \end{align} $$     (Eq 2.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} f_1 =  \frac {dF}{dy^{(1)}}  =  f_p q + g_p  =  0 + x \end{align} $$     (Eq 2.14)
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 * <p style="text-align:right">
 * }


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$$ \begin{align} f_2 =  \frac {dF}{dy^{(2)}}  =  f  =  x^2 \end{align} $$     (Eq 2.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using our values for $$\displaystyle f_0 $$, $$\displaystyle f_1 $$ ,and $$\displaystyle f_2 $$ , in the exactness relation give


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$$ \begin{align} x^2 - \upsilon^2 - 1 + 2 = 0 \end{align} $$     (Eq 2.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} x^2 - \upsilon^2 + 1 \neq 0 \end{align} $$     (Eq 2.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the Bessel Equation fails this exactness condition.

PART 2: Finding an Integrating Factor

Since our equation is in a power form, the integrating factor $$\displaystyle h(x,y) $$ should be in the form


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$$ \begin{align} h(x,y) = x^m y^n \end{align} $$     (Eq 2.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the second equation in the second exactness condition, our integrating factor is multiplied by each of our functions $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$


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

$$ \begin{align} h(x,y) f(x,y,p) = (x^m y^n) * x^2 = x^{m + 2} y^n \end{align} $$     (Eq 2.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$ \begin{align} h(x,y) g(x,y,p) = (x^m y^n) * ((x^2 - \upsilon^2)y + xp) = x^{m + 2} y^{n + 1} - x^m y^{n + 1} \upsilon^2 + x^{m + 1} y^n p\end{align} $$     (Eq 2.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * $$\displaystyle

\begin{array}{*{20}l} {f_x = (m + 2)x^{m + 1} y^n} & {f_{xp} = 0}\\ {f_y = n x^{m + 2} y^{n - 1}}  & {f_{yp} = 0}\\ {g_p = x^{m + 1} y^n} & {g_{pp} = 0}\\ \end{array} $$

Substituting these values into the second equation for the exactness condition yields


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$$ \begin{align} 0 + p(0) + 2n x^{m + 1} y^n = 0 \end{align} $$     (Eq 2.20)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }


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$$ \begin{align} 2n x^{m + 1} y^n = 0 \end{align} $$     (Eq 2.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

In order for this to be true,


 * $$\displaystyle n = 0 $$

Our integrating factor is now


 * $$\displaystyle x^m y^0 = x^m $$

Using this in combination with the first equation for the second exactness condition


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

$$ \begin{align} h(x,y) f(x,y,p) = x^m * x^2 = x^{m + 2} \end{align} $$     (Eq 2.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} h(x,y) g(x,y,p) = x^m * ((x^2 - \upsilon^2)y + xp = x^{m + 2} y - x^m y \upsilon^2 + x^{m + 1} p \end{align} $$     (Eq 2.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Finding the derivatives of these two equations


 * $$\displaystyle

\begin{array}{*{20}l} {f_x = (m + 2)x^{m + 1} } & {f_{xx} = (m + 2)(m + 1) x^m }\\ & {f_{xy} = 0}\\ {f_y = 0}  & {f_{yy} = 0}\\ {g_x = (m + 2)x^{m + 1} y - m x^{m - 1} y \upsilon^2 + (m + 1)x^m p} & {g_{xp} = (m + 1)x^m}\\ {g_y = x^{m + 2} - \upsilon^2 x^m } & {g_{yp} = 0} \end{array} $$

Substituting these values into our equation yields


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$$ \begin{align} (m + 2)(m + 1)x^m + 2p (0) + p^2 (0) = (m + 1)x^m + p (0) - x^{m + 2} + \upsilon^2 x^m\end{align} $$     (Eq 2.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} (m + 2)(m + 1) = (m + 1) - x^2 + \upsilon^2 \end{align} $$     (Eq 2.25)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} m^2 + 3m + 2 = m + 1 - x^2 + \upsilon^2 \end{align} $$     (Eq 2.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} m^2 + 2m + 1 = \upsilon^2 - x^2 \end{align} $$     (Eq 2.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} m + 1 = \pm \sqrt{\upsilon^2 - x^2} \end{align} $$     (Eq 2.28)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \begin{align} m = \pm \sqrt{\upsilon^2 - x^2} - 1 \end{align} $$     (Eq 2.29)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since $$\displaystyle m $$ must be a real number, and not a function, then there is no integrating factor which will make the Bessel Equation exact.

128.227.57.129 21:25, 19 October 2010 (UTC) EGM6321.f10.team4.osentowski - co-author &  Editor

Egm6321.f10.team4.Yoon 09:09, 20 October 2010 (UTC) - Co-Author & Editor

= Problem 3 -  Euler-Bernoulli Beam Equation =

From Meeting 25, p 25-2.

Given
The equation of motion of a Euler-Bernoulli Beam from Meeting 24, p 24-2:


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

$$\displaystyle -EI\frac{\partial^4u }{\partial x^4}+f=m\frac{\partial^2u }{\partial t^2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-1)
 * }
 * }

Find

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

$$\displaystyle X(x) $$
 * }
 * }

In terms of


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

$$\displaystyle \sin, \cos, \sinh, \cosh $$
 * }
 * }

Solution
In order to solve for $$\displaystyle X(x) $$ in terms of $$\displaystyle \sin, \cos, \sinh, \cosh $$, we must make a series of assumptions.

