User:Egm6321.f10.team03.sigillo/Hwk5

=Problem 4: Find a particular solution using the method "Variatoin of Constants"=

Given
The following equation is given from page 11-2 out of the class notes:
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$$\,{y}'+xy=2x+3$$
 * $$\,(Eq. 4.1) $$
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Find
The particular solution $${{y}_{p}}$$ has to be found using the method of variation of constants after having found the homogeneous solution $${{y}_{h}}$$.

Solution
First we find the homogeneous solution to the homogeneous equation out of Eq. 4.1, that is:
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$$\,{y}'+xy=0$$ With the separation of variables we get:
 * $$\,(Eq. 4.1) $$
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$$\frac{1}{y}dy=-x\,dx$$ The result of the integration on both sides of the equation yields to:
 * $$\,(Eq. 4.2) $$
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$$\int{\frac{1}{y}dy}=\int{-x\,dx+C\Leftrightarrow \log \left( y \right)=\frac{-{{x}^{2}}}{2}+C}$$ hence the homogeneous solution $${{y}_{h}}$$ is: The result of the integration on both sides of the equation yields to:
 * $$\,(Eq. 4.3) $$
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$${{y}_{h}}={{e}^{\frac{-{{x}^{2}}}{2}}}\cdot A$$ with
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 * $$\,(Eq. 4.4) $$
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$$\,A={{e}^{C}}$$ The next Step is the variation of the constant $$\, A$$ that now becomes a function of the variable $$\, x$$. The solution $$\, y$$ and its derviation can thus be written as:
 * $$\,(Eq. 4.5) $$
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$$y=A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}$$
 * $$\,(Eq. 4.6) $$
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$${y}'={A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x$$ Inserting Eq. 4.5 and Eq. 4.6 into Eq. 4.1 the given equation can be written as:
 * $$\,(Eq. 4.7) $$
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$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x+A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x=2x+3$$ The second and third term of the left-hand-side cacel out and Eq. 4.7 can be written as:
 * $$\,(Eq. 4.8) $$
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$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}=2x+3$$ To find the particular solution for $$y$$ the second last step is to find an expression for $$A(x)$$. This can be obtained by integrating $${A}'(x)$$ of Eq. 4.9:
 * $$\,(Eq. 4.9) $$
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$$A(x)=\int{\left( 2x+3 \right)}\cdot {{e}^{\frac{2}}}\,dx+k$$
 * $$\,(Eq. 4.10) $$
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$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,\cdot s\,ds}+k$$
 * $$\,(Eq. 4.11) $$
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$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot {{e}^{\frac{2}}}+k$$
 * $$\,(Eq. 4.12) $$
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$$A(x)=3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+2{{e}^{\frac{2}}}+k$$ Inserting Eq. 4.13 into Eq. 4.6 gives the particular soluton $${{y}_{p}}$$:
 * $$\,(Eq. 4.13) $$
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$${{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+{{e}^{\frac{-{{x}^{2}}}{2}}}+k$$
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 * $$\,(Eq. 4.14) $$
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Contributing Members
Solved and posted byEgm6321.f10.team03.sigillo 01:33, 31 October 2010 (UTC)

=Problem 6: Equivalence of particular solutions=

Given
The particular solution for a L2-ODE-VC developed in class during Mtg 30 is given as:
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$${{y}_{p}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}$$ Where $$h(x)$$ is equal to:
 * $$\,(Eq. 6.1) $$
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$$h(x)=\frac{{{u}_{1}}^{2}(x)}{{{u}_{2}}^{\prime }(x)\cdot {{u}_{1}}(x)-{{u}_{2}}(x)\cdot {{u}_{1}}^{\prime }(x)}$$
 * <p style="text-align:right;">$$\,(Eq. 6.2) $$
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Find
Show that the developed particular solution is equal to the particular solution in King, Billingham and Otto page 8 Eq. 1.6 as:
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$${{y}_{p}}=\oint{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)}\,ds$$ Where $$W(s)$$ is the Wronskian and given as:
 * <p style="text-align:right;">$$\,(Eq. 6.3) $$
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$$W(s)={{u}_{1}}(s)\cdot {{u}_{2}}^{\prime }(s)-{{u}_{1}}^{\prime }(s)\cdot {{u}_{2}}(s)=\left| \begin{matrix} {{u}_{1}}(x) & {{u}_{2}}(x) \\ {{u}_{1}}^{\prime }(x) & {{u}_{2}}^{\prime }(x) \\ \end{matrix} \right|$$
 * <p style="text-align:right;">$$\,(Eq. 6.4) $$
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Solution
The right hand side of Eq. 6.1 can be integrated using the partial integration method:
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$$\int(t)\cdot r(t)\,dt=r(t)\cdot v(t)-\int{{r}'(t)\cdot v(t)}\,dt$$ Where in the specific case of Eq. 6.1, $$r(t)$$ and $$v(t)$$ can be found as:
 * <p style="text-align:right;">$$\,(Eq. 6.5) $$
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$${{u}_{1}}\left( x \right)\int{\underbrace{\frac{1}{h\left( x \right)}}_\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{r}}\,dx$$ Applying the partial integration method to Eq. 6.1 the right hand side can be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.6) $$
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$${{u}_{1}}(x)\cdot \left[ \underbrace{\int{h(x)f(x)\,dx}}_{r}\cdot \underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}-\int{\underbrace{h(x)\cdot f(x)}_\cdot }\underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}\,dx \right]$$ Applying Eq. 6.2, $$h(x)$$ can be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.7) $$
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$${{u}_{1}}(x)\cdot \left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{{{u}_{1}}(s)}\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Multiplying $${{u}_{1}}$$ into the squared brackets we get:
 * <p style="text-align:right;">$$\,(Eq. 6.8) $$
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$$\left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{1}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{1}\cdot \frac{{{u}_{1}}(x)\cdot {{u}_{1}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Which is equal to the contour integral:
 * <p style="text-align:right;">$$\,(Eq. 6.9) $$
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$$\oint\limits_{x}{f(s)\cdot }\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)} \right)\,ds=\oint\limits_{x}{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)\,ds}$$ '''Eq. 6.10, thus agrees with Eq. 1.6 in King, Billingham and Otto page 8.'''
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 * <p style="text-align:right;">$$\,(Eq. 6.10) $$
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Contributing Members
Solved and posted byEgm6321.f10.team03.sigillo 19:42, 31 October 2010 (UTC)