User:Egm6321.f12.team5.kim/Report 2

= Report 2 Problems =

Given

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$$ L_2(y) := (1-x^2)y'' - 2xy' + 2y = 0 $$     (2.1.1)
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$$ y^1_H(x) = x =: P_1(x) $$     (2.1.2)
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$$ y^2_H(x) = \frac{x}{2} log \left(\frac{1+x}{1-x}\right) - 1 =: Q_1(x) $$     (2.1.3)
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Find
Verify that


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$$ L_2(y^1_H) = L_2(y^2_H) = 0 $$     (2.1.4)
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Solution
Substituting equation (2.1.2) into (2.1.1) gives


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$$ L_2(y^1_H) := (1-x^2)\cancelto{0}{\frac{d^2(x)}{dx^2}} - 2x\cancelto{1}{\frac{d(x)}{dx}} + 2(x) = 0 $$     (2.1.5)
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$$ L_2(y^1_H) := - 2x + 2x = 0 $$     (2.1.6)
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which shows that $$y^1_H$$ is a homogeneous solution to $$L_2(\cdot).$$

Since the expression for $$y^2_H$$ is a little more complicated, it is useful to determine the first and second derivatives w.r.t. $$x$$ before substituting equation (2.1.3) into (2.1.1).

The first derivative of $$y^2_H$$ w.r.t. $$x$$ is


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$$ \frac{d(y^2_H)}{dx} := \frac{d}{dx} \left(\frac{x}{2}log \left(\frac{1+x}{1-x}\right) - 1 \right) $$     (2.1.7)
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$$ \frac{d(y^2_H)}{dx} := \frac{1}{2}log \left(\frac{1+x}{1-x}\right) + \frac{x}{2} \left(\frac{1-x}{1+x}\right) \left(\frac{(1-x) + (1+x)}{(1-x)^2}\right) $$     (2.1.8)
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$$ \frac{d(y^2_H)}{dx} := \frac{1}{2}log \left(\frac{1+x}{1-x}\right) + \frac{x}{1-x^2} $$     (2.1.9)
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The second derivative of $$y^2_H$$ w.r.t. $$x$$ is


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$$ \frac{d^2(y^2_H)}{dx^2} := \frac{d^2\left(\displaystyle \frac{1}{2}log \left(\frac{1+x}{1-x}\right)\right)}{dx^2} + \frac{d^2\left(\displaystyle \frac{x}{1-x^2}\right)}{dx^2} $$     (2.1.10)
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$$ \frac{d^2(y^2_H)}{dx^2} := \frac{1}{2} \left(\frac{1-x}{1+x}\right) \left(\frac{(1-x) + (1+x)}{(1-x)^2}\right) + \frac{(1-x^2) - x(-2x)}{(1-x^2)^2} $$     (2.1.11)
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$$ \frac{d^2(y^2_H)}{dx^2} := \frac{1}{1-x^2} + \frac{1+x^2}{(1-x^2)^2} $$     (2.1.12)
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Substituting equations (2.1.2), (2.1.9), and (2.1.12) into (2.1.1) gives


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$$ L_2(y^2_H) := (1-x^2)\left(\frac{1}{1-x^2} + \frac{1+x^2}{(1-x^2)^2}\right) - 2x\left(\frac{1}{2}log \left(\frac{1+x}{1-x}\right) + \frac{x}{1-x^2} \right) + 2\left(\frac{x}{2}log \left(\frac{1+x}{1-x}\right)\right) = 0 $$     (2.1.13)
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$$ L_2(y^2_H) := 1 + \frac{1+x^2}{1-x^2} - x\;log\left(\frac{1+x}{1-x}\right) - \frac{2x^2}{1-x^2} + x\;log\frac{1+x}{1-x} - 2 $$     (2.1.14)
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$$ L_2(y^2_H) := 1 + \cancelto{1}{\left(\frac{1+x^2}{1-x^2} - \frac{2x^2}{1-x^2}\right)} + \cancelto{0}{\left(x\;log\left(\frac{1+x}{1-x}\right) - x\;log\left(\frac{1+x}{1-x}\right)\right)} - 2 $$     (2.1.15)
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$$ L_2(y^2_H) := 2 - 2 = 0 $$     (2.1.16)
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which shows that $$y^2_H$$ is a homogeneous solution to $$L_2(\cdot).$$

Given

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$$ p(x) = k_1e^{-x} + x - 1 $$     (2.2.1)
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Find
Verify that equation (2.2.1) is indeed the solution for


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$$ p' + p = x $$ (2.2.2)
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Solution
Taking the derivative of equation (2.2.1) w.r.t. $$ x $$ gives


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$$ p(x)' = \frac{d (k_1e^{-x})}{dx} + \frac{d (x)}{dx} - \frac{d (-1)}{dx} $$     (2.2.3)
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$$ p(x)' = -k_1e^{-x} + 1 $$     (2.2.4)
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Adding equations (2.2.1) and (2.2.4) gives


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$$ p(x)' + p(x) = \underbrace{(-k_1e^{-x} + 1)}_{\displaystyle p'(x)} + \underbrace{(k_1e^{-x} + x - 1)}_{\displaystyle p(x)} $$     (2.2.5)
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$$ p(x)' + p(x) = \cancelto{0}{(-k_1e^{-x} + k_1e^{-x})} + x + \cancelto{0}{(1 - 1)} $$     (2.2.6)
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$$ p(x)' + p(x) = x $$ (2.2.7)
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Which shows that equation (2.2.1) is indeed the solution for equation (2.2.2).

