User:Egm6321.f10.team5.abc/f09.h1

= Problem 1 - Maglev First and Second Derivative Derivations =

Given
The equation of motion for a Maglev train can be modeled by the function :

 Nice contribution Maglev train. Egm6321.f09 12:47, 22 September 2009 (UTC)


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f(Y^1 (t),t) $$
 * $$\displaystyle
 * $$\displaystyle


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Find
Derive the first and second time derivatives for the given equation shown by Equation 1 and 2 respectively.


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\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)


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\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 2)


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Solution
 Solving for Equation 1

Taking the time derivative:

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

For ease, a dummy variable
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s = Y^1(t) $$
 * $$\displaystyle
 * $$\displaystyle
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is used and substuted into the previous equation. The chain rule is now applied:
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\frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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 * }

Note that
 * $$\frac{\partial s}{\partial t}=\dot Y^1(t)$$

Equation 3 can be rewriten to the form:


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$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$
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Solving for Equation 2

Note that

\frac{d^2 }{dt^2} f= \frac{d}{dt}\left(\frac{d}{dt}f\right) $$ Substituting in Eq. 1 yields

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = \frac{d}{dt}\left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \right) $$ Now, we can apply the chain rule so that

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \frac{\partial t}{\partial t} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\frac{\partial t}{\partial t} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \frac{\partial t}{\partial t} $$ Collecting terms in the above equation

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + 2\frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2} +  \frac{\partial^2 f}{\partial t^2} $$ To compare with the given notation, the derivates in the equations above are replaced with subscripts. For example, dX/dy would be simply X,y. The following equation shows this transformation.

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\frac{\partial s}{\partial t}\right)^2 + 2f_{,st} \frac{\partial s}{\partial t} + f_{,s}\frac{\partial^2 s}{\partial t^2} + f_{,tt} $$ Finally, the dummy variable is replaced and the desired solution remains.
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$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
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= Problem 2 - Method of Integrating Factor =

Given
A Linear 1st order ODE is given by


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$$
 * $$\displaystyle y' + y = x$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)


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Find
Solve Eq. 4 for y(x) and show
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 * $$\displaystyle y(x) = A e^{-x} + x - x$$
 * $$\displaystyle y(x) = A e^{-x} + x - x$$


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Solution
To find the solution, we need to perform the exactness test which requires two steps:

1. Writing the ODE in the form of
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 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$

$$
 * $$\displaystyle (Eq. 5)
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2. Checking the condition given by
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 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

$$
 * $$\displaystyle (Eq. 6)
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 * }.

Checking Condition 1

Re-writing Eq. 4 in the form of Eq. 5 gives


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 * $$\displaystyle (y-x)+ y' = 0$$
 * $$\displaystyle (y-x)+ y' = 0$$


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where $$\displaystyle M=(y-x)$$ and $$\displaystyle N=1$$. It is shown that condition 1 is met.

Checking Condition 2

Now, we have to test if $$\displaystyle M_y=N_x$$


 * $$\displaystyle M_y = 1$$ and $$\displaystyle N_x = 0$$

Therefore, Eq. 4 is not an exact differential equation and an integrating factor is needed to force the ODE to be exact. The definition of the integrating factor is


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 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$


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where


 * $$\displaystyle f(x)=\frac{1}{N}\left[ M_p - N_x \right ]$$

Substiuting the values of M and N yield


 * $$\displaystyle f(x)== \frac{1}{1}\left[ 1 - 0 \right ] = 1 $$

Therefore the integrating factor is
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 * $$\displaystyle h(x)= e^x$$
 * $$\displaystyle h(x)= e^x$$


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Multiplying Eq. 4 by the integrating factor gives


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 * $$\displaystyle e^x y + e^x y' = xe^x$$
 * $$\displaystyle e^x y + e^x y' = xe^x$$


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Now it can be seen that the left hand side is in the form of a derivative product. Recognizing this yields:


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 * $$\displaystyle (e^xy)' = xe^x$$
 * $$\displaystyle (e^xy)' = xe^x$$


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Integrating both sides results


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 * $$\displaystyle (e^xy) = e^x(x-1)+A$$
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$

Finally, dividing by e^x results in
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$$\displaystyle y = Ae^{-x}+(x-1)$$
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Problem #3: Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution

Define a linear operator $$D\left ( \cdot \right )$$ such that $$D(\cdot )=2x^2+\sqrt{\left (\cdot  \right ) }+x^5\left (\cdot  \right ) ^3\frac{d}{dx}\left ( \cdot  \right )$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition gives:

$$D(\alpha u + \beta v)=2x^2+\sqrt{\alpha u + \beta v}+x^5\left ( \alpha u + \beta v \right )^3\frac{d}{dx}\left ( \alpha u+\beta v \right )$$

