User talk:Egm6321.f10.team5.oh/hw2

= Problem #1 =  Verify Non-linear 1st Order ODE(N1-ODE)

Given

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$$\displaystyle M(x,y) + N(x,y)y' = 0 =: F(y) $$ $$\displaystyle (1-1) $$
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Find
1) Verify the above equation is in General N1-ODE

2) Give an example for L1-ODE

Solve
== > (1-1) is Ordinary Differential Equation(ODE)
 * (1-1) is the equation that has one independent variable and its derivatives with respect to the variable.
 * Highest order of derivative of (1-1) is 1 == > (1-1) is 1st Order Ordinary Differential Equation(ODE)
 * To be a Linear Equation, (1-1) must be satisfied with following two conditions
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 * To be a Linear Equation, (1-1) must be satisfied with following two conditions
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$$\displaystyle F(u+v) = F(u) + F(v) $$ $$\displaystyle (1-2) $$
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$$\displaystyle F(\alpha u) = \alpha F(u) $$ $$\displaystyle (1-3) $$
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Step 1
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$$\displaystyle F(u+v) = M(x,u+v) + N(x,u+v)(u+v)' $$ $$\displaystyle (1-4) $$
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$$\displaystyle F(u) + F(v) = M(x,u) + N(x,u)u' + M(x,v) + N(x,v)v' $$ $$\displaystyle (1-5) $$
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$$\displaystyle F(u+v) {\neq} F(u) + F(v) $$ $$\displaystyle (1-6) $$
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Step 2
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$$\displaystyle F(\alpha u) = M(x,\alpha u) + N(x,\alpha u)(\alpha u)' $$ $$\displaystyle (1-7) $$
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$$\displaystyle \alpha F(u) = \alpha (M(x,u) + N(x,u)u') $$ $$\displaystyle (1-8) $$
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$$\displaystyle F(\alpha u) {\neq} \alpha F(u) $$ $$\displaystyle (1-9) $$
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(1-6) and (1-9) ==> (1-1) is a Non-linear equation 


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 * Therefore, (1-1) is a General Non-linear 1st Order Ordinary Differential Equation(N1-ODE)
 * Therefore, (1-1) is a General Non-linear 1st Order Ordinary Differential Equation(N1-ODE)


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Solution of 2)

example :
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$$\displaystyle xy + xy' = 0 =: F(y) $$ $$\displaystyle (1-10) $$
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Step 1
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$$\displaystyle F(u+v) = x(u+v) + x(u+v)' $$ $$\displaystyle (1-11) $$
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$$\displaystyle F(u) + F(v) = xu + xu' + xv + xv' $$ $$\displaystyle (1-12) $$
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$$\displaystyle F(u+v) = F(u) + F(v) $$ $$\displaystyle (1-13) $$
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Step 2
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$$\displaystyle F(\alpha u) = x(\alpha u) + x(\alpha u)' $$ $$\displaystyle (1-14) $$
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$$\displaystyle \alpha F(u) = \alpha (xu + xu') $$ $$\displaystyle (1-15) $$
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$$\displaystyle F(\alpha u) = \alpha F(u) $$ $$\displaystyle (1-16) $$
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(1-13) and (1-16) ==> (1-10) is a linear equation 


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 * Therefore, (1-10) is an example of L1-ODE
 * Therefore, (1-10) is an example of L1-ODE

<Author> Egm6321.f10.team5.oh
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= Problem #2 = Verify (7) from p. 7-1 is N1-ODE

Given

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\underbrace{(4x^7 + sin y)}_{M(x,y)} + \underbrace{(x^2y^3)}_{N(x,y)}y' = 0 =: F(y) $$ <p style="text-align:right;">$$\displaystyle (2-1) $$
 * $$\displaystyle
 * $$\displaystyle
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Find
 Verify (2-1) is N1-ODE

