User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework1

= Problem 1 =

Solution
= Problem 2 =

Solution
= Problem 3 =

Solution
= Problem 4 =

Solution
= Problem 5 - Verification of Homogenous Solutions of Legendre Differential Equation for n = 1=

Given
Legendre differential equation:
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{{L}_{2}}\left( y \right) = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ Homogenous solutions of the above equation for $$\displaystyle  n = 1 $$ [[media:2010_09_02_14_58_46.djvu|(Mtg 6-page1)]], are given by:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
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$$ $$
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$\displaystyle (Eq. 2)
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$$ $$ Where $$y_{H}^{1}\left( x \right)$$ is the first homogenous solution and $$y_{H}^{2}\left( x \right)$$ is the second homogenous solution
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$\displaystyle (Eq. 3)
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Find
We need to prove that:
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 * $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$
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Solution
Let us consider first $${{L}_{2}}\left( y_{H}^{1} \right)$$:


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$$ Substitute Eq.2 for $${{L}_{2}}\left( y_{H}^{1} \right)$$ in Eq.4 =>
 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{1} \right)}^{''}}-2x{{\left( y_{H}^{1} \right)}^{'}}+2\left( y_{H}^{1} \right)$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)
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 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( x \right)}^{''}}-2x{{\left( x \right)}^{'}}+2\left( x \right)$$
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$$ {{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right)\left( 0 \right)-2x\left( 1 \right)+2\left( x \right)=0$$ $$
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 * $$\displaystyle (Eq. 5)
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Now consider $${{L}_{2}}\left( y_{H}^{2} \right)$$:


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$$ Determine first derivative of $$y_{H}^{2}$$ (Eq.2) using Product rule, Logarithmic derivative rule ($$ (\ln f)'= \frac{f'}{f} \quad$$) and chain rule of differentiation:
 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{2} \right)}^{''}}-2x{{\left( y_{H}^{2} \right)}^{'}}+2\left( y_{H}^{2} \right)$$
 * $$\displaystyle (Eq. 6)
 * $$\displaystyle (Eq. 6)
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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}.\frac{d}{dx}\left( \log \left( \frac{1+x}{1-x} \right) \right)+\frac{d}{dx}\left( \frac{x}{2} \right).\log \left( \frac{1+x}{1-x} \right)-\frac{d}{dx}\left( 1 \right)$$
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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1}{\left( \frac{1+x}{1-x} \right)}.\frac{d}{dx}\left( \frac{1+x}{1-x} \right) \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)+0$$
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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
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$${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$ $$
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 * $$\displaystyle (Eq. 7)
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Second derivative of $$y_{H}^{2}$$ (Eq.2) is:


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 * $${{\left( y_{H}^{2} \right)}^{''}}={{\left( {{\left( y_{H}^{2} \right)}^{'}} \right)}^{'}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
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 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}} \right)+\frac{d}{dx}\left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
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 * $${{\left( y_{H}^{2} \right)}^{''}}=\left( \frac{\left( 1-{{x}^{2}} \right)-x.\left( -2x \right)} \right)+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)$$
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 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{2} \right)$$
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 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}$$
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$${{\left( y_{H}^{2} \right)}^{''}}=\frac{2}$$ $$ Substituting Eqs.3,7,8 into Eq.6 yields,
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 * $$\displaystyle (Eq. 8)
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 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right)\left( \frac{2} \right)-2x\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)+2\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)$$
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 * $${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2}{\left( 1-{{x}^{2}} \right)}-\frac{2{{x}^{2}}}{1-{{x}^{2}}}-x\log \left( \frac{1+x}{1-x} \right)+x\log \left( \frac{1+x}{1-x} \right)-2$$
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$${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2-2{{x}^{2}}-2\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)}=0$$ $$
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 * $$\displaystyle (Eq. 9)
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Eqs.5 and 9 verify that :{| style="width:10%" border="0" $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$ Egm6321.f10.team3.Sudheesh 18:11, 14 September 2010 (UTC)
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=Notes and references=