User:Egm6321.f09.team3.Kuljevan.Francisco/HW2

=Problem 1=

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Given
For case #2

$$ h_xN = 0 \! $$

Find
Complete the details on how to obtain h(y) for case #2

Solution
General form



\begin{align} &h_xN - h_yM + (N_x-My) = 0 \qquad \text{investigating the case for} \qquad h_xN = 0 \\ &-h_yM =h(N_x - M_y) \qquad \text{factoring out}\\ &\frac{h_y}{h} = \frac{1}{M}\left(N_x-M_y\right) \qquad \text{integrating both sides gives}\\ &\int\left(\frac{h_y}{h}\right)dy = \int\frac{1}{M}\left(N_x-M_y\right)dy \qquad \text{evaluating the integral}\\ &\ln h =\int\frac{1}{M}\left(N_x-M_y\right)dy \end{align} $$

We know that
 * $$ h_x \underbrace{N}_{\neq 0} = 0 $$

which means that $$h(x,y) = h(y)$$

and that,


 * $$\frac{1}{M(x,y)}\left(N_x(x,y)-M_y(x,y)\right) = -g(y)$$

since

$$\ln h = \underbrace{\int\frac{1}{M}\left(N_x-M_y\right)dy}_{-g(y)}$$

then we arrived to $$h(y)$$


 * $$h(y)=\exp \int g(s)ds$$


 * }

=Problem 2=

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Given
A non-homogeous L1-ODE-VC


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

Find
Show that


 * $$y(x)=\frac{x^{3}}{4}+\frac{C}{x}$$

Solution
The general form for a non-homogenous L1-ODE-VC is:


 * $$y' +a_0(x)y = b(x) \!$$

The one given in the problem can be re-written as:


 * $$\underbrace{1}_{N(x,y)}\dfrac{dy}{dx} + \underbrace{\frac{1}{x}y}_{M(x,y)}=b(x)$$

We know that:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ 1(0 - a_0(x)) &= -f(x)\qquad \text{which yields}\\ f(x) &= a_0(x)\\ \end{align} $$

Then making use of integrating factor knowledge:



\begin{align} a_0(x) &= \frac{1}{x}\\ h(x)&=\exp\int^{x}a_0(s)ds\qquad \text{replacing values for}\quad a_0(x) \quad \text{and evaluating the integral}\\ h(x)&=\exp(\ln x)\\ h(x) &= x \end{align} $$

Now we can solve for $$y(x)$$



\begin{align} y(x) &= \frac{1}{h(x)}\int^{x} h(s)b(s)ds \qquad \text{plugging known values in this expression yields}\\ y(x)&=\frac{1}{x}\int^{x}s s^2 ds \qquad \text{analyzing this integral and expanding terms out we end up with our desire result}\\ y(x)&=\frac{x^3}{4} + \frac{C}{x} \end{align} $$
 * }

=Problem 3=

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Given
A non-homogenous L1_ODE_VC


 * $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$

Find
Show that the above non-homogenous L1_ODE_VC is exact

Solution
For the non-homogenous L1_ODE_VC to be exact it must staisfy the 2 conditions of exactness



\begin{align} N(x,y)y' + M(x,y) &= 0 \\ M_y(x,y) &= N_x(x,y) \end{align} $$

So the first condition is met by:


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y+10]}_{M(x,y)}= 0 $$

Let's take a look at the 2nd condition:



\begin{align} M_y(x,y) & = x^4 \\ N_x(x,y) & = x \end{align} $$

Clearly $$M_y(x,y) \neq N_x(x,y)$$, so we make use of the integrating factor to see if we can make it exactly integrable.

