User:Egm6321.f10.team2.dube/HW3

=Problem 6:Discuss h=

Given
Example given in MT15-3 as


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$$  \displaystyle x({y}')^{2}+y{y}'+xy{y}''=0 $$     (Eq 6.1)
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$$  \displaystyle g(x,y,p)=x({y}')^{2}+y{y}' $$     (Eq 6.2)
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$$  \displaystyle f(x,y,p)=xy $$     (Eq 6.3)
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$$  \displaystyle h_{x}+h_{y}p=0 $$     (Eq 6.4)
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Find
Without assuming h=const. discuss other solutions.

Solution
For the condition given Eq.6.4 both terms must be zero. So


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$$  \displaystyle h_{x}=0 $$     (Eq 6.5)
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$$  \displaystyle h_{y}p=0 $$     (Eq 6.6)
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And we know that h=h(x,y).Possible outcomes for the derivations;


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$$  \displaystyle h_{x}=h(x,y) $$     (Eq 6.7)
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$$  \displaystyle h_{y}=h(x,y) $$     (Eq 6.8)
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By this way we can not find common h for stisfy Eq.6.4

=Problem 7:SC-L1-ODE-CC=

Given

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$$ \displaystyle \dot{\phi}=\omega $$ (Eq 7.1)
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$$ \displaystyle \dot{\omega}=-\frac{1}{\tau }\omega+\frac{Q}{\tau }\delta $$ (Eq 7.2)
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$$ \displaystyle \dot{\delta}=u $$ (Eq 7.3)
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$$ \displaystyle \dot{x}(t)=\underline{A} \underline{x}(t)+\underline{B} \underline{u}(t) $$ (Eq 7.4)
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Find
Make 7.1-7.3 in the form of 7.4

Solve

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$$ \displaystyle \begin{bmatrix} \dot{\phi }\\ \dot{\omega }\\ \dot{\delta } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi \\ \omega \\ \delta \end{bmatrix}
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+ \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} u(t) $$ (Eq 7.5)
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$$ \displaystyle \underline{A}= \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix}$$ (Eq 7.6)
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$$ \displaystyle \underline{B}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$ (Eq 7.7)
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=Problem 8:Exactness Condition for Second Order=

Given
Second order non-linear function can be shown as follows which was given in MT15-2 ;


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$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=0 $$     (Eq 8.1)
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F is a function that derivative of function that is non-linear first order equation equals constant.


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$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=\frac{\mathrm{d} \phi \left ( x,y,{y}' \right )}{\mathrm{d} x} $$ (Eq 8.2)
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F is exact if there is exist a fuunction $$ \phi (x,y,{y}')$$

If we want to show that an arbitrary second order function to be exact, this arbitrary function has to provide two exactness conditions.

First condition:An arbitrary second order function has to provide this form:


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$$  \displaystyle F\left ( x,y,{y}',{y} \right )=g\left ( x,y,p \right )+f\left ( x,y,p \right ){y} $$     (Eq 8.3)
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where $$p={y}' $$

Second Condition:has two parts


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$$  \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} $$     (Eq 8.4)
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$$  \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} $$     (Eq 8.5)
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where
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$$  \displaystyle g\left ( x,y,p \right )=\phi _{x}+\phi _{y}{y}' $$     (Eq 8.6)
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$$  \displaystyle f\left ( x,y,p \right )=\phi _{p}^{} $$     (Eq 8.7)
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Find
Given function in lecture mtg 17-2 as follows ;


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$$  \displaystyle \left ( 15p^{4}cosx^{2} \right ){y}''+6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ]=0 $$     (Eq 8.8)
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For this equation check exactness.

Solution

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$$  \displaystyle f= 15p^{4}cosx^{2} $$     (Eq 8.9)
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$$  \displaystyle g=6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ] $$     (Eq 8.10)
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$$  \displaystyle f_{xx}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.11)
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$$  \displaystyle f_{xy}=0 $$     (Eq 8.12)
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$$  \displaystyle f_{yy}=0 $$     (Eq 8.13)
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$$  \displaystyle g_{xp}=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.14)
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$$  \displaystyle g_{yp}=12xy $$     (Eq 8.15)
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$$  \displaystyle g_{y}=12xyp+6y^{2} $$     (Eq 8.16)
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$$  \displaystyle f_{xp}=-120p^{3}xsinx^{2} $$     (Eq 8.17)
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$$  \displaystyle f_{yp}=0 $$     (Eq 8.18)
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$$  \displaystyle f_{y}=0 $$     (Eq 8.19)
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$$  \displaystyle g_{pp}=-120p^{3}xsinx^{2} $$     (Eq 8.20)
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By substituting this tools into two equations (8.4 and 8.5) we observe that ;


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$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+2p.0+p^{2}.0=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+p12xy-12xyp-6y^{2} $$     (Eq 8.21)
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$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.22)
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$$  \displaystyle -120p^{3}xsinx^{2}+p.0+0=-120p^{3}xsinx^{2} $$     (Eq 8.23)
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$$  \displaystyle -120p^{3}xsinx^{2}=-120p^{3}xsinx^{2} $$     (Eq 8.24)
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=Problem 9=

Given
For this problem, the equation is the same as which was given in problem 8. We can find first integral $$ \phi \left ( x,y,{y}' \right ) $$ by taking first derivative of f.


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$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+\int f\left ( x,y,{y}' \right )dp $$     (Eq 9.1)
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$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+3p^{5}cosx^{2} $$     (Eq 9.2)
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$$  \displaystyle g=\phi _{y}{y}'+\phi _{x} $$     (Eq 9.3)
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By substituting $$\phi _{x}$$ and$$ \phi _{y}$$ in the definition of g we concluded ;


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$$ \displaystyle (6xy^2)y^{'}+2y^3=h_yy^{'}+h_x $$ (Eq 9.4)
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The following was also given:
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$$ \displaystyle h_yy{'}=2y^3 $$ (Eq 9.5)
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Find
If  $$ \displaystyle h_{x}=6xy^{2}{y}'$$ then what is $$ \displaystyle \phi (x,y,p)$$ (Meeting 18-3)

Solution
For the second order ODEs we assume that :


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$$  \displaystyle \phi =h(x,y)+\int f(x,y,p)dp $$     (Eq 9.6)
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So h(x,y) is only function of x and y. For the second option


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$$  \displaystyle h_{x}=6xy^{2}{y}' $$     (Eq 9.7)
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tells us that after takind partial derivative of h over x we would find that another function depends on y prime.


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$$  \displaystyle z(x,y,{y}')=6xy^{2}{y}' $$     (Eq 9.8)
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So the choice of derivation of h must be depend on x,y,and $${y}'$$ if we assume the choice is true.(?) However we can not expect that. Because h only depends on x and y.The derivation of h over x must yield a function depends on x and y.

=References=