# User:Egm6321.f11.team4/HW6

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# Problem R*6.1 Solve Nonhomogeneous L2-ODE-CC

From Mtg 33-1

## Given

Equation (1) p.32-5 is $\displaystyle a_2y''+a_1y'+a_0y=F(t)$.

Equation (1) p.16-5 is $\displaystyle f_{tt}+2pf_{ty}+p^2f_{yy}=g_{tp}+pg_{yp}-g_y$.

Equation (2) p.16-5 is $\displaystyle f_{tp}+pf_{yp}+2f_y=g_{pp}$.

## Find

$\displaystyle \bold {1)}$ Find the PDEs that govern the integrating factor $\displaystyle h(t,y)$ for equation (1) p.32-5. Can you solve these PDEs for $\displaystyle h(t,y)$?

$\displaystyle \bold {2.1)}$ Find $\displaystyle (\bar a_1, \ \bar a_0)$ in terms of $\displaystyle (a_0, \ a_1, \ a_2)$.

$\displaystyle \bold {2.2)}$ Find the quadratic equation for $\displaystyle \alpha$.

$\displaystyle \bold {2.3)}$ Find the reduced-order equation.

$\displaystyle \bold {2.4)}$ Use the IFM to solve (3) p.33-3.

$\displaystyle \bold {2.5)}$ Show that $\displaystyle \alpha \beta = \frac{a_0}{a_2}, \ \ \alpha + \beta = \frac{a_1}{a_2}$.

$\displaystyle \bold {2.6)}$ Deduce the particular solution $\displaystyle y_p(t)$ for general excitation $\displaystyle F(t)$.

$\displaystyle \bold {2.7)}$ Verify result with table of particular solutions for: $\displaystyle F(t)=t\exp(bt)$.

$\displaystyle \bold {2.8)}$ Solve the nonhomogeneous L2-ODE-CC (1) p.32-5 with $\displaystyle F(t)=\tanh t$.

For the coefficients $\displaystyle (a_0, \ a_1, \ a_2)$, consider two different characteristic equations:

$\displaystyle \bold {2.8.1)}$ $\displaystyle (r+1)(r-2)=0$.

$\displaystyle \bold {2.8.2)}$ $\displaystyle (r-4)^2=0$.

$\displaystyle \bold {2.9)}$ Determine the fundametal period of undamped free vibration, and plot the solution for the excitation $\displaystyle F(t)=\tanh t$ for about 5 periods, assuming zero initial conditions.

## Solution Part 1

$\displaystyle \bold {1)}$ We need equation (1) p.32-5 to satisfy the first exactness condition for N2-ODEs, $\displaystyle g(t,y,p)+f(t,y,p)y''=0$.

Hence, $\displaystyle f(t,y,p)=h(t,y)*a_2$

$\displaystyle g(t,y,p)=h(t,y)*[a_1y'+a_0y-F(t)]$.

Using the second exactness condition {equations (1)-(2) p. 16-5}, we obtain the following PDEs:

$\displaystyle a_2{h_{tt}}+2pa_2h_{ty}+p^2a_2h_{yy}=a_1h_t+\cancel{pa_1h_y}-\cancel{a_1ph_y}-a_0h-a_0yh_y-F(t)h_y$.

$\displaystyle 2a_2h_y=0$.

Combine the two equations to obtain

$\displaystyle a_2h_{tt}-a_1h_t+a_0h=0$.

The PDEs reduce to a single ODE that cannot be solved.

## Solution Part 2.1

We are given the trial solution as $\displaystyle h(t)=\exp(\alpha t)$.

So, $\displaystyle \int \exp(\alpha t)[a_2y''+a_1y'+a_0y]dt=\int \exp(\alpha t) F(t)dt$.

Then, $\displaystyle \int \exp(\alpha t)[a_2y''+a_1y'+a_0y]dt=\exp(\alpha t)[\cancel{\bar a_2}y''+\bar a_1y'+\bar a_0y]$.

Differentiate the RHS to obtain, $\displaystyle \int \exp(\alpha t)[a_2y''+a_1y'+a_0y]dt=\int \exp(\alpha t)[\bar a_1y''+(\bar a_1\alpha+\bar a_0)y'+(\bar a_0\alpha)y]dt$.

Equate the two sides to obtain the three expressions below.

$\displaystyle a_2=\bar a_1$.

$\displaystyle a_1=\bar a_1 \alpha+\bar a_0$.

$\displaystyle a_0=\bar a_0 \alpha$.

Rearrange the second and third equations to yield

$\displaystyle \bar a_1= a_2$.

$\displaystyle \bar a_0=a_1-\alpha a_2$.

$\displaystyle \bar a_0=\frac{a_0}{\alpha}$.

