User:Egm6321.f10.team2.patterson/Homework 2

= Problem 7: Ordinary Differential Equations with Varying Coefficients (ODE_VC) =

Given
Show that the constant, k1, is not necessary when the following expression is evaluated:


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$$ \displaystyle
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y\left( x \right)=\frac{1}{h\left( x \right)}\int_ – ^{x}{h\left( s \right)b\left( s \right)ds}

$$ (eq 7.1) [1]
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Solution
The function h(x), given in Equation 7.1, is further expressed as:


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$$ \displaystyle
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h\left( x \right)={{e}^{\int_ – ^{x}{{{a}_{0}}\left( s \right)ds}}}

$$ (eq 7.2) [2]
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By converting the integral using the following principle:
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$$ \displaystyle
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\int_ – ^{x}{f\left( s \right)ds}=\int{f\left( x \right)dx}+k $$ (eq 7.3) [2]
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The function h can be expressed as follows:
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$$ \displaystyle
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h\left( x \right)={{e}^{\left( \int{{{a}_{0}}\left( x \right)dx} \right)+k}}={{e}^{k}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 7.4)
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Setting the constant ek to k yields the following:


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$$ \displaystyle
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h\left( x \right)={k}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 7.5)
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Plugging Equation 7.4 into both instances of h(x) in Equation 7.1 yields the following:


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$$ \displaystyle
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y\left( x \right)=\frac{1}\int_ – ^{x}{{{k}_{2}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}=\frac{{e}^{-\int{{{a}_{0}}\left( x \right)dx}}}\int_ – ^{x}{{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}

$$ (eq 7.6)
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Let the constant k2=k3k1 which therefore makes k1 unnecessary.

Given
Show that Equation 7.1 agrees with the following:


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$$ \displaystyle
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y\left( x \right)=A{{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)

$$ (eq 7.7) [3]
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Solution
Applying the principle in Equation 7.3 directly into Equation 7.1 yields the following:


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$$ \displaystyle
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y\left( x \right)=h{{\left( x \right)}^{-1}}\left[ \int{h\left( x \right)b\left( x \right)dx}+k \right]

$$ (eq 7.8)
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$$ \displaystyle
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y\left( x \right)=kh{{\left( x \right)}^{-1}}+h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 7.9)
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It is shown that the two equations agree by letting A=k and making the following substitutions:


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$$ \displaystyle
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{{y}_{h}}\left( x \right)=h{{\left( x \right)}^{-1}}

$$ (eq 7.10)
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$$ \displaystyle
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{{y}_{p}}\left( x \right)=h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 7.11)
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Given
Find yH independently by solving the following equation:


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$$ \displaystyle
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y'+{{a}_{0}}y=0

$$ (eq 7.12)
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Solution
Equation 7.12 can be converted into the following form:


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$$ \displaystyle
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\frac{dx}+{{a}_{0}}{{y}_{H}}=0

$$ (eq 7.13)
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In solving for yH, it follows that


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$$ \displaystyle
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{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;=-{{a}_{0}}dx

$$
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$$ \displaystyle
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\int{{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;}=-\int{{{a}_{0}}dx} $$
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$$ \displaystyle
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\ln \left( {{y}_{H}} \right)=-\int{{{a}_{0}}dx}+k $$
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$$ \displaystyle
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{{y}_{H}}=K{{e}^{-\int{{{a}_{0}}dx}}}$$
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= Problem 8: Exact Differential Equations =

Given
Find a N1_ODE that is either exact or can be made exact using the following equation:


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$$ \displaystyle
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\bar{b}\left( x \right)c\left( y \right)y'+a\left( x \right)\bar{c}\left( y \right)=0

$$ (eq 8.1)
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With the following values
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$$ \displaystyle
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a\left( x \right)=\sin \left( {{x}^{3}} \right) $$ (eq 8.2)
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$$ \displaystyle
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b\left( x \right)=\cos \left( x \right) $$ (eq 8.3)
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$$ \displaystyle
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c\left( y \right)={e}^{2y} $$ (eq 8.4)
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Solution
The functions with superscript bars are integrals of the original functions and thus the following equations are derived:


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$$ \displaystyle
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b\left( x \right)=\cos \left( x \right)\Rightarrow \bar{b}\left( x \right)=\sin \left( x \right) $$ (eq 8.5)
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$$ \displaystyle
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c\left( x \right)={{e}^{2y}}\Rightarrow \bar{c}\left( x \right)={}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} $$ (eq 8.6) Inserting all of the values into Equation 8.1 yields the following:
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$$ \displaystyle
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\underbrace{\sin \left( x \right)\left( {{e}^{2y}} \right)}_{N\left( x,y \right)}y'+\underbrace{\sin \left( {{x}^{3}} \right)\left( {}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} \right)}_{M(x,y)}=0 $$ (eq 8.7) Equation 8.7 meets the first condition of exactness which states that:
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$$ \displaystyle
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N\left( x,y \right)y'+M\left( x,y \right)=0 $$ (eq 8.8)
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Divide Equation 8.7 by e2y.

