User:Egm6321.f10.team5/hw2

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

Given

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$$\displaystyle M(x,y) + N(x,y)y' = 0 =: F(y) $$ $$\displaystyle (1-1) $$
 * }
 * }

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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 * }


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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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 * }


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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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 * }


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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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 * }


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$$\displaystyle \alpha F(u) = \alpha (M(x,u) + N(x,u)u') $$ $$\displaystyle (1-8) $$
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 * }


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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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 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * 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)


 * }
 * }
 * }

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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 * }


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$$\displaystyle F(u) + F(v) = xu + xu' + xv + xv' $$ $$\displaystyle (1-12) $$
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 * }


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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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 * }


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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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 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * Therefore, (1-10) is an example of L1-ODE
 * Therefore, (1-10) is an example of L1-ODE

<Author> Egm6321.f10.team5.oh
 * }
 * }
 * }

Egm6321.f2010.team5.riveros

=Problem 2 - Verifying 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
 * }
 * }

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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 * }


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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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 * }


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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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 * }


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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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 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * 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)


 * }
 * }
 * }

<Author> Egm6321.f10.team5.oh

=Problem 3 - Graph nonlinear variables=

Find
Plot:


 * $$y_H^1(x)$$ & $$y_H^2(x)$$

and show that


 * $$y_H^1(x) \ne \alpha y_H^2(x)$$

for any $$x$$ and any $$\alpha$$ that belong to $$\mathbb{R}$$ where


 * $$y_H^1(x) = x$$

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

Solution


In the above figure, the points of $$y_H^1(x)$$ are marked as "x" and the points of $$y_H^2(x)$$ are maked as "o".

To show that $$y_H^1(x)$$ and $$y_H^2(x)$$ are linearly independent, we use Wronskian.
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$$\displaystyle W(y_H^1,y_H^2)(x) = \left| \begin{matrix} {y_H^1} & {y_H^2} \\ {y_H^1}' & {y_H^2}'\end{matrix} \right| = \left| \begin{matrix} x & {\frac x 2 \log \frac {1+x} {1-x} - 1} \\ 1 & {\frac x {1-x^2} + \frac 1 2 \log \frac {x+1} {x-1}} \end{matrix} \right| $$ <p style="text-align:right;">$$\displaystyle (3.1) $$
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 * style="width:92%; padding:10px; border:0px solid #8888aa" |
 * style="width:92%; padding:10px; border:0px solid #8888aa" |
 * $$\displaystyle
 * $$\displaystyle

= x (\frac x {1-x^2} + \frac 1 2 \log \frac {x+1} {x-1}) - (\frac x 2 \log \frac {1+x} {1-x} - 1) 1 \ne 0 $$ <p style="text-align:right;">$$\displaystyle (3.2) $$
 * }

Therefore we show that $$y_H^1(x)$$ and $$y_H^2(x)$$ are linearly independent.

=Problem 4: Derivation and proof of exact N1-ODE=

Find
Find $$F(x,y,y') = \frac {d\phi} {dx} (x,y)$$ where $$\phi(x,y) = x^2 y^{\frac 3 2} + \log (x^3 y^2) = k$$ and verify that $$F$$ is exact N1_ODE and invent 3 more.

Solution
First, differentiate $$\frac {d\phi} {dx} (x,y)$$ To do so, we use the product rule and the chain rule for the first term in $${phi}(x,y)$$ and a Logarithmic derivative for the second term. So then $$F(x,y,y')$$ comes out to be: $$F(x,y,y') = 2 x y^{\frac 3 2} + \frac 3 {x} + [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y] y'=0$$

where the constant $$k$$ goes to zero after differentiation.

Verification
For $$F(x,y,y')$$ to meet the criteria of an exact N1_ODE it must meet three conditions: 1) Have an integral $$\frac {d\phi} {dx} (x,y)$$ that exists, which it clearly does.

2) Fit the form $$M(x,y) + N(x,y) y' = 0$$, which it does if we make
 * $$M(x,y) = 2 x y^{\frac 3 2} + \frac 3 {x}$$

and
 * $$N(x,y) = [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y]$$

3) If $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$, which we can see that it does as well.

Therefore it can be said that $$F(x,y,y')$$ is an exact non-linear first order ordinary differential equation.

Three other examples
Here are three other examples of exact N1_ODE
 * 1) $$F(x,y,y') = 5 x^4 ln(y) + x^5 \frac 1 y y'$$
 * 2) $$F(x,y,y') = 5 x^{\frac 3 2} y^2 + 4 x^{\frac 5 2} y y' $$
 * 3) $$F(x,y,y') = 3 \frac 1 x y^3 + 9 ln(x) y^2 y' $$

=Problem 5 - Prove that an N1-ODE cannot be exact=

Given
A first order nonlinear ordinary differential equation (N1-ODE) may be represented as follows:


 * $$ F(x,y,y') = M(x,y) + N(x,y)f(y') = 0 \! $$

Find
Find $$ f(y') \! $$ such that there is no analytical solution to $$ f(y') = - \frac{M}{N} \! $$ (i.e. such N1-ODE cannot be exact).

