User:Egm6321.f11.team1.colsonfe/HW2

= R*2.5 Conversion to Particular N1-ODE =

Find
Explain the following equations:


 * {| style="width:100%" border="0" align="left"

$$
 * $$\bar M(x,y)+\bar N(x,y)(y')^3=0$$
 * $$\displaystyle (Equation\;2.5.1)
 * $$\displaystyle (Equation\;2.5.1)
 * }
 * }


 * $$M(x,y):=\left [\bar M(x,y)\right]^{1/3}$$


 * $$N(x,y):=\left [\bar N(x,y)\right]^{1/3}$$

Basic Equations
General N1-ODE form:


 * $$G\left(y',y,x\right)=0$$

Particular N1-ODE form:


 * {| style="width:100%" border="0"

$$M\left(x,y\right)+N(x,y)y'=0$$
 * $$\,(Equation\;2.5.2) $$
 * style= |
 * }

Solution
The relation among the three equations we are to explain is illustrated by the ability to convert the first from a GENERAL form N1-ODE to a PARTICULAR form N1-ODE. The following steps show how redefining M and N as given will convert one type of equation into another.

Use definition of M and N and take cube power of each side.

$$ \left[M(x,y)\right]^{1/3}=\bar M(x,y) $$

$$ \left[N(x,y)\right]^{1/3}=\bar N(x,y) $$

Enter this result into Equation 2.5.1.

$$\left[M(x,y)\right]^3+\left[N(x,y)\right]^3(y')^3=0$$

Subtract one term to the right side.

$$\left[M(x,y)\right]^3=-\left[N(x,y)\right]^3(y')^3$$

$$\left[M(x,y)\right]^3=\left[-\left[N(x,y)\right](y')\right]^3$$

Take the cube root and add to the left side.

$$M(x,y)=-\left[N(x,y)\right](y')$$

$$M\left(x,y\right)+N(x,y)(y')=0$$

This now matches Equation 2.5.2. We can see that we were able to convert an apparent general form N1-ODE to a particular form N1-ODE.

 Solution by Egm6321.f11.team1.colsonfe 02:31, 17 September 2011 (UTC)

= R*2.6 Linear Independence of Homogeneous Solutions =

Given
Two homogeneous solutions:


 * $$y^1_H(x)=x$$


 * $$y^2_H(x)=\frac{x}{2}\log\left(\frac{1+x}{1-x}\right)-1$$

Find
Show that $$y^1_H(x)$$ and $$y^2_H(x)$$ are linearly independent.

Plot $$y^1_H(x)$$ and $$y^2_H(x)$$.

Basic Equations
Linear Independence:


 * $$\forall\alpha\in\mathbb R, y^1_H(\cdot)\ne\alpha y^2_H(\cdot)$$

i.e, for any given $$\alpha$$, show that


 * $$\exists\hat x$$ such that $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$

Solution
By definition of linear independance, we merely need to show there is a value $$\hat x$$ where $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$ for some $$\alpha$$.

Let us then choose $$\hat x=0$$.Then

$$y^1_H(\hat x)=\hat x=0$$

$$y^2_H(\hat x)=\frac{\hat x}{2}\log\left(\frac{1+\hat x}{1-\hat x}\right)-1=-1$$

Now lets consider any given $$\alpha$$ for our chosen $$\hat x=0$$,

$$y^1_H(\hat x) \overset{?}{=}\alpha y^2_H(\hat x)$$

$$0 \overset{?}{=}-\alpha$$

It is clear that the above equality IS NOT true for all $$\alpha$$.

Therefore, for $$\hat x=0$$ and $$\alpha=1$$, $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$ for some $$\alpha$$, which means that $$y^1_H(x)$$ and $$y^2_H(x)$$ are linearly independent.


 * HW2_6plots.gif

 Solution by Egm6321.f11.team1.colsonfe 02:31, 17 September 2011 (UTC)

= R*2.8 First Exactness Condition =

Given

 * {| style="width:100%" border="0"

$$M\left(x,y\right)\cos y'+N(x,y)\log y'=0$$
 * $$\,(Equation\;2.8.1) $$
 * style= |
 * }

Find
Does Equation 2.8.1 satisfy the first exactness condition?

Basic Equations
First exactness condition for N1-ODE


 * $$M\left(x,y\right)+N(x,y)y'=0$$

Solutions
Let us define

$$\bar M\left(x,y\right):=M(x,y)$$

$$\bar N\left(x,y\right):=N(x,y)$$

Then Equation 2.8.1 becomes

$$\bar M\left(x,y\right)\cos y'+\bar N\left(x,y\right)\log y'=0$$

$$\bar M\left(x,y\right)+\bar N\left(x,y\right)\frac{\log y'}{\cos y'}=0$$

$$\bar M\left(x,y\right)+\bar N\left(x,y\right)F(y')=0$$

According to notes, if the function $$F$$ has no explicit inverse, then it CANNOT be put in the form which satisfies the First Exactness Condition.

The function $$F(y')=\frac{\log y'}{\cos y'}$$ has no inverse $$F^{-1}\left(y'\right)$$, and so therefore this N1-ODE fails to satisfy the First Exactness Condition.

 Solution by Egm6321.f11.team1.colsonfe 04:30, 17 September 2011 (UTC)

= References=