Substituting for the right side of the linearity condition:

$$\alpha D(u)+\beta D(v)=\alpha\left (2x^2+\sqrt{u}+ x^5u^3\frac{du}{dx} \right )+\beta\left ( 2x^2+\sqrt{v}+x^5v^3\frac{dv}{dx} \right )$$

Simplifying and comparing both sides of the linearity condition:

$$2x^2+\sqrt{\alpha u+\beta v}+\alpha x^5(\alpha u+ \beta v)^3\frac{du}{dx}+\beta x^5(\alpha u + \beta v)^3\frac{dv}{dx}\neq2x^2\alpha+\alpha\sqrt{u}+\alpha x^5u^3\frac{du}{dx}+2x^2\beta+\beta\sqrt{v}+\beta x^5v^3\frac{dv}{dx}$$

The condition is therefore not met, and $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is a nonlinear 1st Order ODE. 

Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

Problem 4:
Show that $$F(x,y,y') = 0$$ is a nonlinear, 1st-order ordinary differential equation, where $$F(x,y,y') = x^2 y^5 + 6(y')^2=0$$.

Order of the ODE:

Since, $$F(x,y,y') = x^2 y^5 + 6(y')^2=0$$, the highest order is $$y'$$, the function is a 1st-order ordinary differential equation.

Proof of Nonlinearity:

First, we define the differential operator, $$D(\cdot)$$ as:

$$D(\cdot)=x^2 (\cdot)^5 + 6\left(\frac{d(\cdot)}{dx}\right)^2 = 0$$

Where $$u$$ and $$v$$ are functions of $$x$$, and $$\alpha$$ and $$\beta$$ are scalars. Next, we define the equation $$D(\alpha u + \beta v)$$ as:

$$D(\alpha u + \beta v) = x^2(\alpha u + \beta v)^5 + 6\left[\alpha\frac{du}{dx} + \beta\frac{dv}{dx}\right]^2$$

We then solve for $$\alpha D(u) + \beta D(u)$$, which is given as

$$\alpha D(u) + \beta D(v) = \alpha\left[x^2 u^5 +6\left(\frac{du}{dx}\right)^2 \right] + \beta \left[x^2 v^5 + 6 \left(\frac{dv}{dx}\right)^2\right]$$

Since, $$x^2(\alpha u + \beta v)^5 + 6\left[\alpha\frac{du}{dx} + \beta\frac{dv}{dx}\right]^2 \neq \alpha\left[x^2 u^5 +6\left(\frac{du}{dx}\right)^2 \right] + \beta \left[x^2 v^5 + 6 \left(\frac{dv}{dx}\right)^2\right]$$

It follows that

$$D(\alpha u + \beta v) \neq \alpha D(u) + \beta D(v)$$

In order for $$F(x,y,y') = 0$$ to be linear, the equation $$D(\alpha u + \beta v) = \alpha D(u) + \beta D(u)$$  $$ \forall u, v,$$   and    $$\forall \alpha, \beta \in \Re$$. This is known as the superposition principle and it must be satisfied in order for a differential equation to be linear.

Since the superposition principle is not satisfied, the function is nonlinear.

 Good solution.--Egm6321.f09.TA 03:41, 24 September 2009 (UTC)

Problem #5:

Part 1. Create an exact nonlinear 1st Order ODE of the form $$\Phi_x\left(x,y\right)+\Phi_y\left(x,y\right)y^{\prime}=0$$ using the equation $$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Solution:

If,

$$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Then

$$\Phi_{x}\left(x,y\right)=24x^{3}=M$$

and

$$\Phi_{y}\left(x,y\right)=3y^{1/2}=N$$

For a nonlinear, 1st order, ODE:

$$M + Ny^{\prime}=0$$

Therefore,

$$24x^3 + 3y^{1/2}y^{\prime}=0$$

We have forced exactness

Part 2. Create three more exact nonlinear 1st Order ODEs by inventing new $$\Phi\left(x,y\right)$$ functions.

Ex. 1

Choose $$\Phi\left(x,y\right)=x^{1/2}+y^{3}\quad\text{:}$$

$$ \begin{align} \Phi_{x}\left(x,y\right)&=0.5x^{-1/2}=M\left(x,y\right) \\ \Phi_{y}\left(x,y\right)&=3y^{2}=N\left(x,y\right) \end{align} $$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = \frac{1}{2}x^{-1/2}+3y^{2}y' = 0$$

Ex. 2

Choose $$\Phi\left(x,y\right)=3y^2+2x^5\quad\text{:}$$

$$\Phi_x\left(x,y\right)=10x^4=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=6y=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 10x^4+(6y)y' = 0$$

Ex. 3

Choose $$\Phi\left(x,y\right)=3x^2y^3\quad\text{:}$$

$$\Phi_x\left(x,y\right)=6xy^3=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=9x^2y^2=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 6xy^3+\left(9x^{2}y^{2}\right)y' = 0$$ 

Nice work.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)