User:Egm6321.f09.Team1/HW1

 See my comments or TA's comments below. Nice organization of wiki page. Egm6321.f09 20:49, 19 September 2009 (UTC)

= 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)


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

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


 * }
 * }

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)


 * }
 * }


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

\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)


 * }
 * }

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

s = Y^1(t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

is used and substuted into the previous equation. The chain rule is now applied:
 * {| style="width:100%" border="0" align="left"



\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)
 * }
 * }

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

Equation 3 can be rewriten to the form:


 * {| style="width:60%" border="0"

$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

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.
 * {| style="width:60%" border="0"

$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

= Problem 2 - Method of Integrating Factor =

Given
A Linear 1st order ODE is given by


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

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


 * }
 * }

Find
Solve Eq. 4 for y(x) and show
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y(x) = A e^{-x} + x - x$$
 * $$\displaystyle y(x) = A e^{-x} + x - x$$


 * }
 * }

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

1. Writing the ODE in the form of
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$

$$
 * $$\displaystyle (Eq. 5)
 * style= |
 * }

2. Checking the condition given by
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 * style="width:95%" |
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

$$
 * $$\displaystyle (Eq. 6)
 * style= |
 * }.

Checking Condition 1

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


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


 * style="width:100%" |
 * $$\displaystyle (y-x)+ y' = 0$$
 * $$\displaystyle (y-x)+ y' = 0$$


 * style= |
 * }

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


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


 * style="width:95%" |
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$


 * style= |
 * }

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


 * style="width:95%" |
 * $$\displaystyle h(x)= e^x$$
 * $$\displaystyle h(x)= e^x$$


 * style= |
 * }

Multiplying Eq. 4 by the integrating factor gives


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


 * style="width:95%" |
 * $$\displaystyle e^x y + e^x y' = xe^x$$
 * $$\displaystyle e^x y + e^x y' = xe^x$$


 * style= |
 * }

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


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


 * style="width:95%" |
 * $$\displaystyle (e^xy)' = xe^x$$
 * $$\displaystyle (e^xy)' = xe^x$$


 * style= |
 * }

Integrating both sides results


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


 * style="width:95%" |
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$

Finally, dividing by e^x results in
 * style= |
 * }
 * {| style="width:60%" border="0"

$$\displaystyle y = Ae^{-x}+(x-1)$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

= Problem 3 - Linearity Test of ODEs =

Given


\left( 2 x^2 + \sqrt{y} \right) + x^5 y^3 y' = 0 $$

Find
Show that the following first order ODE is non-linear

Solution
The definition of a linear function is that it obeys the cumulative and additive properties shown respectively. F is an arbitrary function while C is an arbitrary constant.


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


 * style="width:95%" |
 * $$\displaystyle F(Cx) = CF(x)$$
 * $$\displaystyle F(Cx) = CF(x)$$


 * style= |
 * }


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


 * style="width:95%" |
 * $$\displaystyle F(x+y) = F(x) + F(y)$$
 * $$\displaystyle F(x+y) = F(x) + F(y)$$


 * style= |
 * }

The given equation is defined as follows:

H(y)=\left( 2 x^2 + \sqrt{y} \right) + x^5 y^3 y' $$ Applying the cumlitive property
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle H(Cy) = CH(y)$$
 * $$\displaystyle H(Cy) = CH(y)$$

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


 * style="width:95%" |
 * $$\displaystyle H(C y)=\left( 2 x^2 + \sqrt{C} \sqrt{y} \right) + x^5 C^3 y^3 y' $$
 * $$\displaystyle H(C y)=\left( 2 x^2 + \sqrt{C} \sqrt{y} \right) + x^5 C^3 y^3 y' $$


 * style= |
 * }

and


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


 * style="width:95%" |
 * $$\displaystyle C H(y) = C \left [\left( 2 x^2 + \sqrt{y} \right) + x^5 y^3 y'\right]$$
 * $$\displaystyle C H(y) = C \left [\left( 2 x^2 + \sqrt{y} \right) + x^5 y^3 y'\right]$$


 * style= |
 * }

Because these two equations are clearly not equal, this function in nonlinear.

= Problem 4 - Linearity Test of ODEs=

Given

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


 * style="width:95%" |
 * $$\displaystyle G(x,y,y') = x^2 y^5 + 6 (y')^2 = 0 $$
 * $$\displaystyle G(x,y,y') = x^2 y^5 + 6 (y')^2 = 0 $$


 * style= |
 * }

Find
Show that $$\displaystyle G(x,y,y') = 0$$ is a non-linear 1st order ODE.

Solution
The ODE is first order, since the highest order derivative of $$y$$ in the equation is 1st order.

