User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework6

= Problem 2 - Solution of L2-ODE-VC=

Given

 * Given two $$\displaystyle L2-ODE-VC $$,


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \left( a \right) \left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=f\left( x \right)$$
 * $$\displaystyle (Eq. 2.1)$$
 * }
 * $$\displaystyle (Eq. 2.1)$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \left( b \right) x{{y}^{''}}+2{{y}^{'}}+xy=f\left( x \right)$$
 * $$\displaystyle (Eq. 2.2)$$
 * }
 * $$\displaystyle (Eq. 2.2)$$
 * }
 * }

Find

 * (1) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=0$$


 * (2) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=\sin x$$

(1.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=0$$ (homogenous)

 * For the case of $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.1$$ can be written as:


 * {| style="width:100%" border="0" align="left"


 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=0$$
 * $$\displaystyle (Eq. 2.3)$$
 * }
 * }
 * }


 * Consider a trial solution,


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$
 * $$\displaystyle (Eq. 2.4)$$
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.


 * Taking derivative,


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 * $$\displaystyle{{y}^{'}}=c{{x}^{c-1}}{{e}^{rx}}+r{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{{y}^{''}}=c\left( c-1 \right){{x}^{c-2}}{{e}^{rx}}+2cr{{x}^{c-1}}{{e}^{rx}}+{{r}^{2}}{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }


 * Now substitute $$\displaystyle y$$ and whose derivatives in $$\displaystyle (Eq. 2.3)$$,implies:


 * {| style="width:100%" border="0" align="left"


 * $$\left( x-1 \right){{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]-x{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c}}{{e}^{rx}}=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}-c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c}}-2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c+1}}-{{r}^{2}}{{x}^{c}}-c{{x}^{c}}-r{{x}^{c+1}}+{{x}^{c}}=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\Rightarrow {{x}^{c+1}}\left[ {{r}^{2}}-r \right]+{{x}^{c}}\left[ 2cr-{{r}^{2}}-c+1 \right]+{{x}^{c-1}}\left[ {{c}^{2}}-c-2cr \right]-{{x}^{c-2}}\left[ {{c}^{2}}-c \right]=0$$
 * $$\displaystyle (Eq. 2.5)$$
 * }
 * }
 * }


 * Solving $$\displaystyle (Eq. 2.5)$$ using Wolfram Alpha give constants $$\displaystyle c$$ and $$\displaystyle r$$


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} i) & c=0 & r=1\\  ii) & c=1 & r=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Now homogenous solutions are obtained by substituting $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$


 * First Homogenous solution is


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 * $${{y}_}={{x}^{0}}{{e}^{1x}}={{e}^{x}}$$
 * }
 * }
 * }


 * and second homogenous solution is


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 * $${{y}_}={{x}^{1}}{{e}^{0x}}=x$$
 * }
 * }
 * }


 * Therefore solution of $$\displaystyle (Eq. 2.3)$$ is the linear combination of two homogenous solution


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$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 2.6)$$
 * }
 * }

.

(1.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=0$$ (homogenous)

 * For $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.2$$ reduces to:


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 * $$\displaystyle x{{y}^{''}}+2{{y}^{'}}+xy=0$$
 * $$\displaystyle (Eq. 2.7)$$
 * }
 * }
 * }


 * Considering a trial solution for y $$\displaystyle (Eq. 2.4)$$, and substituting y and its derivatives in $$\displaystyle (Eq. 2.7)$$,yields:


 * {| style="width:100%" border="0" align="left"


 * $$x{{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]+2{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c+1}}{{e}^{rx}}=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2r{{x}^{c}}+{{x}^{c+1}}=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\Rightarrow {{x}^{c+1}}\left( {{r}^{2}}+1 \right)+{{x}^{c}}\left( 2cr+2r \right)+{{x}^{c-1}}\left( {{c}^{2}}+c \right)=0$$
 * $$\displaystyle (Eq. 2.8)$$
 * }
 * }
 * }


