User:EGM6321.F10.TEAM1.HW2

Given
Eq5) from pg 6-3:

Find
Verify the equation given is a N1_ODE

Solution
The largest number of a differentiation is 1. (1.1) equation is 1st order. The differential equation consists of one dependent variable and its derivatives. (1.1) equation is Ordinary Differential Equation. To be a linear differential equation, (1.2) equation must satisfy superposition (1.3).

(1.2) equation does not satisfy superposition. In general (1.1) equation is N1-ODE. In a particular case this equation is L1-ODE.

Example :

$$\therefore $$ (1.12) equation is Linear.

Author EGM6321.f10.team1.Kim.TH

Given
Eq.7 from pg 7-1:

Find
Verify this equation is N1_ODE

Solution
The largest number of a differentiation is 1. (2.1) equation is 1st order. The differential equation consists of one dependent variable and its derivatives. (2.1) equation is Ordinary Differential Equation. To be a linear differential equation, (2.2) equation must satisfy superposition (2.3).

(2.2) equation does not satisfy superposition. (2.1) equation is N1-ODE.

Author EGM6321.f10.team1.Kim.TH

Given
and

Find
1) Show plot of $$ y^1_H(x) $$ and $$ y^2_H(x) $$ and 2) show that there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x

Solution
Plot of $$ y^1_H(x) $$ and $$ y^2_H(x) $$



To prove that there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x The following equation was used:

$$ y^1_H(\tilde{x}) - \alpha\ y^2_H(\tilde{x}) \ne \ 0 $$

This equation would only be satisfied for $$ \alpha = 0 $$.

Then, using $$ \tilde{x}=5 $$ and the newly defined $$ \alpha = 0$$ yields

$$5-0 \ \left( \frac{5}{2}\log \left( \frac{1+5}{1-5} \right)-1 \right)= \ 0$$ which reduces to $$5= 0\ $$ within the realm of complex numbers, but $$5 \ne \ 0\ $$

$$\therefore$$ there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x

Author EGM6321.F10.TEAM1.WILKS

Given

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$$ \varnothing(x,y)= x^2y^\frac{3}{2}+log(x^3y^2)=k$$ (4.1)
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Find
1)$$ F(x,y,y')=\frac{d\varnothing}{dx}(x,y)$$ 2) Verify that F is exact N1_ODE 3) Invent three more N1_ODE's

Solution
This problem asks for F(x,y,y') and for the verification that F is exactly N1_ODE. In other words, verify that there exists a function $$ \phi(x,y)\ $$ such that: 
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$$ F = \frac{d\phi}{dx}$$ (4.2)
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In other terms, verify that:


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$$F(x,y,y') = \frac{d}{dx}\phi(x,y) = \frac{d}{dx}k$$ (4.3, 4.4)
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Solving for the partial derivatives in equations 2 and 3 yields:


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$$F(x,y,y') = \frac{d\phi}{dx}(x,y) + \frac{d\phi}{dx}(x,y)\frac{dy}{dx} = M(x,y) + N(x,y)\frac{dy}{dx} = 0$$ (4.5, 4.6 and 4.7)
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Where:


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$$M(x,y) = \frac{d\phi}{dx}(x,y) = (2)(y^{3/2})x + \frac{3}{x}$$ (4.8, 4.9)
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$$N(x,y) = \frac{d\phi}{dx}(x,y) = \frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y}$$ (4.10, 4.11)
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1) and 2) Substituting equations 4.9 and 4.11 into equation 4.5 yields the final answer for F(x,y,y') (4.12) and also verifies that it is exactly N1_ODE (4.13)


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$$F(x,y,y') = ((2)(y^{3/2})x + \frac{3}{x}) + (\frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y})y' = 0$$ (4.12, 4.13)
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3) Below are 3 examples of N1_ODEs


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$$ F(x,y,y') = 22x^2y + y^5y'\ $$ (4.15) 
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$$ F(x,y,y') = sin(x^{7/2}) - \frac{4}{3}y^6y'\ $$ (4.16) <span id="(1)">
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$$ F(x,y,y') = x^8 + y^4y'\ $$ (4.17)
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Author: User:EGM6321.f10.team1.russo

Find
Find a particular function for f(y') such that there is no analytical solution to the given equation. (i.e. find f(y') such that the N1_ODE cannot be exact)

Solution
Begin with an equation in the form of eq.5.1 Then one possible definition of f'(y) would be:

in which case eq. 5.1 would become:

Although, this definition of f(y') appears to be a simple polynomial, the roots of y' cannot be solved analytically. This is, in general, true of polynomials of order higher than four as is discussed here:

Author

Bryan Allison Egm6321.f10.team1.allison 20:26, 21 September 2010 (UTC)

Given
where k is a constant

Find
Verify eq6.1 is a valid solution to the eq6.2

Solution
Start with eq6.2

then by differentiating both sides with respect to x and solving for y'(x) we get:

which can then be substituted into eq6.1 to yield: which is easily reduced to the trivial equation $$ 0=0\ $$

$$\therefore$$ eq6.1 is a valid solution to the eq6.2

Author Bryan Allison Egm6321.f10.team1.allison 20:25, 21 September 2010 (UTC)

Find
1) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming

2) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming $$ a_1(x)\ne \ 0\ $$ for all x 3) Find Solution in terms of $$ a_0\ $$ , $$ a_1 \ $$  and $$ b \ $$  for $$ y(x)\ $$ Assuming

Solution
1) and 2) Substitute equations 7.2, 7.3 and 7.4 into 7.1 to get:

The integrating factor is shown and solved for below.

