User:Egm6321.f12.team4/Report2

Statement
Verify that

where $$ y_H^1(x) $$ and $$ y_H^2(x) $$ are homogeneous solutions of the Legendre Differential Operator (for n=1). $$ L_2(y):= (1-x^2)y''-2xy'+n(n+1)=0 $$

Given
{| style="width:100%" border="0" The Legendre Differential Operator for n=1 is given as
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Solution
To solve this problem, one must substitute the linearly independent homogeneous solutions and their first and second derivatives back into the Legendre Differential Operator for n=1 and verify that they are equal to zero.

Looking at the first homogeneous solution (Equation 2.1.2), its first an d second derivatives are $$ y'=\frac{d}{dx}(x)=1 $$ (2.1.4) $$ y''=\frac{d^2}{dx^2}(x)=\frac{d}{dx}(1)=0 $$ (2.1.5)
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Substituting Equations 2.1.4 and 2.1.5 back into Equation 2.1.1, one can see that $$ 0=(1-x^2)(0)-2x(1)+2x $$

Thus $$ L_2(y_H^1)=0 $$

Looking at the second homogeneous solution (Equation 2.1.3), its first derivative is

$$\displaystyle y'=\frac{d}{dx}[\underbrace{\frac{x}{2}\underbrace{ln(\frac{1+x}{1-x})}_{\frac{d}{dx}(ln(u(x)))=\frac{1}{u(x)}u'(x)}}_{differentiate\,by\,parts}-1] $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})\underbrace{(\frac{d}{dx}(\frac{1+x}{1-x}))}_{ \frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v(x)u'(x)-u(x)v'(x)}{v(x)^2}} $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})[\frac{(1-x)(1)-(1+x)(-1))}{(1-x)^2}] $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})(\frac{2}{(1-x)^2}) $$

$$ \displaystyle y'=\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2} $$
 * (2.1.6)

The second derivative is

$$ \displaystyle \frac{d^2}{dx^2}[\frac{x}{2}ln(\frac{1+x}{1-x})-1]=\frac{d}{dx}[\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2}] $$

Using the same derivation rules listed above

$$ \displaystyle \frac{1}{2}(\frac{1-x}{1+x})\frac{d}{dx}(\frac{1+x}{1-x})+\frac{(1-x^2)(1)-x(-2x)}{(1-x^2)^2} $$

$$ \displaystyle \frac{1}{2}(\frac{1-x}{1+x})(\frac{2}{(1-x)^2})+\frac{1+x^2}{(1-x^2)^2} $$

$$ \displaystyle \frac{2}{(1-x^2)^2} $$
 * (2.1.7)

Substituting Equations 2.1.6 and 2.1.7 back into Equation 2.1.1, one can see that $$ \displaystyle 0=(1-x^2)(\frac{2}{(1-x^2)^2})-2x(\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2})+2(\frac{x}{2}ln(\frac{1+x}{1-x})-1) $$

$$ \displaystyle \frac{2}{1-x^2}-(xln(\frac{1+x}{1-x})+\frac{2x^2}{1-x^2})+(xln(\frac{1+x}{1-x})-2) $$ $$ \displaystyle \frac{2}{1-x^2}-\frac{2x^2}{1-x^2}-2(\frac{1-x^2}{1-x^2})$$ $$ \displaystyle \frac{2-2x^2-2+2x^2}{1-x^2}=0 $$

Thus $$ L_2(y_H^2)=0 $$

Therefore it is proven that $$ L_2(y_H^1)=L_2(y_H^2)=0. $$

Author and References

 * Solved and Typed by -- Kaitlin Harris
 * Reviewed by -- Seong Hyeon Hong

Problem R*2.2 - Verify the Solution for L1-ODE-CC Problem
sec7-5

Statement
Verify that Eq.(2.2.1) is indeed the solution for Eq.(2.2.2)

Solution
Taking derivative of Eq (2.2.1),

It follows that,

Substituting $$\displaystyle p' $$ from (2.2.4) in the left hand side (LHS) of (2.2.2), one will get

Hence, p(x) is indeed the solution for Eq. (2.2.2)

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Pushkar Mishra

Problem R*2.3
sec7-6

Statement
Show that (2.3.1) is affine in $$ y' $$, and that (2.3.1) is in general an non-linear first order ordinary differential equation(N1-ODE). However, (2.3.1) is not the most general N1-ODE as represented by (2.3.2). Give an example of a more general N1-ODE.

i) Show that (2.3.1) is affine in $$ y' $$
Manipulating Equation (2.3.1) such that $$ M(x,y) $$ is moved to the right hand side and both sides are divided through by $$ N(x,y) $$

one can see that

Additionally

Now,

Thus it is shown that equation 2.3.1 is affine in y'.