## Solution Part 2.2

Using the last two equations, we can obtain the quadratic equation for $\displaystyle \alpha$.

$\displaystyle a_1-\alpha a_2=\frac{a_0}{\alpha}$.

$\displaystyle \alpha*[a_1-\alpha a_2]=a_0$

$\displaystyle \alpha a_1-\alpha^2 a_2=a_0$

$\displaystyle a_2 \alpha^2 - a_1 \alpha + a_0 = 0$

## Solution Part 2.3

The reduced-order equation is obtained using equations (1) p.33-2 and (2) p.33-1.

$\displaystyle \exp(\alpha t)[\bar a_1y'+\bar a_0y]=\int \exp(\alpha t)F(t)dt$

$\displaystyle \bar a_1y'+\bar a_0y=\exp(-\alpha t)\int \exp(\alpha t)F(t)dt$

## Solution Part 2.4

Next, we will use the IFM to solve the reduced-order equation.

$\displaystyle \bar h(t)=\exp[\int \beta dt]=\exp(\beta t)$

where $\displaystyle \beta=\frac{\bar a_0}{\bar a_1}$

$\displaystyle y(t)=\frac{1}{\bar h(t)}\int^t \bar h(s)B(s)ds$

Divide the result of Part 2.3 by $\displaystyle \bar a_1$ to obtain B(s).

$\displaystyle y'+\beta y = \frac{\exp(-\alpha t)}{\bar a_1} \int \exp(\alpha t)F(t)dt = B(s)$

Then we can combine $\displaystyle h(t), \ B(s)$ into $\displaystyle y(t)$ to find

$\displaystyle y(t)=\frac{1}{[\exp(\beta t)]}\int^t [\exp(\beta s)] [\frac{\exp(-\alpha s)}{\bar a_1} \int \exp(\alpha t)F(t)dt] ds$.

$\displaystyle y(t)=\frac{1}{\bar a_1[\exp(\beta t)]}\int^t [\exp((\beta-\alpha) s)] \left[ \int \exp(\alpha t)F(t)dt \right] ds$

## Solution Part 2.5

Rearrange $\displaystyle \beta=\frac{\bar a_0}{\bar a_1}$ to $\displaystyle \bar a_0=\beta\bar a_1$.

Substitute $\displaystyle a_2=\bar a_1$ to $\displaystyle \bar a_0=a_2\beta$.

That's equal to $\displaystyle \frac{a_0}{\alpha}$ and after rearranging one obtains

$\displaystyle \alpha\beta=\frac{a_0}{a_2}$.

To reach the other equality, just set the equation equal to $\displaystyle a_1-\alpha a_2$ instead of $\displaystyle \frac{a_0}{\alpha}$.

$\displaystyle a_2\beta=a_1-\alpha a_2$.

Rearranging yields $\displaystyle \beta+\alpha=\frac{a_1}{a_2}$.

## Solution Part 2.6

Referencing R5.7, the below equation is given for an Euler L2-ODE-CC.

$\displaystyle y(x) = C_1 x e^{\lambda x} + C_2 e^{\lambda x}$.

After careful inspection, one can deduce that the particular solution is the last equation for Part 2.4 WITHOUT the constants of integration.

$\displaystyle y(t)=\frac{1}{\bar a_1[\exp(\beta t)]}\int^t [\exp((\beta-\alpha) s)] \left[ \int \exp(\alpha t)F(t)dt \right] ds$

## Solution Part 2.7

$\displaystyle \int e^{(\alpha t)} F(t) dt$

$\displaystyle F(t) = t e^{bt}$

$\displaystyle \int t e^{(\alpha+b)t} dt$

$\displaystyle \frac{e^{(\beta-\alpha)t} ( b t + \alpha t - 1 )}{(\alpha + b)^2}$

$\displaystyle \int e^{(\beta-\alpha)s} \left[ e^{(\alpha+b)s} \frac{(bs+\alpha s-1)}{(\alpha + b)^2} \right] ds$

$\displaystyle \int e^{(\beta+b)s} \frac{(bs+\alpha s-1)}{(\alpha + b)^2} ds$

$\displaystyle e^{(\beta+b)t} \frac{\frac{s(\alpha+b)}{b+\beta} - \frac{\alpha+2b+\beta} {(b+\beta)^2}} {(\alpha + b)^2}$

$\displaystyle C_3 = \frac{\alpha+b}{b+\beta}$

$\displaystyle C_4 = \frac{\alpha+2b+\beta}{(b+\beta)^2}$

$\displaystyle C_5 = (\alpha+b)^2$

$\displaystyle e^{(\beta+b)t} \frac{C_3t - C_4}{C_5}$

$\displaystyle y_P(t) = e^{bt} \frac{1}{a_2} \frac{C_3t - C_4}{C_5}$

$\displaystyle y_P(t) = C_{P1} t e^{bt} + C_{P2} e^{bt}$

$\displaystyle C_{P0} = \frac{C_3}{a_2 C_5}$

$\displaystyle C_{P1} = \frac{-C_4}{a_2 C_5}$

NOTE: The following two integrals were performed using Wolfram|Alpha.

integrate t*exp((\alpha+b)*t) dt

integrate (exp((\beta+b)*s)*((b*s+\alpha*s-1)/((\alpha+b)^2)) ds

Now compare to the Table.

