User:EGM6321.f09.Team2.gaddone/HW1

 Clearly delineate where the problem statement ends, and your solution begins. --Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

Problem 1
Derive the 1st order and 2nd order Total Time Derivatives of $$f$$ where $$f=(Y^1(t),t)$$.1

$$Y^1(t)$$ is defined as the nominal motion

Take the 1st derivative d/dt of $$f$$

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

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

Take the 2nd derivative d/dt of $$f$$

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

$$={\partial f\over\partial s}(Y^1(t),t)\ddot{Y^1}(t)+\dot{Y^1}(t)\Bigg({\partial \over\partial s}{\partial f\over\partial s}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial s}(Y^1(t),t)\Bigg)+{\partial \over\partial s}{\partial f\over\partial t}(Y^1(t),t)\dot{Y^1}(t))+{\partial \over\partial t}{\partial f\over\partial t}(Y^1(t),t)$$

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

 Use some more steps to show this result. --Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

Problem 2
Derive $$p(x)=Ae^{-x}+x-1$$ using the method of integrating factors.1

Consider a linear, first order differential equation in the form $$p'+p=x$$.

Using the method of Integrating Factors, derive: $$p(x)=Ae^{-x}+x-1$$ from $$y''+y'=x$$ (Eq.(1) from Pg. 4-2)

Let $$p(x)=y'(x)$$, so $$y''+y'=x$$ becomes $$p'+p=x$$.

Find the multiplying factor. Since $$p$$ is positive with a constant equal to (1), the integrating factor will be:

$$exp\int_{0}^{x} (1)\, dx = e^x$$

Multiply $$p'+p=x$$ by the multiplying factor to obtain:

$$e^xp'+e^xp=e^xx\Rightarrow {d\over dx}(e^xp)=e^xx$$

Integrate: $$\int_{}^{}{d\over dx}(e^xp)\,=\int_{}^{}e^x$$

$$e^xp(x)=(x-1)e^x+A\Rightarrow e^xp(x)=e^xx-e^x+A$$, divide through by $$e^x:$$

$$p(x)=x-1+{A\over e^x}$$, rearrange to obtain:

$$\Big[p(x)=Ae^{-x}+x-1\Big]$$ 1Lecture 4

 Show your integration steps. The footnotes are a nice touch, tying the material back to the lecture for reference.--Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

Problem 3
Prove that (2x2+$$\sqrt{y}$$)+(x5y3y')=0 is non-linear.1

An equation is linear if each term within the equation has only 1 order of derivative, therefore allowing superposition. In other words, for an equation to be linear, where y is a function of x, the coefficients of y and its derivatives must depend on x and coefficients only; they cannot contain any other functions of y.

Mathematically, this means:

L($$\alpha\,$$ u +  $$\beta\,$$ v)' = ($$\alpha\,$$L'(u) + $$\beta\,$$ L'(v)__AND__ L($$\alpha\,$$ u) = $$\alpha\,$$ L(u)

By observation, it is obvious that (2x2+$$\sqrt{y}$$)+(x5y3y')=0 is non-linear due to the presence of >+$$\sqrt{y}$$ in the first term and y3y' in the second term. Mathematically, if we assume:

$$\alpha\,$$ = 2

u = x2 + $$\frac{1}{2}$$$$\sqrt{y}$$

$$\beta\,$$ = 1

v = x5y3y'

L($$\alpha\,$$ u +  $$\beta\,$$ v)' = $$\frac{d}{dx}$$(2x2 + $$\sqrt{y}$$ + x5y3y' = $$\frac{d}{dx}$$(2x2+$$\sqrt{y}$$) + $$\frac{d}{dx}$$(x5y3y')

$$\frac{d}{dx}$$(2x2 + $$\sqrt{y}$$ + x5y3y') = 4x + 5x4y3y'

2$$\frac{d}{dx}$$(x2+$$\frac{1}{2}$$$$\sqrt{y}$$) + $$\frac{d}{dx}$$(x5y3y') = 2 (2x + $$\frac{1}{2}$$$$\frac{d}{dx}$$($$\sqrt{y}$$)) + (5x4y3y')$$\frac{d}{dx}$$(y3)(y")

As the derivative has complex functions of y, the original equation is NON-LINEAR.

1Lecture 4  You should define a differential operator for this problem and show that it does not exhibit linear behavior. You do not need to pick values for $$\alpha,\beta$$--Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

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

Since $$F(x,y,y')$$ contains the derivatives of only one variable, $$y$$, it is an ordinary differential equation (ODE). Since it contains only the first derivative of $$y$$, it is a 1st order ODE. However, since it contains a combination of the unknown variable $$y$$ and it's derivatives to powers other than $$1$$, it is nonlinear.

We can show that $$F(x,y,y')$$ is nonlinear by using the linear differential operator, $$D=(x^2)+6(\tfrac{d}{dx})^2$$. Evaluating,

$$D(y^5)=(x^2)y^5+6(\tfrac{d}{dx})^2y^5$$

This does not equal our original function $$F$$. In fact, there is no way to factor $$y$$ to generate an appropriate linear differential operator. Therefore, $$F$$ is nonlinear.

1Lecture 5  Incorrect differential operator used for this problem. Show that $$D(u+v)\neq D(u)+D(v)$$--Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

Problem 5
'''Complete details of equation on pg. 6-1 and invent 3 additional exact non-linear equations.1'''

Let $$\Phi(x.y)=6x^{4}+2y^{3/2}$$

$$ M=\Phi_{x}$$ $$ N=\Phi_{y}$$

$$ M=\Phi_{x}=\frac{\partial \Phi(x,y)}{\partial x}$$

$$ N=\Phi_{y}=\frac{\partial \Phi(x,y)}{\partial y}$$

M+Ny'=0

$$\scriptstyle F={d\over dt}\Phi(x, y(x))$$

$$\Phi(x)= 6x^4$$ $$\scriptstyle{d\over dt}(\Phi(x))=24x^3$$

$$\Phi(y)= 2y^{3/2}$$ $$\scriptstyle {d\over dt}(\Phi(y))=(3/2)2y^{1/2}=3\sqrt{y}$$

$$\Big[24x^3+3\sqrt{y}y'=0\Big]$$

Examples

$$\Phi(x,y)= 3x^7 + y^5$$

$$\Phi(x,y)= 20x^{100} + y^{7/3}$$

$$\Phi(x,y)= 136x^{137} + 70y^{100/3}$$

1Lecture 6  Should not be $$\frac{d}{dt}$$ operators. Differentiations are with respect to $$x$$. For the three additional examples, calculate $$\frac{\partial \Phi}{\partial x},\frac{\partial \Phi}{\partial y}$$ and formulate the exact differential equation.--Egm6321.f09.TA 02:28, 24 September 2009 (UTC)

Contributing Team Members
Joe Gaddone 14:46, 16 September 2009 (UTC)

Kumanchik 19:30, 16 September 2009 (UTC)

Egm6321.f09.Team 2.walker 20:50, 16 September 2009 (UTC) (Matthew J Walker)

Egm6321.f09.Team2.sungsik 15:13, 23 September 2009 (UTC)