End of proof

ii) Show that (2.3.1) is in general an N1-ODE
Equation (2.3.1) is a 1st order differential equation since the highest order of derivative is 1.

Equation (2.3.1) is an Ordinary Differential equation because the differential equation includes one dependent variable and its derivatives.

We can define $$ f(x,y) $$

Substitute $$ y=\alpha y_1 + \beta y_2 $$

Since, in general

it follows that

Thus it is shown that equation 2.3.1 is nonlinear.

End of proof

iii) Show that equation (2.3.1) is not the most general N1-ODE as represented by (2.3.2)
Recall equations (2.3.10) and (2.3.11)

If and only if $$ M(x,y) $$ is linear, then (2.3.10) will be

If and only if $$ N(x,y) $$ is equal to a constant, C, then (2.3.11) will be

If both of these conditions hold true, (2.3.12) will be

Thus it is shown that equation 2.3.1 can be manipulated such that it is affine with $$ y' $$. As such, it is not the most general form of the N1-ODE.

End of proof

iv) Give an example of a more general N1-ODE
An example of a more general N1-ODE is the following. where

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Che Rui


 * Loc Vu-Quoc., [[media:Pea1.f12.mtg7.djvu|Mtg 8-2]]

Statement
$$ y_{H}^{1}(x) = x $$: $$ y_{H}^{2}(x) = \frac{x}{2}log(\frac{1+x}{1-x})-1 $$
 * To show that $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x) $$ are linearly independent where,

Solution
From the definition of linear independence, it must be shown that for all x, $$ y_{H}^{1}(x) \neq \alpha y_{H}^2(x)$$ for some $$ \alpha (\neq 0) $$

Since the above should hold for all value of x, it can be assumed that x is equal to 0.

Evaluating $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x) $$ at $$x=0$$, $$ y_{H}^{1}(0) = 0 $$  (2.4.1) $$ y_{H}^{2}(0) = \frac {0}{2}log(\frac{1+0}{1-0})-1 = -1 $$  (2.4.2)

For linear dependence, $$ y_{H}^{1}(x) $$ = $$\alpha y_{H}^{2}(x) $$ For this solution $$0 = -\alpha $$

Thus it can be seen that for any value of $$\alpha (\neq 0)$$, the above will not hold true. This proves the linear independence of the two functions.

ii) Plot of $$y_H^{1}(x)$$ & $$ y_H^{2}(x)$$ is shown below.


 * [[File:Graph.jpg]]

Author and References
Solved and typed by -- Pushkar Mishra Reviewed by -- Jinchao Lu http://rechneronline.de/function-graphs/ was used to plot the graph

Statement
Consider the following function $$ \phi (x,y)=x^{2}y^{3/2}+log(x^3y^2)=k $$

Find $$ G(y',y,x) = \frac{d}{dx} \phi(x,y)=0 $$

Solution
We start by differentiating $$ \phi(x,y) $$ with respect to x. Since the function is a product of two functions dependent on x, we use differentiation by parts. $$ \frac{d}{dx}\phi(x,y)=(\frac{d}{dx}x^2)y^{3/2}+x^2(\frac{d}{dx}y^{3/2})+\frac{1}{x^3y^2}((\frac{d}{dx}x^3)y^2+x^3\frac{d}{dx}y^2)= \frac{d}{dx}k $$  (2.5.1)

It follows that, $$ 2xy^{2/3} + x^2\frac{3}{2}y^{1/2}\frac{dy}{dx} + \frac{1}{x^{3}y^{2}} (3x^{2}y^{2}+2x^{3}y\frac{dy}{dx}) =0 $$  (2.5.2) Simplifying the equation, $$ 2xy^{2/3} + \frac{3}{2}y^{1/2}x^2\frac{dy}{dx} + \frac{3}{x}+ \frac{2}{x}\frac{dy}{dx} =0 $$  (2.5.3) It can be seen that the above is a function of x,y,y'.Arranging the terms we get, $$ G(x,y,y')=(2xy^{2/3} + \frac{3}{x})+ (\frac{3}{2}y^{1/2}x^2 + \frac{2}{x})y' =0 $$ (2.5.4) which can be written as , $$ G(x,y,y') = M(x,y) + N(x,y)y'=0 $$ where , $$ M(x,y) = 2xy^{2/3} + \frac{3}{x} $$ and $$ N(x,y) = \frac{3}{2}y^{1/2}x^2 + \frac{2}{x} $$


 * i) The above equation is a 1st order differential equation since the highest order of derivative is 1 (y').
 * ii) It is an Ordinary Differential Equation because the differential equation includes one dependent variable(y) and its derivatives(y') with respect to the independent variable.


