User:Egm6321.f09.team3.Kuljevan.Francisco/HW1

=Homework #1 PEA1 Fall 2009=

Problem #1
Derive the 1st and 2nd order total time derivative for the function $$f$$ related to the Maglev, transrapid trains

Equation (1)
This equation is obtain as follows:

\dfrac{df}{dt} = \dfrac{\partial f}{\partial s} \dfrac{ds}{dt} + \dfrac{\partial f}{\partial t} \dfrac{dt}{dt} $$

With more details shown below:



\begin{align} \dfrac{d}{dt}\left[f\left(Y^{1}(t),t\right)\right] &= \dfrac{\partial}{\partial s}\left[f\left(Y^{1}(t),t\right)\right] \dfrac{d}{dt}\left[Y^{1}(t)\right] + \dfrac{\partial}{\partial t}\left[f\left(Y^{1}(t),t\right)\right]\\ &=\dfrac{\partial}{\partial s}\left[f\left(Y^{1}(t),t\right)\right] \dfrac{d}{dt}\left[Y^{1}(t)\right] + \dfrac{\partial} {\partial t}\left[f\left(Y^{1}(t),t\right)\right]\\ \end{align} $$

Equation (2):
Analogously we can follow the same procedure to obtain the $$2{nd}$$ order total derivative of $$f$$



\begin{align} \dfrac{d^2}{dt^2}\left[f\left(Y^{1}(t),t\right)\right] &= \dfrac{d}{dt}\left[\dfrac{\partial}{\partial s}\left(f\left(Y^{1}(t),t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) + \dfrac{\partial}{\partial s} \left(f\left(Y^{1}(t),t\right)\right)\right]\\ &=\dfrac{\partial}{\partial s} \left[(\dfrac{\partial}{\partial s} \left(f\left(Y^{1}(t),t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) +\dfrac{\partial}{\partial t}\left(f\left(Y^{1}(t),t\right)\right)\right] \dfrac{d}{dt}\left(Y^{1}(t)\right)+ \dfrac{\partial}{\partial t} \left[\dfrac{\partial}{\partial s}\left(f\left(Y^{1}(t),t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) + \dfrac{\partial}{\partial t}\left(f\left(Y^{1}(t),t\right)\right)\right]\\ &=\dfrac{\partial^2}{\partial s^2}\left(f(Y^{1},t)\right) \left[\dfrac{d}{dt}\left(Y^{1}(t)\right)\right]^2 + \dfrac{\partial^2}{\partial s \partial t}\left(f\left(Y^{1},t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) + \dfrac{\partial^2}{\partial s \partial t}\left(f\left(Y^{1},t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) + \dfrac{\partial^2}{\partial t^2}\left(f(Y^{1},t)\right) + \dfrac{\partial}{\partial s}\left(f(Y^{1},t)\right) \dfrac{d}{dt}\left[\dfrac{d}{dt}\left(Y^{1}(t)\right)\right]\\ &=\dfrac{\partial^2}{\partial s^2}\left(f(Y^{1},t)\right)\left[\dfrac{d}{dt}\left(Y^{1}(t)\right)\right]^2 +2\cdot \dfrac{\partial^2}{\partial s \partial t}\left(f\left(Y^{1},t\right)\right) \dfrac{d}{dt}\left(Y^{1}(t)\right) + \dfrac{\partial^2}{\partial t^2}\left(f(Y^{1},t)\right) + \dfrac{\partial}{\partial s}\left(f(Y^{1},t)\right) + \dfrac{d^{2}}{dt^{2}}\left(Y^{1}(t)\right) \end{align} $$

Problem #2
Derive the following equation using the integration factor knowledge:


 * $$ y(x) = \int^{x}q(z)dz = -Aexp^{-x} + \frac{x^{2}}{2} - x +B$$

To illustrate the use of integrating factor the following problem was posed:


 * $$y'' + y' = x\!$$

Let
 * $$ p(x) = y'(x)\! $$

which gives the necessary form to utilize the integrating factor method. The standard form is :


 * $$a_{1}p' + a_{0}p = b(x)\!$$

therefore our case yields the values $$ a_{0} = 1$$, $$b(x) = x$$. From this result we can say that



\begin{align} N(N_{x} - M_{y}) &=-f(x)\\ N(N_{x} - M_{p}) &=-f(x)\\ 1(0 - a_{0}) & = -f(x)\\ -a_{0} &= -f(x) \end{align} $$

The integration factor then becomes:


 * $$ h(x)=exp\int f(s)ds = exp(x)$$

Using the equation derived in class:


 * $$p(x) =\dfrac{1}{h(x)}\int h(s)b(s)ds$$

Now plugging the expression for $$h(x)$$ and $$b(s)$$ in the equation above



\begin{align} p(x) &=\dfrac{1}{exp(x)}\int exp(s)sds\\ &=\dfrac{1}{exp(x)}\left[exp(x) x - exp(x) +A\right]\\ &=x-1 + A \cdot exp(-x) \end{align} $$

Integrating once we can obtain $$ y(x)$$:



\begin{align} y(x) &= \int p(x)\\ & = \int \left[A \cdot exp(-s) + s -1\right]ds\\ &= -A \cdot exp(-x) + \frac{x^{2}}{2} -x + B \end{align} $$

Therefore we obtain the equation derived on class:


 * $$y(x)=-A \cdot exp(-x) + \frac{x^{2}}{2} -x + B$$

Problem #3
Show that the following $$1^{st}$$ order ODE is non-linear:


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

To identify the above expression is non-linear we will test the linearity with the differential operator. The differential operator for the above equation is the follwoing:


