User:Egm6321.f10.team2.oztekin/hw3.13

Given
1st relation in 2nd exactness condition for N2-ODEs
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(3.13.1)
 * $$ \displaystyle f_{xx} + 2pf_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$
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2st relation in 2nd exactness condition for N2-ODEs


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(3.13.2)
 * $$ \displaystyle f_{xp} + pf_{yp} + 2 f_y = g_{pp} $$
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N2-ODE equation for verification,


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 * $$ \displaystyle x\left( y^{\prime} \right)^2 + yy^{\prime} + \left( xy \right) y^{''} = 0 $$
 * $$ \displaystyle x\left( y^{\prime} \right)^2 + yy^{\prime} + \left( xy \right) y^{''} = 0 $$

(3.13.3)
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Find
1. Derive the 2nd relation in the 2nd exactness condition. 2. Derive the 1st relation in the 2nd exactness condition. 3. Verify that the Eq. (3.13.3) satisfies the 1st and 2nd relations in the 2nd exactness condition.

Solution Part C
The N1-ODE-VC: $$\displaystyle \phi(x,y,p)=xp+(2x^{1/2}-1)y = k$$ so we can rewrite, $$\displaystyle xy'+(2x^{1/2}-1)y=k $$ If we rearrange it, we get

$$\displaystyle y'+(2x^{1/2}-\frac{1}{x})y=\frac{k}{x} $$. From the equation (3) p.11-4, we can solve by using integrating factor, $$\displaystyle h(x)=\exp[\int^x a_0(s)ds+k_1] $$

We need to substitute $$\displaystyle a_0(x)=2x^{1/2}-\frac{1}{x} $$ into the h(x)equation $$\displaystyle h(x)=\exp[\int^x (2s^{1/2}-\frac{1}{s})ds+k1] $$ $$\displaystyle h(x)=\exp[\frac{4}{3}x^{3/2}-\ln{x}+k1] $$ $$\displaystyle h(x)=\frac{1}{x}\exp[{\frac{4}{3}x^{3/2}}+k1] $$ From the equation (1) p. 11-5 to find out $$\displaystyle y(x) $$ $$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds + k2\right] $$ $$\displaystyle b(x)=\frac{k}{x} $$ $$\displaystyle y(x)=\frac{1}{\frac{1}{x}\exp[{\frac{4}{3}x^{3/2}}+k1]}*\left[\int^x (\frac{1}{s}\exp[{\frac{4}{3}s^{3/2}}+k1])(\frac{k}{s})ds+k2 \right]$$

Solution
1. Derivation of the 2nd relation in the 2nd exactness condition We know that


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

(3.13.4)
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If we differentiate Eq. 3.13.4 with respect to p, we obtain


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 * $$\displaystyle g_p = \phi_{xp} + \phi_y + p\phi_{yp} $$
 * $$\displaystyle g_p = \phi_{xp} + \phi_y + p\phi_{yp} $$

(3.13.5)
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One more differentiation with respect to p, we obtain
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 * $$ \displaystyle   g_{pp} = \phi_{xpp} + 2\phi_{yp} + \phi_{ypp} p $$
 * $$ \displaystyle   g_{pp} = \phi_{xpp} + 2\phi_{yp} + \phi_{ypp} p $$

(3.13.6)
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Substituting the $$ \displaystyle f :=\phi_p $$ into the Eq. 3.13.6,


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(3.13.7)
 * $$ \displaystyle  g_{pp} = f_{xp} + 2 f_{y} + f_{yp} p $$
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2. Derivation of the 1nd relation in the 2nd exactness condition

i. We know that  $$ \displaystyle \phi_{xp} = \phi_{px} $$


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(3.13.8)
 * $$ \displaystyle ( \phi_{x} )_{p} =( \phi_{p})_{x} $$
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\begin{align} ( g - \phi_y p )_{p} &= f_x \\ g_p -\phi_{yp} p- \phi_y &= f_x \\ g_p - f_y p - \phi_y &= f_x \end{align} $$     (3.13.9) We can get $$\displaystyle \phi_y $$ with rearranging,
 * $$ \displaystyle
 * $$ \displaystyle
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(3.13.10)
 * $$ \displaystyle     \phi_y = g_p - f_y p - f_x $$
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Differentiating $$ \displaystyle \phi_y $$ with respect to x,
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(3.13.11)
 * $$ \displaystyle ( \phi_y )_x = g_{px} - f_{yx} p - f_{xx} $$
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ii. We know that $$ \displaystyle (\phi_{y})_{p} =( \phi_{p} )_{y}$$  and $$ \displaystyle f :=\phi_p $$

Also


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\begin{align} g :=&\phi_x + \phi_y p \\ \phi_y=& \left( \frac {g-\phi_x}{p} \right) \end{align} $$     (3.13.12)
 * $$ \displaystyle
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Combining them together, we get
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\begin{align} \left( \frac{g-\phi_x}{p} \right)_{p} &= f_{y} \\ -\frac{g-\phi_x}{p^2} + \frac{g_p - \phi_{xp}}{p} &= f_y \\ -\frac{g-\phi_x}{p^2} + \frac{g_p - f_{x}}{p} &= f_y \end{align} $$     (3.13.13)
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We can get $$\displaystyle \phi_x $$ with rearranging the equation above,
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(3.13.14)
 * $$ \displaystyle \phi_x = g - p \left( g_p - f_x \right) + p^2 f_y $$
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Differentiating $$ \displaystyle \phi_x $$ with respect to y,


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(3.13.15)
 * $$ \displaystyle (\phi_x)_y = g_y - p (g_{py} - f_{xy}) + p^2 f_{yy} $$
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iii. We know that $$ \displaystyle (\phi_{x})_{y} =( \phi_{y} )_{x}$$

If we substitute Eq 3.13. and Eq.3.13 into equity given above, we obtain


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 * $$ \displaystyle    g_y - p (g_{py} - f_{xy}) + p^2 f_{yy} = g_{px} - f_{yx} p - f_{xx} $$
 * $$ \displaystyle    g_y - p (g_{py} - f_{xy}) + p^2 f_{yy} = g_{px} - f_{yx} p - f_{xx} $$

(3.13.16)
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Rearraning the terms on both sides, we get


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 * $$ \displaystyle    f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$
 * $$ \displaystyle    f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$

(3.13.17) 3. Verify
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From (1) 16-6: $$ \displaystyle x(y')^2+yy'+(xy)y''=0 $$ From (2) 16-6:

$$ \displaystyle g(x,y,y')=x(y')^2+yy' = x p^2 + y p $$ From (3) 16-6:


 * $$ \displaystyle f(x,y,y')=xy $$

Therefore, it satisfies the 1st exactness condition.

For 1nd relation in the 2nd exactness condition,


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\begin{align} f_{xx} + 2p f_{xy} + p^2 f_{yy} &= g_{xp} + p g_{yp} - g_y \\ 0 + 2p (1)   + p^2(0)     &= 2p + p(1) - p \\ 2p &= 2p \\ \end{align} $$     (3.13.18)
 * $$ \displaystyle
 * $$ \displaystyle
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For 2nd relation in the 2nd exactness condition,


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\begin{align} g_{pp} &= f_{xp} + 2 f_{y} + f_{yp} p \\ 2x   &= 0      + 2x      + (0)p \\ 2x &= 2x \end{align} $$
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(3.13.19) It also satisfies 2nd exactness conditions as it was showed in Eq. 3.13.18 and Eq. 3.13.19.
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