 * Hence Eq. (2.5.4) is a N1-ODE.

Author and References
Solved and typed by -- Pushkar Mishra Reviewed by -- Jinchao Lu

Statement
Review calculus and find the minimum degree of differentiability of the function such that $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$ is satisfied. State the full theorem and provide a proof.

Solution
Theorem
 * $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$

when, $$ \phi(x,y), \phi_{x}, \phi_{y}, \phi_{xy}, \phi_{yx} $$ are defined throughout an open and are all continuous throughout an open region.

Proof Equality of this theorem can be established by mean value theorem. Let h, k be arbitrary real numbers,

Let's define a function and apply the mean value theorem,

From the above equations,

This time let there be a function similar to the second equation but instead of y being the variable, let x be the variable.

Degree of Differentiability

''Looking at the theorem and the proof above, in order to satisfy the equality, the function must be differentiated at least three times. Therefore the minimum Degree of Differentiability for Mixed Derivative Theorem is three.''

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reference -- Thomas's Calculus, International Edition (George B. Thomas, Jr): pg. AP-23 ~ AP-25
 * Reviewed by -- Jason Alphonso

Statement
Solve the N1-ODE $$ \frac{d}{dx}\phi(x,y)=75x^{4}+ (\cos y){y}^'= 0 $$ and verify the solution is $$ y(x)= sin^{-1}(k-15x^{5})$$

Solution
-

The solution to the above problem is to first verify if the N1-ODE meets the 1st and 2nd exactness conditions,

1st Exactness Condition - Verifying if the N1-ODE is affine, that is it can be represented in the form $$ M(x,y)+N(x,y)\,y'=0 $$

By observation, the 1st Exactness Condition is met, where $$ M(x,y)= 75x^4 $$ and $$ N(x,y)= cos y $$

2nd Exactness Condition - Verify that $$ M_{y}(x,y)= 0 $$ and $$ N_{x}(x,y)= 0 $$ which by calculation is also true.

Hence there exist a function $$ \phi \left ( x,y \right )$$. Once we confirmed the existence of $$ \phi \left ( x,y \right )$$, we will now solve to find it's value.

Integrating N1-ODE,

where k = sum total of the constants of integration.

which is the required solution.

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Kaitlin Harris

Lecture Notes Reference:

Statement
--

Explain why it is so challenging to solve for Euler's integrating factor -> h(x,y)

Reasoning
-

Understanding the complexity of solving for the integrating factor h(x,y) requires an understanding of the origin of the factor.

For an N1-ODE that is affine (meets the 1st exactness condition) but does not satisfy the 2nd exactness criterion, Euler proposed the use of a factor h(x,y) which satisfies the following equation

Now applying 2nd exactness criterion to the equation above,

After solving as per the 2nd exactness criterion above,

Important Notes:

1. As noted above the values of $$h_x$$ and $$h_y$$ are partial derivatives of h with respect to x and y respectively. While h(x,y) is unknown, the presence of two partial derivatives in one equation adds complexity.

2. While N(x,y) and M(x,y) are known, the product of these terms with $$h_x$$ and $$h_y$$ respectively adds complexity.

3. $$h$$is nonlinear and x and y are both independent variables, hence has varying coefficients.

This complex equation can be solved by making assumptions, such as driving certain terms to zero and studying the case individually.

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Kaitlin Harris

Lecture Notes Reference:

Statement
Suppose $$ h_{x}(x,y)=0,$$thus h is a function of y only; then $$ h_{x}N - h_{y}M + h(N_{x} - M_{y})=0 $$ becomes $$ \frac{hy}{h}=\frac{1}{M}(N_{x}-M_{y})=:m(y) $$ Find h using the above equation.

Solution
From the class note 11-2 it is written that,

If $$ h_{y}(x,y) = 0 $$ then,

In the same manner, this problem can be solved.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Jason Alphonso

Statement
The non-homogeneous L1-ODE-VC is



y'+\frac{1}{x}y=x^2 $$

Show that



h(x)=x $$ and $$ y(x)=\frac{x^3}{4}+\frac{k}{x} $$

Solution
We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=\frac{1}{x}y-x^2$$ and $$N(x,y)=1$$. So the first exactness condition holds.

For exactness condition 2:

Then $$M_y \neq N_x$$. So the second exactness condition does not hold.

Assume a factor $$h(x,y)$$. Let

and

Take the derivative of $$\overline{M}(x,y)$$ and $$\overline{N}(x,y)$$ with respect to $$y$$ and $$x$$ respectively:

Let the above two terms equal to each other:

We further assume that $$h=h(x)$$, namely $$h$$ is only the function of $$x$$. Then after plugging the expression of $$M_y$$ and $$N_x$$ into the equation, we have:

Recall that

Then the equation becomes:

We integrate both side of the equation:

where $$C$$ is the integral constant.