First, let's assume the beam is experiencing free vibration. This would mean the distributive load, $$\displaystyle f $$, in equation 3-1 would be zero.

Thus,


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

$$\displaystyle -EI\frac{\partial^4u }{\partial x^4}=m\frac{\partial^2u }{\partial t^2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-2)
 * }
 * }

If we go on to assume the form of $$\displaystyle u $$ to be


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

$$\displaystyle u(x,t)=X(x)T(t) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-3)
 * }
 * }

then substituting equation 3-3 into equation 3-2 and using the method of separation of variables yields:


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

$$\displaystyle -EIX^{(4)}T=mX\ddot{T} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-4)
 * }
 * }

Next let's assume $$\displaystyle T $$ to be in the form:


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

$$\displaystyle T(t)=e^{iwt} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-5)
 * }
 * }

Taking the the derivatives of equation 3-5 would result in


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$$\displaystyle \ddot{T}=(iw)^2e^{iwt}=-w^2T $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-6)
 * }
 * }

Substituting equation 3-6 into equation 3-4:


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$$\displaystyle -EIX^{(4)}T=-mw^2XT $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-7)
 * }
 * }

Equation 3-7 can be further reduced to:


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

$$\displaystyle \frac{EI}{mw^2}X^{(4)}=X $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-8)
 * }
 * }

Next we can set the a variable $$\displaystyle K $$ to be to help simply equation 3-8


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

$$\displaystyle K^4:=\frac{EI}{mw^2} $$
 * }
 * }
 * }

equation 3-8 can be rewritten as


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

$$\displaystyle K^4X^{(4)}=X $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-9)
 * }
 * }

Note: equation 3-9 is a linear, 4th order ordinary differential equation with constant coefficients (L4_ODE_CC)

Using Euler's method of trial solution, one can assume


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

$$\displaystyle X(x)=e^{rx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-10)
 * }
 * }

Taking the the derivatives of equation 3-10 would result in


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

$$\displaystyle X^{(4)}=r^4e^{rx}=r^4X $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-11)
 * }
 * }

Substituting equation 3-11 into equation 3-9:


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

$$\displaystyle K^4r^4X=X $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-12)
 * }
 * }

Equation 3-12 can be further reduced to:


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

$$\displaystyle r^4=\frac{1}{K^4} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-13)
 * }
 * }

The 4 roots of $$\displaystyle r $$ are


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

$$\displaystyle r=\frac{1}{K},\frac{-1}{K},\frac{i}{K},\frac{-i}{K} $$ where
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle i=\sqrt{-1} $$
 * }
 * }

$$\displaystyle X $$ can now be written as


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

$$\displaystyle X(x)=C_1 e^{\frac{1}{K}x}+C_2 e^{\frac{-1}{K}x}+C_3 e^{\frac{i}{K}x}+C_4 e^{\frac{-i}{K}x} $$ $$ where $$\displaystyle C_1, C_2, C_3, $$ and $$\displaystyle C_4 $$ are constants.
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-14)
 * }
 * }

First, let's use Euler's formula to convert the exponential terms with imaginary powers to $$\displaystyle \sin $$ and $$\displaystyle \cos $$ terms:


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

$$\displaystyle e^{ix} = \cos x + i\sin x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-15)
 * }
 * }

Equation 3-14 can now be written as:


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

$$\displaystyle X(x)=C_1e^{\frac{1}{K}x}+C_2e^{\frac{-1}{K}x}+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(-\frac{1}{K}x\right)} + i\sin {\left(-\frac{1}{K}x\right)}\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-16)
 * }
 * }

Using Trigonometric Symmetry Identities, Equation 3-16 can be rewritten as:


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

$$\displaystyle X(x)=C_1e^{\frac{1}{K}x}+C_2e^{\frac{-1}{K}x}+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(\frac{1}{K}x\right)} - i\sin {\left(\frac{1}{K}x\right)}\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-17)
 * }
 * }

The remaining exponential terms can be converted to $$\displaystyle \sinh $$ and $$\displaystyle \cosh $$ terms using properties of Hyperbolic functions:


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

$$\displaystyle e^x = \cosh x + \sinh x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-18)
 * }
 * }

and


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

$$\displaystyle e^{-x} = \cosh x - \sinh x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-19)
 * }
 * }

Thus Equation 3-17 can be written as:


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

$$\displaystyle X(x)=C_1 \left[\cosh {\left(\frac{1}{K}x\right)} + \sinh {\left(\frac{1}{K}x\right)}\right]+C_2\left[\cosh {\left(\frac{1}{K}x\right)} - \sinh {\left(\frac{1}{K}x\right)}\right]+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-20)
 * }
 * }

Finally, combing the like trigonometric terms, we can write:

$$\displaystyle X(x)=\left(C_1+C_2 \right) \cosh \left(\frac{1}{K}x\right) + \left(C_1-C_2 \right) \sinh \left(\frac{1}{K}x\right)+ \left(C_3+C_4 \right) \cos \left(\frac{1}{K}x\right)+\left(C_3-C_4 \right) i\sin \left(\frac{1}{K}x\right) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3-21)
 * }
 * }

Egm6321.f10.team4.Auerbach 04:27, 20 October 2010 (UTC) - Primary Author

Egm6321.f10.team4.ejm 16:45, 20 October 2010 (UTC) - Editor