User:Egm6321.f09.Team6/HW1

Problem 1
Derive the Equation 1 (Lecture 1) 1st order total time derivative of f where f is a function of (Y1(t),t).

Y1(t) is the Nominal motion

f(s,t)

s=X1 = axial co-ordinate

t=time

Evaluating at s=Y1(t), we have f(Y1(t),t)

To find the 1st derivative of function f(Y1(t),t)

$${d\over dt}f(Y^1(t),t)=$$$$\frac{\partial f}{\partial s}+\frac{\partial f}{\partial t}={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)(1)$$

1st Derivative $$=\Bigg[{d\over dt}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)\Bigg]$$

Problem 2
Derive the Equation 2 (Lecture 1) 2nd order total time derivative of function f.

1st Derivative $$=\Bigg[{d\over dt}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t)+\frac{\partial f}{\partial t}(Y^1(t),t)\Bigg]$$

$$={\partial f\over\partial s}(Y^1(t),t)\ddot{Y^1}(t)+\dot{Y^1}(t)\Bigg({\partial \over\partial s}{\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial s}(Y^1(t),t)\Bigg)+{\partial \over\partial s}{\partial f\over\partial t}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial t}(Y^1(t),t)$$  You have a second derivative expression as a continued "equal to" statement with the first derivative expression. Your steps need to be clear.--Egm6321.f09.TA 03:41, 24 September 2009 (UTC)

2nd Derivative $$=\Bigg[{d^2\over dt^2}f(Y^1(t),t)={\partial f\over\partial s}(Y^1(t),t)\ddot{Y^1}(t)+{\partial f\over\partial ss}(Y^1(t),t)\dot{Y^1}^2(t)+2{\partial f\over\partial st}(Y^1(t),t)\dot{Y^1}(t)+{\partial f\over\partial tt}(Y^1(t),t)\Bigg]$$

 also, $$\frac{\partial}{\partial tt}$$ is bad notation. It should be $$\frac{\partial^{2}}{\partial t^{2}}$$--Egm6321.f09.TA 03:41, 24 September 2009 (UTC)  This is still part of problem #1.--Egm6321.f09.TA 03:41, 24 September 2009 (UTC)

Problem 3:
Solve the 2nd-order differential equation: $$y+y' = x$$ or $$y+y'-x=0$$     (3.1)

It is missing the dependent variable, $$y$$.

So, we can use reduction of order by setting $$P(x)=y'$$    (3.2)

This reduces Equation 3.1 to a 1st-order ODE: $$P'+P=x$$    (3.3)

Next, use the method of integrating factor to solve for $$P(x)$$ :

Equation 3.3 is already in the form: $$P'+ a_0 P=b$$ where $$a_0 =1$$ and $$b=x$$.

The solution for $$P(x)$$ is given by: $$P(x)=\frac{1}{h(x)}\int h(s)b(s)ds$$    (3.4)

First, we need to find $$h(x)$$, which is given by $$h(x) =exp \int a_0 (s)ds$$    (3.5)

Plugging in $$a_0=1$$, we get that $$h(x)=e^x$$     (3.6)

Plugging in Equation 3.6, and $$b=x$$ into Equation 3.4, we get $$P(x)=\frac{1}{e^x}\int e^x xdx$$.

This results in $$P(x)=\frac{(x-1)e^x + A}{e^x}$$    (3.7),

where $$A$$ is a constant. Rearranging Equation 3.7, we find that $$P(x)=Ae^{-x} + x - 1$$.

So $$P(x)= Ae^{-x} + x - 1$$

Integrating $$P(x)$$ to get $$y(x)$$, we get $$y(x)=-Ae^{-x}+\frac{1}{2}x^2 - x + B$$ , where $$B$$ is a constant.

 Try centering some of your equation results - this will make your report easier to read.--Egm6321.f09.TA 03:41, 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:
Show that the 1st-order ordinary differential equation $$F(x,y,y') = 0$$ is nonlinear, where $$F(x,y,y') = (2x^2 + \sqrt{y}) + x^5 y^3 y' = 0$$.

Proof of Nonlinearity:

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

$$D(\cdot) = (2x^2 + \sqrt{(\cdot)}) + x^5 (\cdot)^3 \frac{d(\cdot)}{dx} = 0$$.

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:

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

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

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

Since,

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

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 6:
Generating exact nonlinear 1st order ODEs: Let $$ \Phi(x,y)=6x^{4}+2y^{3/2}$$, then $$ M=\Phi_{x}$$ $$ N=\Phi_{y}$$ Complete the details and invent 3 more examples.

a) First, it is known that: $$ M=\Phi_{x}=\frac{\partial \Phi(x,y)}{\partial x}$$ and  $$ N=\Phi_{y}=\frac{\partial \Phi(x,y)}{\partial y}$$

Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ for the following $$ \Phi(x,y)=6x^{4}+2y^{3/2}$$, yields: $$ M=\Phi_{x}=24x^3$$ and $$ N=\Phi_{y}=3y^{1/2}$$

Putting into the form $$ M+Ny'=0$$ so that the nonlinear 1st order ODE is in the desired form yields: $$ 24x^3+3y^{1/2}y'=0$$ b) Let $$ \Phi(x,y)=\frac{1}{3}x^{5}-4y^{3}$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields:  $$ M=\Phi_{x}=\frac{5}{3}x^4$$ and  $$ N=\Phi_{y}=-12y^{2}$$ Putting into the form  $$ M+Ny'=0$$ yields:  $$ \frac{5}{3}x^4-12y^{2}y'=0$$ c) Let $$ \Phi(x,y)=x^6y^7+\frac{2}{5}y^{4}$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields: $$ M=\Phi_{x}=6x^5y^7$$ and $$ N=\Phi_{y}=7x^6y^6+\frac{8}{5}y^3$$ Putting into the form $$ M+Ny'=0$$ yields: $$ 6x^5y^7+(7x^6y^6+\frac{8}{5}y^3)y'=0$$ d) Let $$ \Phi(x,y)=\frac{1}{2}xy^{3/2}+x^2y^5$$ Taking the partial derivative for $$ \Phi_{x}$$ and $$ \Phi_{y}$$ yields:  $$ M=\Phi_{x}=\frac{1}{2}y^{3/2}+2xy^5$$ and  $$ N=\Phi_{y}=\frac{3}{4}xy^{1/2}+5x^2y^4$$ Putting into the form  $$ M+Ny'=0$$ yields:  $$ (\frac{1}{2}y^{3/2}+2xy^5)+(\frac{3}{4}xy^{1/2}+5x^2y^4)y'=0$$

 Very nice.--Egm6321.f09.TA 03:41, 24 September 2009 (UTC) Egm6321.f09.Team6.PaulMoore 18:09, 15 September 2009 (UTC)

Egm6321.f09.team6.breaux 19:33, 15 September 2009 (UTC)

Egm6321.f09.Team6.dianafoster 20:20, 15 September 2009 (UTC)

Egm6321.f09.Team06.sada 21:20, 15 September 2009 (UTC)