Given
General N1-ODEs:


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$$ G(y',y,x) = 0 $$     (2.3.1)
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Particular Class of N1-ODEs:


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$$ M(x,y) + N(x,y)\underbrace{y'}_{\displaystyle \frac{dy}{dx}} = 0 $$     (2.3.2)
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Find
Show that equation (2.3.2) is affine in $$y'$$, and that in general an N1-ODE, even if it is not the most general N1-ODE as represented by equation (2.3.1). Give an example of a more general N1-ODE.

Show $$M(x,y) + N(x,y)y' = 0$$ is affine in $$y'$$
Affine function is defined as a linear function that has displacement added onto it.

$$N(x,y)y'$$ in equation (2.3.2) is linear since it satisfies the requirement as shown below:


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$$ N(x,y)(\alpha \nu + \beta u) = N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.3)
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Since equation (2.3.2) also has the $$M(x,y)$$ term, which displaces the term $$N(x,y)y'$$ that is linear in $$y'$$, the equation is affine in $$y'.$$

Show $$M(x,y) + N(x,y)y' = 0$$ is in general an N1-ODE
It is clear from looking at equation (2.3.2) that it is a function of $$x, y,$$ and $$y'$$. Since $$y'$$ is the highest degree derivative of $$y,$$ the equation is a 1-ODE.

To show that equation (2.3.2) is nonlinear, we substitute $$\alpha \nu + \beta u$$ for $$y',$$ as was done for equation (2.3.3).


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$$ M(x,y) + N(x,y)(\alpha \nu + \beta u) = M(x,y) + N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.4)
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It is clear that the right side of equation (2.3.4) does not equal


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$$ M(x,y) + N(x,y)(\alpha \nu) + M(x,y) + N(x,y)(\beta u) = 2M(x,y) + N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.4)
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Therefore, equation (2.3.2) is in general an N1-ODE.

Given

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$$ y^1_H(x) = x =: P_1(x) $$     (2.4.1)
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$$ y^2_H(x) = \frac{x}{2} log \left(\frac{1+x}{1-x}\right) - 1 =: Q_1(x) $$     (2.4.2)
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Find
Show that equations (2.4.1) and (2.4.2) are linearly independent, i.e.,


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$$ \forall \alpha \in \mathbb R, \, y^1_H(\cdot) \ne \alpha \, y^2_H(\cdot) $$     (2.4.3)
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$$\left[ \right.$$i.e., for any given $$\alpha$$, show that $$\exist \hat x \; $$ such that $$ y^1_H(\hat x) \ne \alpha \, y^2_H(\hat x)\left.\right]$$

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

Given

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$$ \phi(x,y) = x^2y^\frac{3}{2} + log(x^3y^2) = k $$ (2.5.1)
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Find
Verify that


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$$ G(y',y,x) = \frac{d}{dx}\phi(x,y) = 0 $$     (2.5.2)
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and show that equation (2.5.2) is an N1-ODE.

Given
Assuming that $$ \phi(x,y) $$ is smooth, we have


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$$ \phi_{xy}(x,y) = \phi_{yx}(x,y) \; \; \; \; \; \; \; \left(i.e., \frac{\partial{^2\phi(x,y)}}{\partial{x}\partial{y}} =  \frac{\partial{^2\phi(x,y)}}{\partial{y}\partial{x}} \right) $$     (2.6.1)
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Find
Review calculus, and find the minimum degree of differentiability of the function $$ \phi(x,y) $$ such that equation (2.6.1) is satisfied. State the full theorem and provide a proof.