 * $$D(\cdot) = \left(2x^{2} +\sqrt{(\cdot)}\right) + \left(x^{5}(\cdot)^{3}\right)\dfrac{d(\cdot)}{dx}$$

For the ODE to be linear it must follow the following condition:


 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Let's find each of the conditions stated above:



\begin{align} D\left(\alpha u + \beta v \right) &= \left[2x^{2} + \sqrt{\left(\alpha u + \beta v \right)}\right] + \left[x^{5}\left(\alpha u + \beta v \right)^{3}\right] \dfrac {d}{dx}\left(\alpha u + \beta v \right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left[2x^{2} + \sqrt{\left(\alpha u \right)}\right] + \left[x^{5}\left(\alpha u \right)^{3}\right] \dfrac{d(\alpha u )} {dx} + \left[2x^{2} + \sqrt{\left(\beta v\right)}\right] + \left[x^{5}\left(\beta v\right)^{3}\right] \dfrac {d(\beta v)} {dx} \end{align} $$

Since the above conditions do not equal to each other, then we conclude that:


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

is non-linear.

Problem #4
Show that the following ODE is $$1^{st}$$ order and non-linear:


 * $$ x^{2}y^{5} + 6(y')^{2} = 0\!$$

To identify the above expression is non-linear we will test the linearity with the differential operator. The differential operator for the above equation is the follwoing:


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

For the ODE to be linear it must follow the following condition:


 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Let's find each of the conditions stated above:



\begin{align} D\left(\alpha u + \beta v \right) &= x^{2} \left(\alpha u + \beta v \right)^{5} + 6\left[ \dfrac {d}{dx} \left(\alpha u + \beta v \right)\right]^2\\ D\left(\alpha u \right) + D\left(\beta v \right) &= x^{2} \left(\alpha u \right)^{5} + 6\left[ \dfrac {d( \alpha u )} {dx}\right]^{2} + x^{2} (\beta v )^{5} + 6\left[ \frac {d ( \beta v )} {dx}\right]^2 \end{align} $$

Since the above conditions do not equal to each other and that since the highest order of the derivative is first order, then we conclude that:


 * $$ x^{2}y^{5} + 6(y')^{2} = 0\!$$

is a $$1^{st}$$ order and non-linear ODE.

Part (I)
For the following $$\phi(x,y)$$ find $$M$$, $$N$$, and $$F(x,y,y')$$


 * $$ \phi = 6x^4 + 2y^{3/2}\!$$

$$M$$ and $$N$$ represent derivative expressions of $$\phi$$ the equations for their solutions are as follows:


 * $$ M = \phi_x = 24x^3\! $$


 * $$ N = \phi_y = 3y^{1/2}\! $$

With the results above we can find $$ M_y$$ and $$ N_x $$


 * $$ M_y = 0\!$$


 * $$ N_x = 0\!$$

Since, $$ M_y = N_x $$ it proves that the equation exact.

$$ F(x, y, y') = M + N \cdot y' = 24x^3 + 3y^{/frac{1}{2}} \cdot y' = 0 \! $$

Part (II)
Create 3 exact non-linear $$1^{st}$$ ODEs:

#1



\begin{align} \phi_1 &= 2x^3y^4 + 4x^2y^5 \!\\ M = \phi_x &= 6x^2y^4 + 8xy^5 \!\\ N = \phi_y &= 8x^3y^3 + 20x^2y^4\! \\ M_y &= 24x^2y^3 + 40xy^4\! \\ N_x &= 24x^2y^3 + 40xy^4\! \end{align} $$

therefore the equation is exact!


 * $$ F1(x, y, y') = M + N y' = (6x^2y^4 + 8xy^5) + (8x^3y^3 + 20x^2y^4) y' = 0 \!$$

#2



\begin{align} \phi_2 &= 12x^{\frac{7}{4}}y^4 + 5x^2y^{\frac{9}{2}}\!\\ M = \phi_x &= 21x^{\frac{3}{4}}y^4 + 10xy^{\frac{9}{2}}\!\\ N = \phi_y &= 48x^{\frac{7}{4}}y^3 + 22.5x^2y^{\frac{7}{2}}\!\\ M_y &= 84x^{\frac{3}{4}}y^4 + 45xy^{\frac{7}{2}}\!\\ N_x &= 84x^{\frac{3}{4}}y^3 + 45xy^{\frac{7}{2}}\! \end{align} $$

therefore the equation is exact!


 * $$ F2(x, y, y') = M + N y' = \left(21x^{\frac{3}{4}}y^4 + 10xy^{\frac{9}{2}}\right) + \left(48x^{\frac{7}{4}}y^3 + 22.5x^2y^{\frac{7}{2}}\right) y' = 0 \!$$

#3



\begin{align} \phi_3 &= 4x^{-3}y^7 + 7x^6y^{\frac{1}{2}}\\ M = \phi_x &= -12x^{-4}y^7 + 42x^5y^{\frac{1}{2}} \\ N = \phi_y &= 28x^{-3}y^6 + 3.5x^6y^{\frac{-1}{2}} \\ M_y &= -84x^{-4}y^6 + 21x^5y^{\frac{-1}{2}}\\ N_x &= -84x^{-4}y^6 + 21x^5y^{\frac{-1}{2}} \end{align} $$

therefore the equation is exact!


 * $$ F3(x, y, y') = M + N y' = \left(-12x^{-4}y^7 + 42x^5y^{\frac{1}{2}}\right) + \left(28x^{-3}y^6 + 3.5x^6y^{\frac{-1}{2}}\right) y' = 0 $$