Next we can check to see if the equation passes the second condition of exactness in which My must equal Nx
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$$ \displaystyle
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{{M}_{y}}\left( x,y \right)=0 $$ (eq 8.9)
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$$ \displaystyle
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{{N}_{x}}\left( x,y \right)=\cos \left( x \right) $$ (eq 8.10) From the previous equations it is evident that
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$$ \displaystyle
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{{M}_{y}}\left( x,y \right)\ne {{N}_{x}}\left( x,y \right) $$ (eq 8.11) Therefore, in order to make the equation exact, Euler's integrating factor, h(x) must be derived as follows:
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$$ \displaystyle h\left( x \right)={{e}^{-\int{\frac{1}{N}\left( {{N}_{x}}-{{M}_{y}} \right)dx}}} $$ (eq 8.12)4
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$$ \displaystyle ={{e}^{-\int{\frac{1}{\sin \left( x \right)}\left( \cos \left( x \right)-0 \right)dx}}} $$
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$$ \displaystyle ={{e}^{-\ln \left( \sin \left( x \right) \right)}} $$
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$$ \displaystyle h\left( x \right)=\frac{1}{\sin \left( x \right)} $$ (eq 8.13)
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In order to verify that the derived integrating factor is correct it is necessary to multiply Equation 8.8 by h(x) and denote the new values for M and N as follows:
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$$ \displaystyle \underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \sin \left( x \right) \right)}_{\bar{N}=hN\left( x,y \right)}y'+\underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \frac{\sin \left( {{x}^{3}} \right)}{2} \right)}_{\bar{M}=hM\left( x,y \right)}=0 $$ (eq 8.14)
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Again we check to see if the second condition of exactness has been met as follows:


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$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)=0 $$ (eq 8.15)
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$$ \displaystyle {{\bar{N}}_{x}}\left( x,y \right)=0 $$ (eq 8.16)
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The N1_ODE has been made exact since


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$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)={{\bar{N}}_{x}}\left( x,y \right) $$ (eq 8.17)
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Given
Find the first integral of the N1_ODE used in the previous problem so that


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$$ \displaystyle
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\phi \left( x,y \right)=k

$$ (eq 8.18)
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Solution
Considering the following equations:
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$$ \displaystyle
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M\left( x,y \right)={{\phi }_{x}}\left( x,y \right) $$ (eq 8.19)5
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$$ \displaystyle
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N\left( x,y \right)={{\phi }_{y}}\left( x,y \right) $$ (eq 8.20)5
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The following equations can be derived:
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$$ \displaystyle
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\int{M\left( x,y \right)dx}+w\left( y \right)=\phi \left( x,y \right) $$ (eq 8.21)6
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$$ \displaystyle
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\int{N\left( x,y \right)dy}+v\left( x \right)=\phi \left( x,y \right) $$ (eq 8.22)6
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By setting equations 8.21 and 8.22 equal to each other, the functions w(y) and v(x) can be found. From there, equation 8.13 can be solved.
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$$ \displaystyle \int{M}\left( x,y \right)dx+w\left( y \right)=\int{N}\left( x,y \right)dy+v\left( x \right) $$ (eq 8.23)
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$$ \displaystyle \int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+w\left( y \right)=y+v\left( x \right) $$
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$$ \displaystyle \phi \left( x,y \right)=\int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+y=k $$
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= References = 1Vu-Quoc,L. EGM 6321.F10, 10-3, Equation (6)

2Vu-Quoc,L. EGM 6321.F10, 11-1:11-2

3King, A.C. et al. Differential Equations: Linear, Nonlinear, Ordinary, Partial. Cambridge: Cambridge University Press

4Vu-Quoc,L. EGM 6321.F10, 10-1, Equation (6)&(7)

5Vu-Quoc,L. EGM 6321.F10, 8-2, Equations (1)&(2)

6Zill, D., and Cullen, M. Differential Equations with Boundary-Value Problems. Pacific Grove, CA: Brooks/ Cole,