Solve
== > (2-1) is Ordinary Differential Equation(ODE)
 * (2-1) is the equation that has one independent variable and its derivatives with respect to the variable.
 * Highest order of derivative of (2-1) is 1 == > (2-1) is 1st Order Ordinary Differential Equation(ODE)
 * To be a Linear Equation, (2-1) must be satisfied with following two conditions
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 * To be a Linear Equation, (2-1) must be satisfied with following two conditions
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$$\displaystyle F(u+v) = F(u) + F(v) $$ <p style="text-align:right;">$$\displaystyle (2-2) $$
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$$\displaystyle F(\alpha u) = \alpha F(u) $$ <p style="text-align:right;">$$\displaystyle (2-3) $$
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Step 1
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$$\displaystyle F(u+v) = 4x^7 + sin(u+v) + x^2(u+v)^3(u+v)' $$ <p style="text-align:right;">$$\displaystyle (2-4) $$
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$$\displaystyle F(u) + F(v) = 4x^7 + sin u + x^2u^3u' + 4x^7 + sin v + x^2v^3v' $$ <p style="text-align:right;">$$\displaystyle (2-5) $$
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$$\displaystyle F(u+v) {\neq} F(u) + F(v) $$ <p style="text-align:right;">$$\displaystyle (2-6) $$
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Step 2
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$$\displaystyle F(\alpha u) = 4x^7 + sin(\alpha u) + x^2(\alpha u)^3(\alpha u)' $$ <p style="text-align:right;">$$\displaystyle (2-7) $$
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$$\displaystyle \alpha F(u) = \alpha (4x^7 + sin u + x^2u^3u') $$ <p style="text-align:right;">$$\displaystyle (2-8) $$
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$$\displaystyle F(\alpha u) {\neq} \alpha F(u) $$ <p style="text-align:right;">$$\displaystyle (2-9) $$
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(2-6) and (2-9) ==> (2-1) is a Non-linear equation 


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 * Therefore, (2-1) is a Non-linear 1st Order Ordinary Differential Equation(N1-ODE)
 * Therefore, (2-1) is a Non-linear 1st Order Ordinary Differential Equation(N1-ODE)


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<Author> Egm6321.f10.team5.oh

= Problem #3 =

Solve
= Problem #4 =

Solve

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$$ F(x,y,y') = {\frac {d{\phi}(x,y)}{d x}} = {\frac} + {\frac} {\frac{dy}{dx}}$$

(4.1)
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$$\begin{align} & M(x,y):={\phi}_{x}(x,y): =\frac{\partial \phi(x,y)}{\partial x} \\ & N(x,y):={\phi}_{y}(x,y): =\frac{\partial \phi(x,y)}{\partial y} \\ \end{align}$$

(4.2)
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So, that is
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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' $$

(4.3)
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From the given $${{\phi}\left(x,y\right)}$$, we can get M(x,y) and N(x,y)


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$$\begin{align} &M\left(x,y\right) :={\phi}_{x}(x,y)= 2 x y^{\frac{3}{2}} + {\frac{3}{x}}\\ &N(x,y) := {\phi}_{y}(x,y)= {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}}\\ \end{align} $$      (4.4)
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Follow the $$ 1^{st} $$ condition of exactness,


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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' = \left( 2 x y^{\frac{3}{2}} + {\frac{3}{x}} \right) + \left( {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}} \right) y' = 0 $$      (4.5)
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$$ y' = - \frac{M\left(x,y\right)}{N(x,y)} $$      (4.6)
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And then, apply $$ 2^{nd} $$ condition of exactness,


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$$ \underbrace{M_{y}\left(x,y\right)}_{\frac{\partial M(x,y)}{\partial y}}= \underbrace{N_{x}(x,y)}_{\frac{\partial N(x,y)}{\partial x}} $$      (4.7)
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$$ {\frac {{\partial}M(x,y)} {\partial y} } = {\frac {{\partial} \left( 2 x y^{\frac{3}{2}}+{\frac{3}{x}}\right)} {\partial y}} = 3 x y^{\frac{1}{2}} = {\frac {{\partial}\left( {\frac{3}{2}} y^{\frac{1}{2}} x^{2} + {\frac{2}{y}}\right)} {\partial x}} = {\frac {{\partial} N(x,y)}{\partial x}} $$      (4.8)
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So, F is exact N1_ODE.