Solution
The first condition of exactness is defined as follows.

For $$ F(x,y,y') = 0 \! $$ to be exact, it must be in the form $$ F(x,y,y') = M(x,y) + N(x,y)y' = 0 \! $$.

From this condition, it is understood that if $$ f(y') \not= y' \!$$, then $$ f(y') \! $$ must be solved explicitly for $$ y' \!$$ in order to force the equation into the form shown in the first condition definition above.

For the present problem, we are interested in finding a function $$ f(y') \! $$ that cannot be solved explicitly. To state this another way, we desire $$ f(y') \!$$ such that $$ f^{-1}(y') \!$$ cannot be determined analytically.

Let $$ f(y') = y'(3+y') \!$$, for example.


 * $$ F(x,y,y') = M(x,y) + N(x,y)f(y') = M(x,y) + N(x,y)y'(3+y') = 0 \! $$


 * $$ y'(3+y') = - \frac{M}{N} \! $$

Clearly, this expression cannot be solved explicitly for $$ y' \! $$. Furthermore, the N1-ODE is not exact because this particular $$ f(y') \!$$ cannot satisfy the first condition of exactness.

=Problem #6 - Verify solution to N1_ODE=

Given

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y(x)=sin^{-1}(k-15x^5)  (4) $$ 75x^4+cos(y)y'=0        (1) $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Verify that equation 4 satisfies equation 1.

Solve
First determine $$\frac{d}{dx}y(x)$$ for equation (4).


 * $$\begin{align}

&\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\\ &y'(x) = \frac{-75x^4}{\sqrt{1-(k-15x^5)^2}}\\ \end{align}$$ solve eq 1 for y'(x)
 * $$\begin{align}

&cos(sin^{-1}(x))=sin(cos{-1}(x))=\sqrt{1-x^2}\\ &75x^4+cos(y)y'=75x^4+cos(sin^{-1}(k-15x^5))y'=> y'(x)=\frac{-75x^4}{\sqrt{1-(k-15x^5)^2}}\\ \end{align}$$

$$y(x)'$$ from equation 4 is equal to $$y(x)'$$ from equation 1. Thus equations 4 satisfies equation 1.

=Problem 7 - Solving Non-Homog. L1-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 $$\displaystyle y(x) $$
 * }
 * }

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
 * }
 * }

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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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }


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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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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }


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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;">$$\displaystyle (7-12) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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;">$$\displaystyle (7-14) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }


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

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
 * }
 * }


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

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
 * }
 * }


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

<3> \ u := s^2 + 1 $$
 * $$\displaystyle
 * $$\displaystyle

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

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

$$\displaystyle = 2 \sqrt{ \left(x^2 + 1 \right)} $$
 * }
 * }
 * }


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

$$\therefore \ y(x) = 2 + \underbrace{\frac{k}{\sqrt{ \left(x^2 + 1 \right)}}}_{k \ : \ int. \ const} $$ <p style="text-align:right;">$$\displaystyle (7-17) $$
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

<Author> Egm6321.f10.team5.

=Problem 8 - Show that k1 is not necessary=

Find: $$\left(\alpha \right)$$ show that k1 is not necessary
If we integrate left side of (6) pg 10-3(lecture) it will reveal how k1 cancels during integration.


 * $$ \int_{}^{x} \frac{ {h}_{x}}{h}dx= \int_{}^{s}n \left(s \right)ds$$


 * $$ \left[ \ln \left(h \right)+ {k}_{1}\right]- \left[ \ln \left(h \left(0 \right) \right)+ {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * $$\ln \left(h \right)+ \ln \left(h \left(0 \right) \right)+\left[{k}_{1}- {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$
 * $$\ln \left(h \right)+ \ln \left(h \left(0 \right) \right)+\left[{k}_{1}- {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$

k1 term cancels on left side
 * }
 * }
 * }

$$\left( \beta \right)$$ Show that equation (6) pg 10-3(lecture) agrees with King pg 512
The linear, 1st order, ODE in King is written in the form:


 * $$\displaystyle \frac{dy}{dx}+P \left(x \right)y=Q \left(x \right)$$

The structure of the solution in King is written as:


 * $$\displaystyle y={y}_{h}+ {y}_{p}$$

where the particular solution is in the form:


 * $$ {y}_{p}= \exp \left\{- \int_{}^{x}P \left(x \right)dt \right\} \int_{}^{x}Q \left(s \right) \exp \left\{ \int_{}^{s}P \left(t \right)dt \right\}ds$$

However, equation 6 pg10-3 (lecture) is a particular solution written in the form:


