User:Egm6321.f09.Team6.PaulMoore

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

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

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

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.

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.

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:  $$ (\frac{1}{2}y^{3/2}+2xy^5)+(\frac{3}{4}xy^{1/2}+5x^2y^4)y'=0$$ = Make-Up Problem 1.3 d =

Problem Statement
Find the general solution of

$$\frac{d^2y}{dx^2}+y=f(x), y(0)=y'(0)=0$$

Solution
First, we solve for the homogeneous solution.

$$y_H(t)=c_1cos(t)+c_2sin(t)$$

So we have

$$y_1(t)=cos(t)$$ $$y_2(t)=sin(t)$$

The Wronskian of these two functions is

$$W=\left| \begin{array}{ c c } cos(t) & sin(t) \\ -sin(t) & cos(t) \end{array} \right|=cos^2(t)+sin^2(t)=1 $$

Therefore, the particular solution is

$$y_p(t)=-y_1\int\frac{y_2f(t)}{W(y_1,y_2)}dt+y_2\int\frac{y_1f(t)}{W(y_1,y_2)}dt=-cos(t)\int sin(t)f(t)dt+sin(t)\int cos(t)f(t)dt$$

Therefore, the general solution is

$$y=c_1cosx+c_2sinx-cosx\int^{x}sin(t)f(t)dt+sinx\int^{x}cos(t)f(t)dt$$

By substitution of the initial conditions, we find the $$c_1=c_2=0$$ which yields the following result for the general solution:

$$y=-cosx\int^{x}sin(t)f(t)dt+sinx\int^{x}cos(t)f(t)dt$$