Clairaut's Theorem
The equality of a function's mixed partial derivatives is theorized by Clairaut's Theorem, which states that, if $$f_xy$$ and $$f_yx$$ are continuous in an open disk containing $$(a,b),$$ then $$f_{xy}(a,b) = f_{yx}(a,b).$$

Proof of Clairaut's Theorem$$^{[1]}$$
Given a small nonzero number $$h,$$ let


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$$ \Delta(h) = (f(a+h, b+h)-f(a+h,b)) - (f(a, b+h) - f(a,b)) $$     (2.6.2)
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Define


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$$ g(x) = f(x,b+h), $$     (2.6.3)
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the change in $$f$$ given a small change in $$y,$$ and let


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$$ \Delta(h) = g(a+h) - g(a), $$     (2.6.4)
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the change in $$g$$ given a small change in $$x.$$

By hypothesis, for small enough $$h, g$$ is differentiable on the interval determined by $$a$$ and $$a+h,$$ and so by the Mean Value Theorem, there exists a number $$c$$ between $$a$$ and $$a+h$$ such that


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$$ g'(c) = \frac{g(a+h)-g(a)}{h} $$     (2.6.5)
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and therefore


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$$ \Delta(h) = g(a+h)-g(a) = g'(c)h = h(f_x(c,b+h)-f_x(c,b)) $$     (2.6.6)
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Also since $$f_x$$ is differentiable w.r.t. to $$y,$$ we can apply the Mean Value Theorem again, and so there exists a number $$d$$ between $$b$$ and $$b+h$$ such that


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$$ \frac{\partial}{\partial y}(f_x) = f_{xy}(c,d) = \frac{f_x(c,b+h)-f_x(c,b)}{h} $$     (2.6.7)
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That is,


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$$ f_x(c,b+h)-f_x(c,b) = hf_{xy}(c,d) $$     (2.6.8)
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Combine equations (2.6.6) and (2.6.8) to get


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$$ \Delta(h) = h(hf_{xy}(c,d)) = h^2f_{xy}(c,d) $$     (2.6.9)
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As $$h \rightarrow 0, (c,d) \rightarrow (a,b),$$ and so since $$f_{xy}$$ is continuous at $$(a,b),$$


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$$ \lim_{h\to0}\frac{\Delta(h)}{h^2} = \lim_{(c,d)\to(a,b)}f_{xy}(x,d) = f_{xy}(a,b) $$     (2.6.10)
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Similarly we can write


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$$ \Delta(h) = (f(a+h,b+h)-f(a,b+h)) - (f(a+h,b)-f(a,b)) $$     (2.6.11)
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and use the same procedure as above to obtain $$f_{yx}$$ at $$(a,b)$$ to get


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$$ \lim_{h\to0}\frac{\Delta(h)}{h^2} = f_{yx}(a,b) $$     (2.6.12)
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Thus $$f_{xy}(a,b) = f_{yx}(a,b),$$ which proves equation (2.6.1).

The theorem does not set any condition for minimum differentiability; only that the function be continuous in an open disk containing (a,b) [i.e. the point where equation (2.6.1) is being tested].

Given

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$$ \frac{d\phi(x, y(x))}{dx} = 75x^4 + (\cos y)y' = 0 $$     (2.7.1)
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and


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$$ y(x) = \sin ^{-1}(k - 15x^5) $$     (2.7.2)
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Find
Verify that the equation (2.7.2) is indeed a solution of the non-linear 1st order differential equation (N1 - ODE) represented by the equation (2.7.1)

Given
Consider an N1-ODE with the particular form


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$$ M(x,y) + N(x,y)y' = 0 $$     (2.9.1)
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which satisfies the 1st exactness condition, but does not satisfy the 2nd exactness condition,


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$$ M_y(x,y) \ne N_x(x,y) $$     (2.9.2)
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Using the Euler integrating factor method, it is possible to find an integrating factor so that the resulting N1-ODE satisfies both exactness conditions.


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$$ h(x,y)[M(x,y) + N(x,y)y'] = 0 $$     (2.9.3)
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$$ \underbrace{hM}_{\displaystyle \bar M} + \underbrace{hN}_{\displaystyle \bar N}y' = 0 $$     (2.9.4)
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Applying the 2nd exactness condition to equation (2.9.4) gives


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$$ \bar M_y = \bar N_x $$     (2.9.5)
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$$ \begin{cases} & \bar M_y = h_yM + hM_y \\ & \bar N_x = h_xN + hN_x \end{cases} $$     (2.9.6)
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Where substituting for $$\bar M_y$$ and $$\bar N_x$$ in equation (2.9.5) and rearranging all the terms to the left side of the equation yields,


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$$ h_xN -h_yM + h(N_x - M_y) = 0 $$     (2.9.7)
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If it assumed that $$h_x(x,y) = 0,$$ so that $$h$$ is a function of $$y$$ only, then we can rearrange equation (2.9.7) so that it becomes


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$$ \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) $$     (2.9.8)
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Find
Using equation (2.9.8), solve for $$h.$$

Solution
Integrating both sides of equation (2.9.8) w.r.t. $$y$$


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$$ \int \frac{h_y}{h}dy = \int \frac{1}{M}(N_x - M_y)dy $$     (2.9.9)
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$$ log \;h(y) = \int \frac{1}{M}(N_x - M_y)dy $$     (2.9.10)
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Solving for $$h$$ applying the exponential function on both sides of the equation,


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$$ h(y) = exp\left[\int \frac{1}{M}(N_x - M_y)dy\right] $$     (2.9.11)
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