User:Egm6321.f09.team4/HW1

Homework #1: Team 4 (Complete) Problem #1: Given a function $$f\left(y(t),t\right)$$, find: $$\frac{df}{dt}$$ and $$\frac{d^{2}f}{dt^2}$$.

Solution:

By the chain rule for partial differentiation, the first derivative can be decomposed into: $$\begin{align} \frac{df}{dt} &= \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial t} \\ &= \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t}     \end{align} $$  Need to have $$\frac{d}{dt}$$ differentials. Specifically, $$\frac{df}{dt}=\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}$$. This accounts for both explicit dependence on $$ t$$ with the partial differential, and implicit dependence with the first term. --Egm6321.f09.TA 02:58, 24 September 2009 (UTC) The second derivative can be found by repeating the same process on the result found for the first derivative: $$\begin{align} \frac{d^{2}f}{dt^{2}} &= \frac{d}{dt}\left(\frac{df}{dt}\right) = \frac{d}{dt}\left(\frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial t}\right) \\ &= \frac{d}{dt}\left(\frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial t} + \frac{d}{dt}\left(\frac{\partial y}{\partial t}\right) \frac{\partial f}{\partial y} + \frac{d}{dt}\left(\frac{\partial f}{\partial t}\right) \qquad \text{(by product rule)} \\ &= \left( \frac{\partial^{2}f}{\partial y\partial t} \frac{\partial t}{\partial t} + \frac{\partial^{2}f}{\partial y^{2}} \frac{\partial y}{\partial t}\right) \frac{\partial y}{\partial t} + \left( \frac{\partial^{2}y}{\partial t^{2}} \frac{\partial t}{\partial t}\right)\frac{\partial f}{\partial y} + \left( \frac{\partial^{2}f}{\partial t^{2}}\frac{\partial t}{\partial t} + \frac{\partial^{2}f}{\partial t \partial y}\frac{\partial y}{\partial t}\right) \qquad \text{(by chain rule)} \\ &= \frac{\partial^{2}f}{\partial y\partial t}\frac{\partial y}{\partial t} + \frac{\partial^{2}f}{\partial y^{2}}\left(\frac{\partial y}{\partial t}\right)^{2} + \frac{\partial f}{\partial y} \frac{\partial^{2}y}{\partial t^{2}} + \frac{\partial^{2} f}{\partial t^{2}} + \frac{\partial^{2} f}{\partial y \partial t}\frac{\partial y}{\partial t} \\ &= 2\frac{\partial^{2} f}{\partial y \partial t}\frac{\partial y}{\partial t} + \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{\partial y}{\partial t}\right)^{2} + \frac{\partial f}{\partial y}\frac{\partial^{2} y}{\partial t^{2}} + \frac{\partial^{2} f}{\partial t^{2}} \end{align} $$  Why have you changed notation? The problem was posed differently. It is ok to change notation or variables if it is helpful, however the statement should be phrased in the manner it was posed in class. Then state that you are using different variables. Finally, convert your solution to the class notation so that it is clear that you have arrived at the solution. --Egm6321.f09.TA 03:00, 24 September 2009 (UTC) Both of these results can alternatively be expressed in condensed notation as follows: $$   \begin{align} \frac{df}{dt} &= f_{y}\cdot y' + f_{t} \\ \frac{d^{2}f}{dt^{2}} &= 2\cdot f_{yt}\cdot y' + f_{yy}\cdot \left(y'\right)^{2} + f_{y}\cdot y'' + f_{tt} \end{align} $$

Problem #2: Show that the solution of $$ p\left(x\right)'+p\left(x\right) = x $$ is $$ p\left(x\right)=Ae^{-x}+x-1 $$ using integrating factors.

Solution:

We want to find a function $$h\left(x\right)$$ such that

$$\begin{align} h(x)p^{\prime}(x)+h(x)p(x)=h(x)x&=(p(x)*h(x))^{\prime} \\ &=p^{\prime}(x)h(x)+h^{\prime}(x)p(x) \qquad \text{(by product rule)} \end{align}$$ 

This statement is not rigorous. You have arrived at the proper answer but the steps are not properly connected. The equation needs to be rewritten in terms of $$\overline{M},\overline{N}$$--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

$$h(x)p(x)=h^{\prime}(x)p(x)$$

$$h(x)=h^{\prime}(x)$$

$$\frac{h^{\prime}(x)}{h(x)}=1$$

Integrate:

$$\int 1dx=\int \frac{h{}'(x)}{h(x))}dx=ln(h(x))$$

$$e^{\int 1dx}=e^{ln(h(x))}=>e^{x+c}=C_{1}e^{x}+C_{2}=h(x)$$

Let $$C_{2}=0$$ because it is arbitrary. Plug $$h(x)$$ expression back into original: 

You do not need any constants of integration in solving the separable diffeQ for $$h(x)$$ because you multiply $$h(x)$$ through the entire original equation, and the constants will drop out (as your $$C_{1}$$ does below).--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

$$C_{1}e^{x}p^{\prime}(x)+C_{1}e^{x}p(x)=

(p(x)*C_{1}e^{x})^{\prime}=xC_{1}e^{x}$$

Integrate:

$$p(x)*C_{1}e^{x}=\int{xC_{1}e^{x}}dx$$

Divide through:

$$p(x)=\frac{1}{C_{1}e^{x}}\int{xC_{1}e^{x}}dx$$

From integration tables,

$$p(x)=\frac{1}{C_{1}e^{x}}\int{xC_{1}e^{x}}dx=\frac{C_{1}}{C_{1}e^{x}}\left [ \frac{1}{1^{2}}(x-1)e^{x}+A_{1}\right ]$$

$$p(x)=\frac{1}{e^{x}}\left [ xe^{x}-e^{x}+A_{1} \right ]=x-1+A_{1}e^{-x}$$

Answer.