Then

Note that

Plug it into the first term of the equation to replace $$h(x)$$ with $$xh_x(x)$$:

We reorganize it to be:

Then we can integrate both side of the equation with respect to x:

If we let the constant $$k=\frac{C_2}{C_1}$$, then we can get the required form of $$y(x)$$:

For the required expression of $$h(x)=C_1x$$, we can let the integral constant $$C_1=1$$, then:

Thus, we have shown the expression of both $$h(x)$$ and $$y(x)$$ as required.

Author and References

 * Solved and Typed by -- Rui Che
 * Reviewed by -- Seong Hyeon Hong

Statement
The general L1-ODE-VC is



a_1(x)y'+a_0(x)=b(x) $$


 * (1) $$a_1(x)=1, a_0(x)=x+1, b(x)=x^2+4$$


 * (2) Solve $$y(x)$$ in terms of $$a_1(x),a_0(x)$$ and $$b(x)$$


 * (3) $$a_1(x)=x^2+1, a_0(x)=x^3, b(x)=x^4$$

Solve these two specific L1-ODEs-VC.

Solution(1)
For (1):

We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=(x+1)y-(x^2+4)$$ and $$N(x,y)=1$$. So the first exactness condition holds.

For exactness condition 2:

Then $$M_y \neq N_x$$. So the second exactness condition does not hold.

Assume a factor $$h(x,y)$$. Let

and

Take the derivative of $$\overline{M}(x,y)$$ and $$\overline{N}(x,y)$$ with respect to $$y$$ and $$x$$ respectively:

For the second exactness condition:

We further assume that $$h=h(x)$$, namely $$h$$ is only the function of $$x$$. Then after plugging the expression of $$M_y$$ and $$N_x$$ into the equation, we have:

Recall that

Then the equation becomes:

We integrate both side of the equation:

where $$C$$ is the integral constant.

Then

Note that

Plug it into the first term of the equation to replace $$h(x)$$ with $$xh_x(x)$$:

We reorganize it to be:

Then we can integrate both side of the equation with respect to x:

This integration is difficult to be solved. Finally we use a symbolic integration software to integrate it out:

Solution(2)
The general L1-ODE-VC is

which is in the form:

Dividing Eq (2.11.27) throughout by $$ a_1(x) $$, we get,

Substituting $$\frac{a_0(x)}{a_1(x)}$$ as $$ a_0'(x)$$ and $$\frac{b(x)}{a_0(x)}$$ as $$b'(x)$$ we get,

which is of the same form as Eq. (1)p11.4 [p11.4].

For the second exactness condition to hold true as in Eq 4 p11.2 [p11.2].

Since $$h$$ is a function of x only,

It can be easily seen that, $$\frac{-1}{N}(N_x-M_y)= a_0'(x)$$. Integrating both sides of Eq. (2.11.30)

Since $$ h=ah_x $$ and using the above in Eq. (2.11.28), we get

The above can be integrated to get, the solution $$ y(x) $$

Using the values of $$h(x)$$ and replacing $$a_0'(x)$$ with $$\frac{a_0(x)}{a_1(x)}$$ and $$b'(x)$$ with $$\frac{b(x)}{a_1(x)} $$

Solution(3)
For (3):

We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=\frac{x^3y-x^4}{x^2+1}$$ and $$N(x,y)=1$$. So the first exactness condition holds.

For exactness condition 2:

Then $$M_y \neq N_x$$. So the second exactness condition does not hold.

Assume a factor $$h(x,y)$$. Let

and

Take the derivative of $$\overline{M}(x,y)$$ and $$\overline{N}(x,y)$$ with respect to $$y$$ and $$x$$ respectively:

Let the above two terms equal to each other:

We further assume that $$h=h(x)$$, namely $$h$$ is only the function of $$x$$. Then after plugging the expression of $$M_y$$ and $$N_x$$ into the equation, we have:

Recall that

Then the equation becomes:

We integrate both side of the equation:

where $$C$$ is the integral constant.

Then

Note that

Plug it into the first term of the equation to replace $$h(x)$$ with $$xh_x(x)$$:

We reorganize it to be:

Then we can integrate both side of the equation with respect to x:

At this time, even the software cannot solve this complex integration analytically. We can assume the result to be:

So the final result can be written as:

where $$k=\frac{C_2}{C_1}$$.

Author and References

 * Solved and Typed by -- Rui Che & Pushkar Mishra
 * Reviewed by -- Pushkar Mishra

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.