 * By Solving $$\displaystyle (Eq. 2.8)$$ using Wolfram Alpha yields constants $$\displaystyle c$$ and $$\displaystyle r$$


 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} i) & c=-1 & r=i\\  ii) & c=-1 & r=-i\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * So homogenous solutions are obtained by plugging $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$


 * First Homogenous solution is


 * {| style="width:100%" border="0" align="left"


 * $${{y}_}={{x}^{-1}}{{e}^{ix}}=\frac{x}$$
 * }
 * }
 * }


 * and second homogenous solution is


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 * $${{y}_}={{x}^{-1}}{{e}^{-ix}}=\frac{x}$$
 * }
 * }
 * }


 * Hence solution of $$\displaystyle (Eq. 2.7)$$ is the linear combination of two homogenous solutions


 * {| style="width:100%" border="0" align="left"

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 2.9)$$
 * }
 * }

(2.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For the case of $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.1$$ can be written as:


 * {| style="width:100%" border="0" align="left"


 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=\sin (x)$$
 * $$\displaystyle (Eq. 2.10)$$
 * }
 * }
 * }


 * Above equation can be rewritten as:


 * {| style="width:100%" border="0" align="left"


 * $$\underbrace{1}_\cdot {{y}^{''}}\underbrace{-\frac{x}{\left( x-1 \right)}}_{{y}^{'}}+\underbrace{\frac{1}{\left( x-1 \right)}}_y=\underbrace{\frac{\sin (x)}{\left( x-1 \right)}}_{F\left( x \right)}$$
 * $$\displaystyle (Eq. 2.11)$$
 * }
 * }
 * }


 * Method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.11)$$.


 * Integrating factor for the above equation [[media:2010_10_26_15_07_13.djvu|Page3, Mtg 29 (b)]]is,


 * {| style="width:100%" border="0" align="left"


 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 * $$\displaystyle (Eq. 2.12)$$
 * }
 * }
 * }


 * From 1.1, one homogenous solution is,


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$${{u}_{1}}\left( x \right)={{y}_}={{e}^{x}}$$
 * $$\displaystyle (Eq. 2.13)$$
 * }
 * }


 * $$\displaystyle (Eq. 2.12)$$ becomes:


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$$h\left( x \right)={{e}^{2x}}\cdot \exp \left[ \int{\left( -\frac{x}{\left( x-1 \right)} \right)dx} \right]$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$h\left( x \right)={{e}^{x-\log \left( x-1 \right)}}$$
 * $$\displaystyle (Eq. 2.14)$$
 * }
 * }


 * Particular solution is given by[[media:2010_10_28_14_00_19.djvu|Page1, Mtg 30 (b)]],


 * {| style="width:100%" border="0" align="left"


 * $${{y}_{P}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}}\left( \int{h\left( x \right)F\left( x \right)dx} \right)dx$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{y}_{P}}={{e}^{x}}\int\left( \int{{{e}^{x-\log \left( x-1 \right)}}\cdot \frac{\sin (x)}{\left( x-1 \right)}dx} \right)dx$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.15)$$
 * }
 * }


 * Now complete solution is obtained by combining $$\displaystyle (Eq. 2.15)$$ and $$\displaystyle (Eq. 2.6)$$,


 * {| style="width:100%" border="0" align="left"

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$y={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x+{{e}^{x}}\int\left( \int{{{e}^{x-\log \left( x-1 \right)}}\cdot \frac{\sin (x)}{\left( x-1 \right)}dx} \right)dx$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.16)$$
 * }
 * }

(2.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.2$$ becomes:


 * {| style="width:100%" border="0" align="left"

$$x{{y}^{''}}+2{{y}^{'}}+xy=\sin \left( x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.17)$$
 * }
 * }


 * $$\displaystyle Eq. 2.17$$can be rewritten as:


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$$\underbrace{1}_\cdot {{y}^{''}}+\underbrace{\frac{2}{x}}_{{y}^{'}}+\underbrace{y}_=\underbrace{\frac{\sin (x)}{x}}_{F\left( x \right)}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.18)$$
 * }
 * }


 * As in the previous case, method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.17)$$.