So we re-write the ODE


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at case #1 for the integrating factor we obtain:



\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ \frac{2}{x^2}\left(x - x^4\right) &= -f(x) \qquad \text{applying knowledge from integrating factor we know}\\ h(x)&=\exp \int^{x}f(s)ds \qquad \text{replacing f(x) into the integral and evaluating it, obtain}\\ h(x) &= \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right) \end{align} $$

Now we can try the 2nd condition of exactness to see if by using integrating factor we can make this ODE exact



\begin{align} \bar{M}_y &= \bar{N}_x\\ \underbrace{(hM)}_{\bar{M}}dx&=\underbrace{(hN)}_{\bar{N}}dy\qquad \text{calculating the respective values we obtain}\\ \bar{N}_x &=x^2 \exp\left(\frac{2x^3}{3}\right)\\ \bar{M}_y &=x^2 \exp\left(\frac{2x^3}{3}\right) \end{align} $$

Hence the non-homogenous L1_ODE_VC is exact.


 * }

=Problem 4=

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Given
An aribitrary 1st order ODE:


 * $$ \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)=0$$

Find
Show that the above is non-linear and exact

Solution
So let's determine that the above ODE is non-linear by making use of the $$D(.)\!$$ operator

To be a linear ODE the following must be true:


 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Now we will define the $$D(.)\!$$ operator for the above ODE:


 * $$D(.)= \left(\frac{1}{3}x^{3}\right)(\cdot)^{4}\frac{d(\cdot)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\cdot)^{5}\right)$$

Then let's evaluate the linearity condition to see if it's true:



\begin{align} D\left(\alpha u + \beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u + \beta v)^{4}\frac{d}{dx}(\alpha u + \beta v)+(5x^{3}+2)\left(\frac{1}{5}(\alpha u + \beta v)^{5}\right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u)^{4}\frac{d(\alpha u)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\alpha u)^{5}\right) +  \left(\frac{1}{3}x^{3}\right)(\beta v)^{4}\frac{d(\beta v)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\beta v)^{5}\right) \end{align} $$

Clearly $$D\left(\alpha u + \beta v \right) \neq D\left(\alpha u \right) + D\left(\beta v \right)$$ so the ODE is non-linear

To prove that the ODE is exact, let's first look at the 1st condition of exactness:


 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

hence $$N(x,y)y' + M(x,y) = 0\!$$ so 1st condition is met.

Looking at the 2nd condition where $$M_y = N_x\!$$. However, from above we can clearly see that:


 * $$M_y \neq N_x$$

So we will try to find an integrating factor to find if our ODE can be exactly integrable



\begin{align} \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0 \qquad \text{let us multiply througout by} \qquad \frac{3}{x^3y^4}\\ y' +\frac{3}{x^3y^4}(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0\qquad \text{re-writing the equation obtain}\\ \underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y &=0 \end{align} $$

Now we will make use of integrating factor knowledge:



\begin{align} h(x,y) &= \exp\int^{x}a_0(s)ds\qquad \text{replacing the value of} \quad a_0\\ h(x,y) &= \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \text{evaluating the integral obtain}\\ h(x,y) &= \exp\left(3x - \frac{3}{5x^2}\right) \end{align} $$

Now we will try to evaluate the 2nd condition of exactness which is:



\begin{align} (hM)_{y}&=(hN)_x \qquad \text{the condition says:} \\ \bar{M}_y &= \bar{N}_x \qquad \text{evaluating the derivatives we obtain}\\ \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} } & = \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}}} \end{align} $$

Hence the ODE given is then, exact N1-ODE-VC
 * }

=Problem 5=

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Given
The following 2nd order ODE


 * $$(xy)y''+x(y')^{2}+yy'=0\!$$

Find
Show that the second exactness condition is satisfied for the above equation

Solution


\begin{align} \underbrace{(xy)}_{f(x,y,p)}y''+\underbrace{x(y')^{2}+yy'}_{g(x,y,p)}&=0 \qquad \text{we can re-write the ODE}\\ (xy)y''+xp^{2}+yp&=0 \end{align} $$

The second exactness condition is the following



\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ 0 + 2p(1) +p^2(0) &= 2p(1) + 0 + p - (0) - p \qquad \text{which gives}\\ 2p & = 2p \end{align} $$

Also, the following must be met



\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ 0 + 0 + 2x &= 2x + 0\qquad \text{which gives}\\ 2x &=2x \end{align} $$

Hence the second exactness condition is met for this particular case
 * }

=Problem 6=

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Given
The following condition is given