Multiply both sides by the integrating factor yields

Integration then yields

Divide both sides by $$ \exp(\frac{1}{2}x^2) $$ and

3) Substitute equations 7.5, 7.6 and 7.7 into 7.1 to get: Divide through by $$ (x^2+1)\ $$ to obtain

The integrating factor is shown and solved for below.

Multiply both sides by the integrating factor yields

Integration then yields

Divide both sides by $$ (x^2+1)^\frac{1}{2} $$ gives the final result:

Author

User:Egm6321.f10.team1.allen

Given
Eq6, p.10-3 $$ y(x)=\frac{1}{h(x)} \int_{0}^{x} h(s)b(s)\, ds $$ and Eq1, p.10-3 $$ h(x)=e^{ \int_{0}^{x} a_0(s)\, ds } $$

Find
$$ \alpha\ $$ Show that integration factor, $$k_1\ $$, is unnecessary $$ \beta\ $$ Show that Eq6, p.10-3 agrees with King et.al. p.512 i.e. show that $$ y(x)=Ay_h(x)+y_p(x)\ $$ $$ \gamma\ $$ Find $$ y_h(x)\ $$ independently. i.e. solve $$ y'+a_0y=0\ $$

Solution
$$ \alpha\ $$) Using instead of

gives an integrating factor of

instead of Multiplying both sides by the integrating factor results in the following:

$$\begin{align} \therefore e^{K_1} \end{align}$$ cancels out of the equation 8.6 rendering it unnecessary.

$$ \beta$$) Using:

is derived from:

8.8 is comparable to:

Which is derived from:

As shown in K. et al p. 512.

Combining above statements and eq. 8.13:

$$ \gamma $$) Since $$ e^x\ $$ is repeating after integration, it is ideal for this solution.

Egm6321.f10.team1.allen 03:47, 22 September 2010 (UTC)

Given
Eq1, p.12-1

and

Find
1) Find an N1_ODE that is exact or that can be made to be exact 2) Find $$ \phi\ (x,y)=k \ $$

Solution
First $$\overline{b}(x)$$ and $$ \overline{c}(x)$$ must be found.

Then, substituting into (9.1)

It can be shown that $${{e}^{2y}}\ne 0,\forall y\in \mathbb{R}$$ so these terms can be divided out without reducing the generality of the problem.

Next, check for exactness:

and

$$ \therefore $$ the equation is not exact as shown. Since both M and N are both functions of x only it is safe to assume that the integrating factor, h, must be a function of x.

so that equation (9.1) can then become the exact equation:

This exactness of this equation can be proven by:

$$ \therefore y'+\frac{\sin ({{x}^{3}})}{2\sin (x)}=0 $$ is an exact equation.

The first integral $$\phi$$ can then be found through the following process. Rewrite 9.1 as:

where

and

so N can be integrated with respect to y to find $$ \phi $$

differentiate again with respect to x and set equal to M to solve f'(x)

Then integrating f'(x) one last time to yield $$ \phi $$

$$ \phi =y+\int{f'(x)dx}$$

$$ \therefore \phi =y+\int{\frac{\sin ({{x}^{3}})}{2\sin (x)}dx}$$

which has no analytical solution, but could be solved numerically given boundary conditions.

Author

Egm6321.f10.team1.allison 15:28, 22 September 2010 (UTC)

Signatures
Solved problem 1 -- Taeho Kim (TK)

Solved problem 2 -- Taeho Kim (TK)

Solved problem 3 -- EGM6321.F10.TEAM1.WILKS 05:38, 20 September 2010 (UTC)User:EGM6321,F10.TEAM1.WILKS|David Wilks (DW)

Solved problem 4 -- EGM6321.f10.team1.russo 17:10, 22 September 2010 (UTC)Charles Russo (CR)

Solved problem 5 -- Egm6321.f10.team1.allison 20:24, 21 September 2010 (UTC)Bryan Allison (BA)

Solved problem 6 -- Egm6321.f10.team1.allison 20:24, 21 September 2010 (UTC)Bryan Allison (BA)

Solved problem 7 -- Egm6321.f10.team1.allen 18:24, 22 September 2010 (UTC) Sean Allen (SA)

Solved problem 8 -- Egm6321.f10.team1.allen 18:24, 22 September 2010 (UTC) Sean Allen (SA)

Solved problem 9 -- Egm6321.f10.team1.allison 15:27, 22 September 2010 (UTC)Bryan Allison (BA)