To test for linearity, the same requirements have to be meet as in Problem 3. The differential operator for the ODE is
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle G(y) = x^2 y^5 + 6 (y')^2$$
 * $$\displaystyle G(y) = x^2 y^5 + 6 (y')^2$$


 * style= |
 * }

Unlike in Problem 3, the additive condition will be applied, that is for an arbitary function F
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle F(u+v) = F(u) + F(v)$$
 * $$\displaystyle F(u+v) = F(u) + F(v)$$


 * style= |
 * }

Applying this on the given function G results in
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle G(u+v) = x^2 (u+v) ^5 + 6 ((u+v)')^2$$
 * $$\displaystyle G(u+v) = x^2 (u+v) ^5 + 6 ((u+v)')^2$$


 * style= |
 * }

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


 * style="width:95%" |
 * $$\displaystyle G(u) + G(v) = x^2 u ^5 + 6 (u')^2 + x^2 v ^5 + 6 (v')^2$$
 * $$\displaystyle G(u) + G(v) = x^2 u ^5 + 6 (u')^2 + x^2 v ^5 + 6 (v')^2$$


 * style= |
 * }

Clearly, the two equations are not linear and the function G is nonlinear.

= Problem 5 - Exactness of a Nonlinear 1st Order ODE =

Given
The following functions:

1) $$\displaystyle \phi(x,y)=6 x^4 + 2 y ^{\frac{3}{2}}$$

2) $$\displaystyle \phi(x,y)=xy + x^2y^2$$

3) $$\displaystyle \phi(x,y)=e^{xy}+x$$

4) $$\displaystyle \phi(x,y)=y \sin(x)$$

Find
Create four exact non-linear ODEs.

Solution
A 1st Order ODE is given by the following equation


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


 * style="width:95%" |
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$


 * style= |
 * }

The conditions under and exact 1st Order ODE are that 1) the equation must have the form of the one given above and 2) that


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


 * style="width:95%" |
 * $$\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$
 * $$\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$


 * style= |
 * }

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


 * style="width:95%" |
 * $$\displaystyle M = \frac{\partial \phi}{\partial y}$$
 * $$\displaystyle M = \frac{\partial \phi}{\partial y}$$

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


 * style="width:95%" |
 * $$\displaystyle N = \frac{\partial \phi}{\partial x}$$
 * $$\displaystyle N = \frac{\partial \phi}{\partial x}$$


 * style= |
 * }

Substituting $$M$$ and $$N$$ for the given case into the given form yields a exact 1st order ODE.

Solution 1
 * $$\displaystyle \phi(x,y)=6 x^4 + 2 y ^{\frac{3}{2}}$$


 * $$\displaystyle M=24 x^3$$


 * $$\displaystyle N=3 y ^{\frac{1}{2}}$$


 * {| style="width:60%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\Rightarrow \displaystyle \left (24 x^3 \right) + \left( 3 y ^{\frac{1}{2}} \right) y'=0$$
 * $$\Rightarrow \displaystyle \left (24 x^3 \right) + \left( 3 y ^{\frac{1}{2}} \right) y'=0$$


 * style= |
 * }

Solution 2


 * $$\displaystyle \phi(x,y)=xy + x^2y^2$$


 * $$\displaystyle M=y + 2xy^2$$


 * $$\displaystyle N=x + 2yx^2$$


 * {| style="width:60%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\Rightarrow \displaystyle \left (y + 2xy^2 \right) + \left (x + 2yx^2 \right) y'=0$$
 * $$\Rightarrow \displaystyle \left (y + 2xy^2 \right) + \left (x + 2yx^2 \right) y'=0$$


 * style= |
 * }

Solution 3


 * $$\displaystyle \phi(x,y)=e^{xy}+x$$


 * $$\displaystyle M=ye^{xy}+1$$


 * $$\displaystyle N=xe^{xy}$$


 * {| style="width:60%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\Rightarrow \displaystyle \left (ye^{xy}+1 \right) + \left (xe^{xy} \right) y'=0$$
 * $$\Rightarrow \displaystyle \left (ye^{xy}+1 \right) + \left (xe^{xy} \right) y'=0$$


 * style= |
 * }

 On the other hand, choosing $$\displaystyle \phi$$ as shown above resulted in a L1-ODE-VC (linear), instead of a N1-ODE (nonlinear) as required. Egm6321.f09 12:43, 22 September 2009 (UTC)

Solution 4


 * $$\displaystyle \phi(x,y)=y \sin(x)$$


 * $$\displaystyle M=y \cos(x)$$


 * $$\displaystyle N=\sin(x)$$


 * {| style="width:60%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\Rightarrow \displaystyle \left [y \cos(x) \right] + \left [\sin(x) \right] y'=0$$
 * $$\Rightarrow \displaystyle \left [y \cos(x) \right] + \left [\sin(x) \right] y'=0$$


 * style= |
 * }

 On the other hand, choosing $$\displaystyle \phi$$ as shown above resulted in a L1-ODE-VC (linear), instead of a N1-ODE (nonlinear) as required. Egm6321.f09 20:49, 19 September 2009 (UTC)

=Contributing Team Members= Egm6321.f09.Team1.vasquez 15:29, 15 September 2009 (UTC)

Egm6321.f09.Team1.andy 17:54, 15 September 2009 (UTC)

Egm6321.f09.Team1.sallstrom 22:52, 15 September 2009 (UTC)

Egm6321.f09.Team1.AH 00:23, 16 September 2009 (UTC)

=Notes and references=