 * $$f_{xp} + pf_{yp} + 2f_{y} =g_{pp} \!$$

Find
Derive the above equation by differentiating


 * $$g(x,y,p) = \phi_x +\phi_yp\!$$

with respect to $$p (y')\!$$

Solution
We know that:



\begin{align} g &= \phi_x +\phi_yp \qquad \text{differentiating once wrt p}\\ g_p &= \phi_{xp} +\phi_y +p\phi_{yp} \qquad \text{differentiating again wrt p}\\ g_{pp} &= \phi_{xpp} +\phi_{yp} +\phi_{yp} + p\phi_{ypp} \qquad \text{taking into parenthesis}\quad \phi_p \quad \text{we obtain}\\ g_{pp} &= (\phi_p)_{xp} +(\phi_p)_y +(\phi_p)_y + p(\phi_p)_{yp} \qquad \text{finally arriving to the given equation because}\quad \phi_p = f\\ g_{pp} &=f_{xp} + pf_{yp} + 2f_{y} \end{align} $$

=Problem 7=
 * }

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Given
The following condition is given


 * $$f_{xx} +2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}\!$$

Find
Use $$\phi_{xy}=\phi_{yx}\!$$ to obtain the equation given above

Solution
We know that



\begin{align} \phi_x &= g - p(g_p - f_x) + p^2f_y \qquad \text{and that}\\ \phi_y &= g_p -pf_y - f_x \end{align} $$

So what we will need to do is find $$\phi_{xy}=\phi_{yx} \!$$ respectively and equate them to find the given condition



\begin{align} &\phi_{xy} = g_y - pg_{py} + pf_{xy} + p^2f_{yy} \\ &\phi_{yx} = g_{px} - pf_{yx} - f_{xx} \qquad \text{after re-arranging we obtain the desire given condition}\\ &f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$

=Problem 8=
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Given
The following 2nd order ODE


 * $$(8x^5y')y'' + 2x^2y' +20x^4(y')^2 +4xy = 0\!$$

Find
Verify the second exactness condition on the above ODE

Solution
First let us identify some parameters in the ODE



\begin{align} \underbrace{(8x^5y')}_{f}y'' + \underbrace{2x^2y' +20x^4(y')^2 +4xy}_{g} &= 0 \qquad \text{let us look at the exactness conditions needed to be analyzed}\\ f_{xx} +2pf_{xy} + p^{2}f_{yy} &= g_{xp} + pg_{yp} - g_{y} \qquad \text{evaluating each derivative}\\ 160x^3p +2p(0) + p^2(0) & = 160x^3p +4x +p(0) - 4x \qquad \text{which gives}\\ 160x^3p &= 160x^3p \end{align} $$

Also, the other condition is as follow:



\begin{align} f_{xp} + pf_{yp} + 2f_{y}&=g_{pp} \qquad \text{evaluating each derivative}\\ 40x^4 + p(0) + 2(0) &=40x^4\qquad \text{which gives}\\ 40x^4 &= 40x^4 \end{align} $$

Hence the ODE given is exact!

=Problem 9=
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Given
The following 2nd order ODE is given


 * $$\sqrt{x}y''+2xy'+3y=0$$

Find
Verify the exactness of the above ODE

Solution
For the ODE to be exact it must satisfy the two conditions of exactness


 * $$\underbrace{\sqrt{x}}_{f}y''+\underbrace{2xy'+3y}_{g}=0$$

Which means that the 1st condition of exactness is met.



\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ (0)+ p(0) + 2 (0) &= 0 \\ 0 &= 0 \qquad \text{which satisfies the first condition} \end{align} $$

However, the ODE must satisfy both conditions to be exact. So let's evaluate the next condition



\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ -\dfrac{1}{4x^{\frac{3}{2}}} + 2p(0) + p^2(0) &= 2 + p(0) - 3 \\ -\dfrac{1}{4x^{\frac{3}{2}}} &\neq -1 \end{align} $$

Since the ODE did not comply with the all of the exactness conditions then the ODE is non-exact
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