User:Egm6321.f09.Team1.vasquez/HW4

=Problem 1: Legendre differential equation=

Given
The Legendre differential equation with $$\displaystyle n=0$$
 * {| style="width:100%" border="0" align="left"

(1-x^2)y''-2x y' = 0 $$ $$ such that the first homogeneous solution is given by $$\displaystyle u_1(x)=1$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Use reduction of order method 2 (Undet. Fact.) to find $$\displaystyle u_2(x)$$, the second homogeneous solution.

Solution
Eq. 1 can be re-written as follows:


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

y'' - \frac{2x}{1-x^2} y' = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Then


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

a_1(x)=\frac{-2x}{1-x^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

and


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

a_0(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

The second homogeneous solution, $$\displaystyle u_2(x)$$, can be found directly as follows


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

u_2 (x) = u_1(x) \int_{}^x {\frac{1}} \exp \left[ { - \int_{}^t {a_1 (s)ds} } \right]dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

The integral of $$\displaystyle a_1(s)$$ can be found by using partial fractions:


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

\int_{}^t \frac{-2s}{1-s^2} ds= \int_{}^t \left[\frac{1}{1+s} -\frac{1}{1-s} \right]ds = \log(1+t)+\log(1-t)=\log(1-t^2) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Substituting $$\displaystyle u_1(t)=1$$ and Eq. 6 into Eq. 5 gives


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

u_2 (x) = (1) \int_{}^x {\frac{1}{(1)^2} \exp \left[ - \log(1-t^2) \right]}dt = \int_{}^x {\frac{1}{1-t^2}}dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

The integral in Eq. 7 can be solved by using partial fractions


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

\int_{}^x {\frac{1}{1-t^2}}dt = \int_{}^x \left[\frac{1/2}{1-s} +\frac{1/2}{1+s} \right]ds = -\frac{1}{2}\log(1-x)+\frac{1}{2}\log(1+x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

Therefore, the second homogeneous solution is given by


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


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

u_2(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ $$
 * $$\displaystyle (Eq. 9)
 * }
 * }

=Problem 2: Application of the reduction of order method 2 (King, 2003. Problem 1.1 b )=

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

xy''+2 y'+xy = 0 $$ $$ such that the first homogeneous solution is given by $$\displaystyle u_1(x)=x^{-1}\sin(x)$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Use reduction of order method 2 (Undet. Fact.) to find $$\displaystyle u_2(x)$$, the second homogeneous solution.

Solution
Eq. 1 can be re-written as follows:


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

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

Then


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

a_1(x)=\frac{2}{x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

and


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

a_0(x)=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

The second homogeneous solution, $$\displaystyle u_2(x)$$, can be found directly as follows


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

u_2 (x) = u_1(x) \int_{}^x {\frac{1}} \exp \left[ { - \int_{}^t {a_1 (s)ds} } \right]dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

The integral of $$\displaystyle a_1(s)$$ can be found as follows


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

\int_{}^t \frac{2}{s} ds= 2\log(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Substituting $$\displaystyle u_1(t)$$ and Eq. 6 into Eq. 5 gives


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

u_2 (x) = \frac{\sin(x)}{x} \int_{}^x {\frac{1}{(t^{-1}\sin(t))^2} \exp \left[ - 2 \log(t) \right]}dt = \frac{\sin(x)}{x} \int_{}^x {\frac{t^2}{\sin^2(t)} \exp \left[ \log \left(\frac{1}{t^2} \right) \right]}dt = \frac{\sin(x)}{x} \int_{}^x {\frac{1}{\sin^2(t)}}dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Then,


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

\int_{}^x {\frac{1}{\sin^2(t)}}dt = \int_{}^x {\csc^2(t)}dt = -\cot(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Substituting Eq. 8 into Eq. 7 gives


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

u_2(x)=\frac{\sin(x)}{x}\cdot\frac{-\cos(x)}{\sin(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Therefore, the second homogeneous solution is given by


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


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

u_2(x)=-\frac{\cos(x)}{x} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

=Notes and references=