And 3 more equations,


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$$\displaystyle {\phi}(x, y) = x^{5} + y^{3} = k $$ (4.9)
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$$ {\phi}(x, y) = x^{4} \ ln \ y = k $$ (4.10)
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$$ {\phi}(x, y) = {\frac{1}{4}}( x^{4} + {6} x^{2} y^{2} + y^{4})= k $$ (4.11)
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= Problem #5 =

Solve
= Problem #6 =

Solve
Total differential of function $$\phi\left(x,y\right)$$,

As$$\phi\left(x,y\right)=k=const$$, then $$ d\mathbf{\phi}=0 $$, We can rewrite the equation 6.5,

For the given problem,

So, we can verify that the equation 6.1 could be converted to the form of the first condition of exactness, equation 6.2.

The mixed derivatives of $$\begin{align}\phi\end{align}$$, $$\begin{align}\phi_{xy},\phi_{yx} \end{align}$$ are both zero, meeting the second condition of exactness. This is clearly seen from equation 6.7 as $$\begin{align}\phi_{x} \end{align}$$ is only a function of $$\begin{align}x\end{align}$$ while $$\begin{align} \phi_{y} \end{align} $$ is just a function of $$\begin{align}y \end{align}$$.

= Problem #7 = '''Solving Non-Homog. L-1-ODE-VC'''

Given

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$$\displaystyle a_1(x)y' + a_0(x)y = b(x) $$ <p style="text-align:right;">$$\displaystyle (7-1) $$
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Find

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1) Assume $$\displaystyle a_1(x) = 1, a_0(x) = x, b(x) = 2x + 3, $$ Find $$\displaystyle h(x) $$ and $$\displaystyle y(x) $$

2) Assume $$\displaystyle a_1(x) {\neq} 0, \ \forall x$$ (7-1) becomes $$\displaystyle y' + \underbrace{\frac{a_0(x)}{a_1(x)}}_{P(x)}y = \underbrace{\frac{b(x)}{a_1(x)}}_{Q(x)},  $$ Find $$\displaystyle y(x)  $$ in terms of $$\displaystyle a_0, a_1, b  $$ <p style="text-align:right;">$$\displaystyle (7-2) $$
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3) Assume $$\displaystyle a_1(x) = x^2 + 1, a_0(x) = x, b(x) = 2x, $$ Find $$\displaystyle h(x) $$ and $$\displaystyle y(x) $$
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Solve
According to Lecture(Mtg 10. Tue, 14 Sep 10), (5) p.10-2, (1) p.10-3 and (6) p.10-3 We can rewrite,


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y' + a_0(x)y = b(x) $$ <p style="text-align:right;">$$\displaystyle (7-3) $$
 * $$\displaystyle
 * $$\displaystyle
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h(x) = exp[\int^{x} a_0(s) ds] $$ <p style="text-align:right;">$$\displaystyle (7-4) $$
 * $$\displaystyle
 * $$\displaystyle
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y(x) = \frac{1}{h(x)} \int^{x} h(s)b(s) ds] $$ <p style="text-align:right;">$$\displaystyle (7-5) $$
 * $$\displaystyle
 * $$\displaystyle
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Solution of 1)
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h(x) = exp[\int^{x} s ds] $$ <p style="text-align:right;">$$\displaystyle (7-6) $$
 * $$\displaystyle
 * $$\displaystyle
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$$h(x) = exp(\frac{1}{2} x^2)$$ <p style="text-align:right;">$$\displaystyle (7-7) $$
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y(x) = \frac{1}{exp(\frac{1}{2} x^2)} \int^{x} exp(\frac{1}{2} s^2)(2s + 3) ds] $$ <p style="text-align:right;">$$\displaystyle (7-8) $$
 * $$\displaystyle
 * $$\displaystyle
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y(x) = \frac{1}{exp(\frac{1}{2} x^2)} [\underbrace{2 \int^{x} exp(\frac{1}{2} s^2)s \ ds}_{<1>} + 3 \int^{x} exp(\frac{1}{2} s^2) ds] $$ <p style="text-align:right;">$$\displaystyle (7-9) $$
 * $$\displaystyle
 * $$\displaystyle
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<1> \ t := \frac{1}{2} s^2 $$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle dt = s \ ds, \ ds = \frac{dt}{s} $$

