User:EGM6321.f10.team6.cook/hw4

=Solve N2_ODE_VC =

==Given ==

Required

 * 1) Show equation (1.1) is exact
 * 2) Find $$\begin{align}\phi\end{align}$$
 * 3) Solve for $$\begin{align}y(x)\end{align}$$

==Solution   ==

Exactness
To be exact the first and second conditions of exactness must be met.

1st condition of exactness
The 1st condition of exactness is shown to be satisfied if the equation can be written in the form of equation (1.2)

where,

$$\begin{align}&p:=y'\\ &g(x,y,p) := {\phi}_{x}+{\phi}_{y}p\\ &f(x,y,p) := {\phi}_{p}(x,y,p)\end{align}$$

Rearranging equation (1.1) into the form of equation (1.2),

where,

$$\begin{align}g(x,y,p)= ({x}^{2}-\mathrm{sin}x)p+2xy \end{align}$$

$$\begin{align}f(x,y,p)=\mathrm{cos}x\end{align}$$

The given equation thus satisfies the 1st condition of exactness.

2nd condition of exactness
There are two relations for the 2nd condition of exactness.


 * first relation


 * second relation

As both of the exactness conditions are satisfied, we can conclude that the equation 1.1 is exact.

Find $$\begin{align}\phi\end{align}$$
As,

using the definition of $$\begin{align}g(x,y,p)\end{align}$$

Which is equivalent to $$\begin{align}g(x,y,p)\end{align}$$ in equation (1.3)

Assume,

Then,

Taking the derivative of equation (1.13) with respect to $$y$$ to find $$\begin{align}f(y)\end{align}$$

and equating it to $$\begin{align}{h}_{y}\end{align}$$ in equation (1.11)

Then,

Substituting back into equation (1.8)

Consolidating constants,

where, $$\begin{align}k={k}_{2}-{k}_{1}\end{align}$$

The solution can be easily verified by finding the partial derivatives with respect to $$x$$, $$y$$, $$p$$

Using the definition of $$\begin{align}F(x,y,y',y'')\end{align}$$

This is equation (1.1), so we know that equation (1.18) is correct.

Solve for $$\begin{align}y(x)\end{align}$$
To solve the first order nonlinear ordinary differential equation with variable coefficients (N1_ODE_VC) for y(x), equation (1.18) is first rearranged.

Finding the integrating factor $$i(x)$$ first.

then, we can find y(x).

Matlab code for solving this N2-ODE-VC:

Which verifies our solution.

Given
The Bessel Equation,
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$$\begin{align} x^{2} y'' + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.1)
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Required
Verify exactness of Bessel Equation(2.1) using 2 methods; first using equations (1) & (2) from the lecture notes on p. 15_3.

The 2nd condition of exactness

Secondly, using equation (2) from the lecture notes on p. 22_4

where $$\begin{align} f_i = \frac{\partial F}{\partial y^{(i)}} ,i= 0,1,2 \end{align}$$

Finally, if equation (2.1) is not exact, see whether it can be made exact using the Integrating Factor Method with $$\begin{align} h(x,y)=x^{m} y^{n}\end{align}$$; See lecture notes p. 19_1, p.22_4

First Method
From equations (2.1), (2.2)

Substituting equation (2.6) into equation (2.2)

if equation (2.7) is met, equation (2.1) is exact.

From equation (2.3)

Substituting equation (2.8) for equation (2.3)

This satisfies the 2nd condition of exactness.

Second Method

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$$\begin{align} F(x,y,y',y) = x^{2} y + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.10)
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$$\begin{align} &f_0 = \frac{\partial F }{\partial y^{(0)}} = x^2 - {\nu}^2\\ &f_1 = \frac{\partial F }{\partial y^{(1)}} = x\\ &f_2 = \frac{\partial F }{\partial y^{(2)}} = x^2\\ \end{align}$$

(2.11)
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$$\begin{align} \frac{\mathrm{d} f_1 }{\mathrm{d} x} =1 {~}{~}{~}{~}{~} \frac{\mathrm{d^2} f_2 }{\mathrm{d} x^2} =2 \end{align}$$

(2.12)
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Therefore,

Which is the same result as the first method; if equation (2.13) is met, equation (2.1) is exact.

Making the Equation Exact
Verifying the exactness of equation (2.1) using the 2nd condition of exactness, equations (2.2), (2.3)

Equation (2.18) substituted into equation (2.3)

To satisfy equation (2.19), n=0.

Substituting equations (2.16), (2.17) into equation (2.2)

Solve the quadratic equation for m,

Thus, when equation (2.22) is met, the equation is exact.

Problem Statement
The Euler-Bernoulli beam equation, given in Lecture 24 Fall 2010, is as follows:

If we assume that $${{u}_{\left( x,t \right)}}={{X}_{\left( x \right)}}{{T}_{\left( t \right)}}$$, we can apply the technique of separation of variables, which yields the following :

If we further assume that T has the form in equation 3.3, we can rewrite 3.2 as equation 3.4.

After rearranging terms and doing some algebra we obtain:

If we apply the method of undetermined coefficients and assume $${{X}_{\left( x \right)}}=\exp \left( bx \right)$$ we obtain the following two sets of roots (p25_2 Lecture 25 ):

It follows that the solution to equation 3.1 is the linear combination of the above roots:

Required
Express equation 3.7 in terms of sines, cosines, hyperbolic sines, and hyperbolic cosines.

Sine and Cosine Terms
We begin by examining the imaginary terms, to which we will apply Euler’s formula in order to obtain sines and cosines. Euler’s formula is given as :

Since sin(-u)=-sin(u) and cos(-u)=cos(u) we can rewrite 3.9 as follows:

Hyperbolic Sine and Cosine Terms
We now consider the real terms. We apply the definition of hyperbolic functions, however, instead of looking at the terms let us look at the hyperbolic functions themselves.

The hyperbolic sine and cosine are defined as:

If we add 3.11 to 3.12 we obtain the following:

If we let a = 1/k, equation 3.13 becomes the positive real root term:

Considering the identities in equation 3.15 and 3.16 we can add the negative sinh and cosh to obtain equation 3.17:

Now consider the following identities:

If we let a=1/k, equation 3.17 becomes the negative real root term:

Combining 3.14 and 3.18 we obtain:

Combining Terms
If we now combine equation 3.10 with 3.19 we obtain the final form for the solution to equation 3.1: