User:Egm6321.f11.team1.zheng.zx/personal place/HW6

= R*6.7 Using variation of parameters method to get the 2nd homogeneous solution $$Q_2(x)$$=

Given
Legengre differential equation:
 * {| style="width:100%" border="0" align="left"

\displaystyle (1-x^2)y''-2xy'+n(n+1)y=0 $$ $$
 * $$\displaystyle (Equation\; 6.7.1)
 * }
 * }

For n=2
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\displaystyle (1-x^2)y''-2xy'+6y=0 $$ $$
 * $$\displaystyle (Equation\; 6.7.2)
 * }
 * }

First homogeneous solution:
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\displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ $$
 * $$\displaystyle (Equation\; 6.7.3)
 * }
 * }

== Find == Show the second homogeneous solution:
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\displaystyle Q_2(x)=\frac{1}{4}(3x^2-1)\log \left ( \frac{1+x}{1-x} \right )-\frac{3}{2}x $$ $$
 * $$\displaystyle (Equation\; 6.7.4)
 * }
 * }

Solution
For a Non-homogeneous L2-ODE-VC:
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\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 * $$\displaystyle (Equation\; 6.7.5)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x).

$$ \displaystyle y(x)=U(x)u_1(x) $$

$$ \displaystyle \begin{align} y&=Uu_1 \\ y'&=Uu_1'+U'u_1 \\ y&=Uu_1+2U'u_1'+U''u_1 \end{align} $$


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\displaystyle a_0y+a_1y'+y=U[a_0u_1+a_1u_1'+u_1]+U'[a_1u_1+2u_1']+U''u_1=f(x) $$ $$
 * $$\displaystyle (Equation\; 6.7.6)
 * }
 * }

Let

$$ \displaystyle Z:=U' $$

$$ \displaystyle Z'+\frac{a_1u_1+2u_1'}{u_1}Z=\frac{f}{u_1} $$

$$ \displaystyle \begin{align} \displaystyle h(x)&=exp\left [ \int \left ( a_1(x)+\frac{2u_1'(x)}{u_1(x)} \right ) dx \right ] \\ \displaystyle &=u_1^2(x)exp\left [\int a_1(x)dx \right ] \end{align} $$


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\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 * $$\displaystyle (Equation\; 6.7.7)
 * }
 * }
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\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 * $$\displaystyle (Equation\; 6.7.8)
 * }
 * }

$$ \displaystyle \begin{align} \displaystyle u_2(x):&=u_1(x)\int \frac{1}{h(x)}dx \\ \displaystyle &=u_1(x)\int \frac{1}{u_1^2(x)}exp \left [ -\int a_1(x)dx \right ]dx \end{align} $$

For

$$ \displaystyle \begin{align} u_1(x)&=P_2(x)=\frac{1}{2}(3x^2-1) \\ \displaystyle a_1(x)&=\frac{-2x}{1-x^2} \end{align} $$


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\displaystyle \begin{align} \displaystyle \frac{1}{h(x)}&=\frac{4}{(3x^2-1)^2}\cdot exp \left [ -\int \frac{-2x}{1-x^2}dx \right ] dx\\ \displaystyle &=\frac{4}{(3x^2-1)^2(1-x^2)} \end{align} $$ $$
 * $$\displaystyle (Equation\; 6.7.9)
 * }
 * }


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\displaystyle \begin{align} \displaystyle Q_2(x)=u_2(x)&=\frac{1}{2}(3x^2-1)\int \frac{4}{(3x^2-1)^2(1-x^2)}dx \\ \displaystyle &=\frac{1}{4}(3x^2-1)\left (-\frac{6x}{3x^2-1}-\log (1-x)+\log (1+x)\right ) \end{align} $$ $$
 * $$\displaystyle (Equation\; 6.7.10)
 * }
 * }

Rearranging:

Computing with the help of WA, please refer to: WalframAlpha

= R 6.9 Non-homogeneous Legengre equation with $$f(x)=1$$=

Given
Non-homogeneous Legengre differential equation:
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\displaystyle (1-x^2)y''-2xy'+2y=f(x) $$ $$
 * $$\displaystyle (Equation\; 6.9.1)
 * }
 * }


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\displaystyle f(x)=1 $$
 * 
 * }
 * }

First homogeneous solution:


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\displaystyle u_1(x)=P_1(x)=x $$ $$
 * $$\displaystyle (Equation\; 6.9.2)
 * }
 * }

== Find == The final solution $$y(x)$$ by variation of parameters

Solution
For a Non-homogeneous L2-ODE-VC:
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\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 * $$\displaystyle (Equation\; 6.9.3)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x).

$$ \displaystyle y(x)=U(x)u_1(x) $$

$$ \displaystyle \begin{align} y&=Uu_1 \\ y'&=Uu_1'+U'u_1 \\ y&=Uu_1+2U'u_1'+U''u_1 \end{align} $$


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\displaystyle a_0y+a_1y'+y=U[a_0u_1+a_1u_1'+u_1]+U'[a_1u_1+2u_1']+U''u_1=f(x) $$ $$
 * $$\displaystyle (Equation\; 6.9.4)
 * }
 * }

Let

$$ \displaystyle Z:=U' $$

$$ \displaystyle Z'+\frac{a_1u_1+2u_1'}{u_1}Z=\frac{f}{u_1} $$

$$ \displaystyle \begin{align} \displaystyle h(x)&=exp\left [ \int \left ( a_1(x)+\frac{2u_1'(x)}{u_1(x)} \right ) dx \right ] \\ \displaystyle &=u_1^2(x)exp\left [\int a_1(x)dx \right ] \end{align} $$


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\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 * $$\displaystyle (Equation\; 6.9.5)
 * }
 * }
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\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.6)
 * }
 * }

For

$$ \displaystyle \begin{align} \displaystyle u_1(x)&=P_1(x)=x \\ \displaystyle a_1(x)&=\frac{-2x}{1-x^2} \\ \displaystyle f(x)&=1 \end{align} $$


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\displaystyle \begin{align} \displaystyle h(x)&=u_1^2(x)exp \left [ \int a_1(x)dx \right ] \\ \displaystyle &=x^2exp \left [ \log (1-x^2)\right ] \\ \displaystyle &=x^2 (1-x^2) \end{align} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.7)
 * }
 * }


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\displaystyle \begin{align} \displaystyle Z(x)&=\frac{1}{x^2(1-x^2)}\left [ k_2+\int \frac{x^2(1-x^2)}{x}dx \right ] \\ \displaystyle &=\frac{1}{x^2(1-x^2)}\left [ k_2+\frac{1}{4}x^2(2-x^2)\right ] \end{align} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.8)
 * }
 * }


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\displaystyle \begin{align} \displaystyle U(x)&=k_1+\int \frac{1}{x^2(1-x^2)}\left [ k_2+\frac{1}{4}x^2 (2-x^2) \right ]dx \\ \displaystyle &= k_1+k_2\left ( \frac{1}{2}\log \left ( \frac{1+x}{1-x}\right ) - \frac{1}{x} \right ) + \frac{1}{8} \left ( 2x+ \log \left ( \frac{1+x}{1-x} \right ) \right ) \end{align} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.9)
 * }
 * }

Final solution:

= References =