User:Egm6321.f12.team6.Hu.M/report3

==Problem *3.2:Compare the solution in lecture notes with that in King's book ==

Problem Statement

 * Show that the solution of
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$$ \displaystyle 1\cdot y{}'+\frac{Q(x)}{P(x)}y^{(0)}=\frac{R(x)}{P(x)} $$
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 * (2.1)
 * }
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in $$ \displaystyle y(x)=\frac{1}{h(x)}\left [ \int_{ }^{x}h(s)b(s)ds+k_{2} \right ]$$
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 * (2.2)
 * }
 * agrees with the result presented in King 2003 p.512, i.e.,
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$$ \displaystyle y(x)=Ay_{H}(x)+y_{P}(x) $$
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 * (2.3)
 * }
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Use $$ \displaystyle h(x)=\exp \left [ \int_{ }^{x}a_{0}(s)ds+k_{1} \right ] $$
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 * (2.4)
 * }
 * and $$ \displaystyle y(x)=\frac{1}{h(x)}\left [ \int_{ }^{x}h(s)b(s)ds+k_{2} \right ]$$
 * to identify $$ \displaystyle A $$, $$ \displaystyle y_{H}(x) $$ and $$ \displaystyle y_{p}(x) $$. Compare your results with those in King 2003 p.512.

Given

 * In King 2003 p.512, he defines L1-ODE-VC as:
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$$ \displaystyle y'+P(x)y=Q(x) $$
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 * (2.5)
 * }
 * And the integrating factor as:
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$$ \displaystyle h(x)=\exp\int_{ }^{x}P(t)dt $$
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 * (2.6)
 * }
 * Also, in King 2004 p.512, $$ \displaystyle y_{H}(x) $$ and $$ \displaystyle y_{P}(x) $$ in E2.3 are defined as:
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$$ \displaystyle y_{H}=\exp\left \{ -\int_{ }^{x}P(t)dt \right \} $$
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 * (2.7)
 * }
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$$ \displaystyle y_{P}=\exp\left \{ -\int_{ }^{x}P(t)dt \right \}\int_{ }^{x}Q(s)\exp\left \{ \int_{ }^{x}P(t)dt \right \}ds$$
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 * (2.8)
 * }

Nomenclature

 * L1-ODE-VC is defined as Linear First-Order Differential Equations with varying coefficients.
 * $$ \displaystyle y{}' $$ is defined as $$ \displaystyle \frac{dy}{dx} $$.
 * $$ \displaystyle \frac{1}{\exp\left \{ a(x) \right \}}=\exp\left \{ -a(x) \right \} $$.

Solution

 * In lecture notes, the General L1-ODE-VC is defined as E2.1 which is agree to King's L1-ODE-VC(E2.5).
 * From E2.4 we can have:
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$$ \displaystyle h(x)=e^{k_{1}}\exp\left [ \int^{x}a_{0}(s)ds \right ] $$
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 * (2.9)
 * }
 * Expand E2.2 with E2.9 we can have:
 * $$ \displaystyle y(x)=\frac{k_{2}}{h(x)}+\frac{1}{h(x)}\int^{x}h(s)b(s)ds $$
 * $$ \displaystyle =\frac{k_{2}}{e^{k_{1}}\exp\left [ \int^{x}a_{0}(s)ds \right ]}+\frac{1}{\exp\left [ \int^{x}a_{0}(s)ds \right ]}\int^{x}b(s)\exp\left [ \int^{x}a_{0}(s)ds \right ]ds $$
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$$ \displaystyle =\frac{k_{2}}{e^{k_{1}}}\exp\left [ -\int^{x}a_{0}(s)ds \right ]+\exp\left [ -\int^{x}a_{0}(s)ds \right ]\int^{x}b(s)\exp\left [ \int^{x}a_{0}(s)ds \right]ds $$
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 * (2.10)
 * }
 * Compare E2.10 with E2.3, we can have:
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$$ \displaystyle A=\frac{k_{2}}{e^{k_{1}}} $$
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 * (2.11)
 * }
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$$ \displaystyle y_{H}=\exp\left [ -\int^{x}a_{0}(s)ds \right ] $$
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 * (2.12)
 * }
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$$ \displaystyle y_{P}=\exp\left [ -\int^{x}a_{0}(s)ds \right ]\int^{x}b(s)\exp\left [ \int^{x}a_{0}(s)ds \right]ds $$
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 * (2.13)
 * }
 * Obviously, E2.10 agrees with E2.3.