 * From 1.2, one homogenous solution is,


 * {| style="width:100%" border="0" align="left"

$${{u}_{1}}\left( x \right)={{y}_}=\frac{x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.19)$$
 * }
 * }


 * Integrating factor for $$\displaystyle Eq. 2.18$$ is,


 * {| style="width:100%" border="0" align="left"


 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.20)$$
 * }
 * }
 * }


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$$h\left( x \right)=\frac\cdot \exp \left[ \int{\left( \frac{2}{x} \right)dx} \right]$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$h\left( x \right)={{e}^{2ix}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.21)$$
 * }
 * }


 * Particular solution is given by,


 * {| style="width:100%" border="0" align="left"


 * $${{y}_{P}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}}\left( \int{h\left( x \right)F\left( x \right)dx} \right)dx$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{y}_{P}}=\frac{x}\int\left( \int{{{e}^{2ix}}\frac{\sin (x)}{x}dx} \right)dx$$
 * }
 * }


 * Simplifying using Wolfram Alpha,


 * {| style="width:100%" border="0" align="left"

$${{y}_{P}}=\frac{x}\int\left( i\left( \frac{Ei\left( ix \right)-Ei\left( 3ix \right)}{2} \right) \right)dx$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{y}_{P}}=\frac{4x}\left( {{e}^{2ix}}Ei\left( -ix \right)-\left( 1+{{e}^{2ix}} \right)Ei\left( ix \right)+Ei\left( 3ix \right) \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.22)$$
 * }
 * }


 * where $$\displaystyle Ei(x)$$ is the exponential integral


 * Therefore complete solution is the combination of $$\displaystyle (Eq. 2.22)$$ and $$\displaystyle (Eq. 2.9)$$,


 * {| style="width:100%" border="0" align="left"

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$y={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}+\frac{4x}\left( {{e}^{2ix}}Ei\left( -ix \right)-\left( 1+{{e}^{2ix}} \right)Ei\left( ix \right)+Ei\left( 3ix \right) \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.23)$$
 * }
 * }

= Problem 9 - Verification of First Homogenous Solutions of Legendre Equation=

Given

 * Legendre Equation,


 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left( 1-{{x}^{2}} \right){{y}^{''}}-2x{{y}^{'}}+n\left( n+1 \right)y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.1)$$
 * }
 * }


 * Legendre polynomials:


 * {| style="width:100%" border="0" align="left"

$$\displaystyle{{P}_{0}}(x)=1$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.2)$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$\displaystyle{{P}_{1}}(x)=x$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.3)$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.4)$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.5)$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.6)$$
 * }
 * }

Find

 * Verify that $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

(1) n=0,$$\displaystyle {{P}_{0}}(x)=1$$

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} y=1 & {{y}^{'}}=0 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:


 * {| style="width:100%" border="0" align="left"

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 0+0\times 1=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.7)$$
 * }
 * }

(2) n=1,$$\displaystyle{{P}_{1}}(x)=x$$

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} y=x & {{y}^{'}}=1 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:


 * {| style="width:100%" border="0" align="left"

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 1+2\times x=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.8)$$
 * }
 * }

(3) n=2,$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} y=\frac{1}{2}\left( 3{{x}^{2}}-1 \right) & {{y}^{'}}=3x & {{y}^{''}}=3\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:


 * {| style="width:100%" border="0" align="left"

$$\left( 1-{{x}^{2}} \right)\times 3-2x\times 3x+6\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.9)$$
 * }
 * }