Problem #3: Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution

Define a linear operator $$D\left ( \cdot \right )$$ such that $$D(\cdot )=2x^2+\sqrt{\left (\cdot  \right ) }+x^5\left (\cdot  \right ) ^3\frac{d}{dx}\left ( \cdot  \right )$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition gives:

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

Substituting for the right side of the linearity condition:

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

Simplifying and comparing both sides of the linearity condition:

$$2x^2+\sqrt{\alpha u+\beta v}+\alpha x^5(\alpha u+ \beta v)^3\frac{du}{dx}+\beta x^5(\alpha u + \beta v)^3\frac{dv}{dx}\neq2x^2\alpha+\alpha\sqrt{u}+\alpha x^5u^3\frac{du}{dx}+2x^2\beta+\beta\sqrt{v}+\beta x^5v^3\frac{dv}{dx}$$

The condition is therefore not met, and $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is a nonlinear 1st Order ODE. 

Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

Problem #4: Given a function $$F(x,y,y^{\prime})=x^2y^5+6(y^{\prime})^2=0$$, show that F is a nonlinear 1st Order ODE.

Solution:

Define the operator $$D\left ( \cdot \right )$$ such that $$D\left ( \cdot  \right )=x^2\left ( \cdot  \right )^5+6\left ( \frac{d}{dx}\left [ \cdot  \right ] \right )^2$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition, we get:

$$D(\alpha u + \beta v)=x^2\left ( \alpha u+\beta v \right )^5+6\left ( \frac{d\left ( \alpha u+\beta v \right )}{dx} \right )^2$$

This can be simplified to:

$$x^2\left ( \alpha u+\beta v \right )^5+6\left ( \alpha \frac{du}{dx}+\beta\frac{dv}{dx}\right )^2$$

The right side of the linearity condition can be written as follows:

$$\alpha D(u)+\beta D(v)=\alpha\left ( x^2u^5+6\left [ \frac{du}{dx} \right ]^2 \right )+\beta\left ( x^2v^5+6\left [ \frac{dv}{dx} \right ]^2 \right )$$

And can be simplified to:

$$\left ( \alpha u^5 + \beta v^5 \right )x^2+6\alpha \left ( \frac{du}{dx} \right )^2+6\beta \left ( \frac{dv}{dx} \right )^2$$

Comparing the left and right hand sides of the linearity condition, we get:

$$x^2\left ( \alpha u+\beta v \right )^5+6\left ( \alpha \frac{du}{dx}+\beta\frac{dv}{dx}\right )^2\neq \left ( \alpha u^5 + \beta v^5 \right )x^2+6\alpha \left ( \frac{du}{dx} \right )^2+6\beta \left ( \frac{dv}{dx} \right )^2$$

This is clearly not an equality, therefore F(x,y,y') is a nonlinear 1st Order ODE.



Good solution.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)

<H4>Problem #5:</H4>

Part 1. Create an exact nonlinear 1st Order ODE of the form $$\Phi_x\left(x,y\right)+\Phi_y\left(x,y\right)y^{\prime}=0$$ using the equation $$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Solution:

If,

$$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Then

$$\Phi_{x}\left(x,y\right)=24x^{3}=M$$

and

$$\Phi_{y}\left(x,y\right)=3y^{1/2}=N$$

For a nonlinear, 1st order, ODE:

$$M + Ny^{\prime}=0$$

Therefore,

$$24x^3 + 3y^{1/2}y^{\prime}=0$$

We have forced exactness

Part 2. Create three more exact nonlinear 1st Order ODEs by inventing new $$\Phi\left(x,y\right)$$ functions.

Ex. 1

Choose $$\Phi\left(x,y\right)=x^{1/2}+y^{3}\quad\text{:}$$

$$ \begin{align} \Phi_{x}\left(x,y\right)&=0.5x^{-1/2}=M\left(x,y\right) \\ \Phi_{y}\left(x,y\right)&=3y^{2}=N\left(x,y\right) \end{align} $$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = \frac{1}{2}x^{-1/2}+3y^{2}y' = 0$$

Ex. 2

Choose $$\Phi\left(x,y\right)=3y^2+2x^5\quad\text{:}$$

$$\Phi_x\left(x,y\right)=10x^4=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=6y=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 10x^4+(6y)y' = 0$$

Ex. 3

Choose $$\Phi\left(x,y\right)=3x^2y^3\quad\text{:}$$

$$\Phi_x\left(x,y\right)=6xy^3=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=9x^2y^2=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 6xy^3+\left(9x^{2}y^{2}\right)y' = 0$$ <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

Nice work.--Egm6321.f09.TA 03:14, 24 September 2009 (UTC)