$$\displaystyle <1> \ --> \ 2 \int^{x} exp(t) dt $$

$$\displaystyle = 2 exp(\frac{1}{2} x^2) $$
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$$\therefore \ y(x) = 2 + \frac{3}{exp(\frac{1}{2} x^2)} \int^{x} exp(\frac{1}{2} s^2) ds $$ <p style="text-align:right;">$$\displaystyle (7-10) $$
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Solution of 2) 


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$$h(x) = exp[\int^{x} \frac{a_0(s)}{a_1(x)} ds] $$ <p style="text-align:right;">$$\displaystyle (7-11) $$
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$$y(x) = \frac{1}{exp[\int^{x} \frac{a_0(s)}{a_1(x)} ds]} \int^{x} exp[\int^{x} \frac{a_0(s)}{a_1(x)} ds] \ \frac{b(s)}{a_1(s)} ds $$ <p style="text-align:right;">$$\displaystyle (7-12) $$
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Solution of 3) 


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h(x) = exp[\underbrace{\int^{x} \frac{s}{s^2 + 1} \ ds}_{<2>}] $$ <p style="text-align:right;">$$\displaystyle (7-13) $$
 * $$\displaystyle
 * $$\displaystyle
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<2> \ t := s^2 + 1 $$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle dt = 2s \ ds $$

$$\displaystyle \therefore ds = \frac{1}{2} \frac{dt}{s} $$

$$\displaystyle h(x) = exp[\int^{x} \frac{1}{2} \frac{1}{t} dt] $$

$$\displaystyle h(x) = exp(\frac{1}{2} log (x^2 + 1) ) $$
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$$h(x) = \sqrt{ \left(x^2+1\right) } $$ <p style="text-align:right;">$$\displaystyle (7-14) $$
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y(x) = \frac{1}{\sqrt{ \left(x^2+1\right)}} \int^{x} \sqrt{ \left(s^2+1\right)} \frac{2s}{s^2 + 1} ds $$ <p style="text-align:right;">$$\displaystyle (7-15) $$
 * $$\displaystyle
 * $$\displaystyle
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y(x) = \frac{1}{\sqrt{ \left(x^2+1\right)}} \underbrace{\int^{x} \frac{2s}{\sqrt{ \left(s^2+1\right)}} \ ds }_{<3>} $$ <p style="text-align:right;">$$\displaystyle (7-16) $$
 * $$\displaystyle
 * $$\displaystyle
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<3> \ u := s^2 + 1 $$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle du = 2s \ ds, \ ds = \frac{dt}{2s} $$

$$\displaystyle <3> \ --> \int^{x} \frac{1}{\sqrt{ \left(u \right)}} du $$

$$\displaystyle = 2 \sqrt{ \left(x^2 + 1 \right)} $$
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$$\therefore \ y(x) = 2 + \underbrace{\frac{k}{\sqrt{ \left(x^2 + 1 \right)}}}_{k \ : \ int. \ const} $$ <p style="text-align:right;">$$\displaystyle (7-17) $$
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<Author> Egm6321.f10.team5.oh

= Problem #8 =

Solve
= Problem #9 =