Problem Statement

 * Show that the N1-ODE
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$$ \displaystyle \bar{b}(x,y)c(y)y'+a(x)\bar{c}(x,y)=0 $$
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 * (5.1)
 * }
 * satisfies the condition
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$$ \displaystyle \frac{h_{x}}{h}=-\frac{1}{N}(N_{x}-M_{y})=:n(x) $$
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 * (5.2)
 * }
 * that an integrating factor $$ \displaystyle h(x) $$ can be found to render it exact, only if $$ \displaystyle k_{1}(y)=d_{1} $$ (constant).
 * Show that
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$$ \displaystyle \bar{b}(x,y)c(y)y'+a(x)\bar{c}(x,y)=0 $$
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 * }
 * includes
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$$ \displaystyle M+Ny'=\left [ a(x)y+k_{2}(x) \right ]+\bar{b}(x)y'=0 $$
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 * (5.3)
 * }
 * as a particular case.

Given

 * In E5.1, we have:
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$$ \displaystyle M(x,y)=a(x)\bar{c}(x,y) $$
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 * <p style="text-align:right">(5.4)
 * }
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$$ \displaystyle N(x,y)=\bar{b}(x,y)c(y) $$
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 * <p style="text-align:right">(5.5)
 * }
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$$ \displaystyle \bar{b}(x,y)=\int^{x}b(s)ds+k_{1}(y) $$
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 * <p style="text-align:right">(5.6)
 * }
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$$ \displaystyle \bar{c}(x,y)=\int^{y}c(s)ds+k_{2}(x) $$
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 * <p style="text-align:right">(5.7)
 * }

Nomenclature

 * N1-ODE is defined as Nonlinear First-Order Differential Equations.
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$$ \displaystyle y'=\frac{dy}{dx} $$.
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 * }
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$$ \displaystyle M_y = \frac{\partial M(x,y)}{\partial y} $$.
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 * }
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$$ \displaystyle N_x = \frac{\partial N(x,y)}{\partial x} $$.
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 * }

Part a):

 * From E5.4 and E5.7 we can have:
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$$ \displaystyle M(x,y)=a(x)\int^{y}c(s)ds+k_{2}(x)a(x) $$
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 * }
 * Then we have:
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$$ \displaystyle M_{y}=a(x)c(y) $$
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 * <p style="text-align:right">(5.8)
 * }
 * From E5.5 we can have:
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$$ \displaystyle N(x,y)=c(y)\int^{x}b(s)ds+k_{1}(y)c(y) $$
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 * }
 * Then we have:
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$$ \displaystyle N_{x}=c(y)b(x) $$
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 * <p style="text-align:right">(5.9)
 * }
 * So E5.2 should be:
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$$ \displaystyle \frac{h_{x}}{h}=-\frac{1}{c(y)\int^{x}b(s)ds+k_{1}(y)c(y)}\left [ c(y)b(x)-a(x)c(y)\right ] $$
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 * }
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$$ \displaystyle =-\frac{b(x)-a(x)}{\int^{x}b(s)ds+k_{1}(y)} $$
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 * <p style="text-align:right">(5.10)
 * }


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Because of the condition defined in E5.2, then E5.10 can only be a function of $$ \displaystyle x $$, therefore $$ \displaystyle k_{1}(y)=d_{1} $$ in E5.10 must be a constant.
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 * }

Part b):

 * If we define a particular case of E5.1, where,
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1:$$ \displaystyle c(y)=1 $$
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 * }
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2:$$ \displaystyle b(x,y)=b(x) $$
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 * }
 * We can substitute these into E5.1 and expand the integral of E5.7:
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$$ \displaystyle \bar{b}(x)y'+a(x)\left [ y+k_{2}(x) \right ]=0 $$
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 * }
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$$ \displaystyle \rightarrow a(x)y+a(x)k_{2}(x)+\bar{b}(x)y'=0 $$
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 * }
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$$ \displaystyle \rightarrow \left [ a(x)y+k(x) \right ] +\bar{b}(x)y'=0 $$
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 * <p style="text-align:right">(5.11)
 * }


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Thus E5.11 is a particular case of E5.1.
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 * }