(4) n=3,$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} y=\frac{1}{2}\left( 5{{x}^{3}}-3x \right) & {{y}^{'}}=\frac{1}{2}\left( 15{{x}^{2}}-3 \right) & {{y}^{''}}=15x\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:


 * {| style="width:100%" border="0" align="left"

$$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.10)$$
 * }
 * }

(5) n=4,$${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} y=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} & {{y}^{'}}=\frac{35}{2}{{x}^{3}}-\frac{15}{2}x & {{y}^{''}}=\frac{105}{2}{{x}^{2}}-\frac{15}{2}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:


 * {| style="width:100%" border="0" align="left"


 * $$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$=\frac{105}{2}{{x}^{2}}-\frac{105}{2}{{x}^{4}}-\frac{15}{2}+\frac{15}{2}{{x}^{2}}-\frac{70}{2}{{x}^{4}}+\frac{30}{2}{{x}^{2}}+\frac{175}{2}{{x}^{4}}-\frac{150}{2}{{x}^{2}}+\frac{15}{2}$$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$=\frac{2}\left( 175-105-70 \right)+\frac{2}\left( 105+15+30-150 \right)+\frac{15}{2}-\frac{15}{2}=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.11)$$
 * }
 * }


 * Hence $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

=problem3=

Rearranging:
 * {| style="width:100%" border="0"

$$ {Z}^{\prime}+ \left[ \frac{2 {{u}_{1}}^{\prime}}{{u}_{1}}- \frac{2x}{ \left(1- {x}^{2} \right)} \right]Z= \frac{1}{ {u}_{1} {\left(1- {x}^{2} \right)}^{2}} $$ The integrating factor is
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.8) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$h\left( x \right)=exp\left[ \int{\left( \frac{2{{u}_{1}}^{\prime }}-\frac{2x}{\left( 1-{{x}^{2}} \right)} \right)}dx \right]$$ Substituting in our homogeneous solution and integrating gives
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.9) $$
 * }
 * }


 * {| style="width:100%" border="0"

$$h\left( x \right)=exp\left[ log\left( 1-{{x}^{2}} \right)+2log\left( x \right) \right]$$
 * }
 * }


 * {| style="width:100%" border="0"

$$h\left( x \right)={{x}^{2}}\left( 1-{{x}^{2}} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.10) $$
 * }
 * }

The solution to equation 3.8 can be written as


 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{h\left( x \right)}\left[ {{k}_{2}}+\int{\left[ h\left( x \right)\frac{f\left( x \right)}{{{u}_{1}}(x)} \right]}dx \right]$$ Substituting in
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.11) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}{x{{\left( 1-{{x}^{2}} \right)}^{2}}} \right]}dx \right]$$
 * }
 * }


 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{x}{\left( 1-{{x}^{2}} \right)} \right]}dx \right]$$
 * }
 * }


 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}-\frac{1}{2}\log \left( 1-{{x}^{2}} \right) \right]$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.12) $$
 * }
 * }

Integrating
 * {| style="width:100%" border="0"

$$U\left( x \right)={{k}_{1}}+\int{Z}\left( x \right)dx$$
 * }
 * }


 * {| style="width:100%" border="0"

$$U\left( x \right)={{k}_{1}}+\int{\left( \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}-\frac{1}{2}\frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.13) $$
 * }
 * }

Therefore,


 * {| style="width:100%" border="0"

$$y\left( x \right)=U\cdot {{u}_{1}}\left( x \right)$$
 * }
 * }


 * {| style="width:100%" border="0"

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\int{\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}dx}-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }


 * {| style="width:100%" border="0"

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\cdot \frac{1}{2}\left( -\frac{2}{x}-\log (1-x)+\log 1+x \right)-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }


 * {| style="width:100%" border="0"

$$y\left( x \right)={{k}_{1}}\underbrace{x}_+{{k}_{2}}\left( \underbrace{\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1}_ \right)-\underbrace{\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx}_$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.14) $$
 * }
 * }