 * $${y}_{p} \left(x \right)= \frac{1}{h \left(x \right)} \int_{0}^{x}h \left(s \right)b \left(s \right)ds$$

Where, from (1) and (5) pg 10-2 (lecture) we get:


 * $$h \left(x\right)= \exp \left[ \int_{}^{x}n \left(s \right)ds\right]$$


 * $$b \left(x\right)= \frac{R \left(x \right)}{P \left(x \right)}=Q \left(x \right)$$


 * $$n \left(x\right)=P \left(x \right)$$

Finally, plugging h(x),b(x), n(x) into the particular solution form given by King we get:


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * $$ {y}_{p}= \exp \left\{- \int_{}^{x}h \left(s \right)ds\right\} \int_{}^{x}n \left(s \right) \exp \left\{ \int_{}^{x}b \left(s \right)ds \right\}ds$$
 * $$ {y}_{p}= \exp \left\{- \int_{}^{x}h \left(s \right)ds\right\} \int_{}^{x}n \left(s \right) \exp \left\{ \int_{}^{x}b \left(s \right)ds \right\}ds$$


 * }
 * }
 * }

$$\left( \gamma \right)$$ Solve the following equation

 * $$\frac{dy}{dx}+ {a}_{o}y=0$$


 * $$\displaystyle {a}_{o}=x $$

Solve by separation of variables:


 * $$ \int_{}^{y} \frac{dy}{y}+ \int_{}^{x}xdx=0$$


 * $$ \ln \left(y \right)+ \frac{ {x}^{2}}{2}=C$$


 * $$\ln \left(y \right)=-\frac{ {x}^{2}}{2}+C$$


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center" |
 * $$y= \exp \left[- \frac{ {x}^{2}}{2}+C \right]= {y}_{h}$$
 * $$y= \exp \left[- \frac{ {x}^{2}}{2}+C \right]= {y}_{h}$$


 * }
 * }
 * }

=Problem 9 - Exactness of an N1-ODE=

Given

 * $$ a(x) = sin(x^3) \! $$


 * $$ b(x) = cos(x) \! $$


 * $$ c(y) = exp(2y) \! $$

Find
1. Using $$ \bar b(x)c(y)y'+a(x) \bar c(y) = 0 \! $$, find a N1-ODE that is either exact or can be made exact.

2. Find $$ \phi(x,y) = k \! $$.

Solution
First, calculate $$\bar b(x)$$ and $$\bar c(x)$$. $$\bar b(x) = \int^x b(s) ds = \int^x \cos s \ ds = \sin x \qquad\qquad\qquad\qquad\qquad(9.1)$$ $$\bar c(x) = \int^y c(y) ds = \int^y \exp(2s) ds = \frac 1 2 \exp(2y) \qquad\qquad\qquad\qquad\qquad(9.2)$$ We express the equation completely. $$\sin x \ \exp(2y) y' + \sin x^3 \ \frac 1 2 \exp(2y) = 0 \qquad\qquad\qquad\qquad\qquad(9.3)$$ Dividing the both sides by $$\exp(2y)$$, we get $$\sin x \ y' + \sin x^3 \ \frac 1 2 = 0 \qquad\qquad\qquad\qquad\qquad(9.4)$$ To determine if the equation is exact or not $$\frac {\partial M} {\partial y} = 0 \qquad\qquad\qquad\qquad\qquad(9.5)$$ $$\frac {\partial N} {\partial x} = \cos x \ y' \qquad\qquad\qquad\qquad\qquad(9.6)$$ The eqauation is not exact since $$\frac {\partial M} {\partial y} \ne \frac {\partial N} {\partial x}$$, so it can be made exact. $$\sin x \ y' = - \sin x^3 \ \frac 1 2 \qquad\qquad\qquad\qquad\qquad(9.7)$$ Since the equation is the first-order derivative with respect to $$y$$, we need to take an integral with respect to y. $$y' = - \frac 1 2 \frac {\sin x^3} {\sin x} \qquad\qquad\qquad\qquad\qquad(9.8)$$ To express in the form of $$\phi(x,y) = k $$, we take the integration on each side. $$y = - \frac 1 2 \int \frac {\sin x^3} {\sin x} dx + k \qquad\qquad\qquad\qquad\qquad(9.9)$$ $$y + \frac 1 2 \int \frac {\sin x^3} {\sin x} dx = k \qquad\qquad\qquad\qquad\qquad(9.10)$$

=References=

=Contributions=

Raul Riveros Solved problem 1, added team member content to team Wiki page, unified report format.

Sang Min Oh Solved problems 1, 2, and 7.

Michael Faraone Solved problem 6. Helped with problems 3, 4, and 5.

Scott Elliot Solved problem 8.

Rob Carroll Provided primary solution for problem 4 and secondary solutions for problems 3 and 8

Jo Solved problem 3, problem 4, and problem 9.

Michael Steinberg Solved problem 5 and problem 9 (partially).