$\displaystyle f(t) = t e^{bt}$

$\displaystyle y_P(t) = x^m (\sum_{i=0}^n c_i t_i) e^{bt}$

$\displaystyle y_P(t) = \cancel{x^m} C_{P0} \cancel{t^0} e^{bt} + \cancel{x^m} t C_{P1} e^{bt}$

$\displaystyle y_P(t) = C_{P0} e^{bt} + t C_{P1} e^{bt}$

Other than the arbitrary constant numbering convention, they match.

## Solution Part 2.8.1

$\displaystyle y(t)=\frac{1}{\cancel{\bar a_1}[\exp(\beta t)]}\int^t [\exp((\beta-\alpha) s)] \left[ \int \exp(\alpha t)F(t)dt \right] ds$

$\displaystyle \int e^{(\alpha t)} F(t) dt$

$\displaystyle F(t) = \tanh{t}$

$\displaystyle \int e^{(\alpha t)} [\tanh{t}] dt$

$\displaystyle y(t)''-y(t)'-2=\tanh(t)$

$\displaystyle a_2y(t)''+a_1y(t)'+a_0=\tanh(t)$

$\displaystyle \alpha=-1, \ \beta=2$

$\displaystyle e^{-t} - 2 \tan^{-1}(e^{-t})$

$\displaystyle \int^t e^{3s} \left[ e^{-s} - 2 \tan^{-1}(e^s) \right] ds$

$\displaystyle \frac{1}{6}[e^{2s}+2\log(e^{2s}+1)-4e^{3s}(tan^{-1}(e^{-s}))]$

$\displaystyle y_P(t) = \frac{1}{6} \frac{[e^{2s}+2\log(e^{2s}+1)-4e^{3s}(tan^{-1}(e^{-s}))]}{e^{-2t}}$

NOTE: Wolfram|Alpha was used to determine the following integrals:

integrate (e^(-t)*tanh(t)) dt

integrate (e^(3s))(e^(-s)-2 tan^(-1)(e^(-s))) ds

## Solution Part 2.8.2

$\displaystyle \alpha=4, \ \beta=4$

$\displaystyle y(t)=\frac{1}{\cancel{\bar a_1}[\exp(\beta t)]}\int^t [\exp((\beta-\alpha) s)] \left[ \int \exp(\alpha t)F(t)dt \right] ds$

$\displaystyle y(t)=e^{-4t} \int^t \cancel{e^{(\beta-\alpha)s}} \left[ \int e^{4t} F(t) dt \right] ds$

$\displaystyle \int e^{4t} F(t) dt$

$\displaystyle F(t) = \tanh{t}$

$\displaystyle \int e^{4t} [\tanh{t}] dt$

$\displaystyle -e^{2t} + \frac{e^{4t}}{4} + \log(e^{2t}+1)$

$\displaystyle \frac{1}{16} \left[ -8[Li_2 (-e^{2t})] - 8 e^{2t} + e^{4t} \right]$

$\displaystyle y_P(t) = \frac{e^{-4}}{16} \left[ -8[Li_2 (-e^{2t})] - 8 e^{2t} + e^{4t} \right]$

where $\displaystyle Li_2(-e^{2t})]$ is the polylogarithm function

NOTE: Wolfram|Alpha was used to determine the following integrals:

integrate ((e^(4t))(\tanh(t))) dt

integrate (-e^(2t)+(e^(4t)/4)+log(e^(2t)+1)) dt

## Author

Egm6341.f11.team4.allen 10:56, 16 November 2011 (UTC)

# Problem R*6.2 - Show (1)p.34-6 agrees with King 2003 (1.6)p.8

From Mtg 35-1

## Given

Equation (1) p.34-6 is $\displaystyle y_P(x)=u_1(x)\int\frac{1}{h(x)}\left[\int h(x)\frac{f(x)}{u_1(x)}\,dx\right]\,dx$.

## Find

$\displaystyle \bold {A)}$ Show that Equation (1) p.34-6 agrees with King 2003 (1.6) p.8 which is

$\displaystyle y_P(x)=\int^x f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds$

with

$\displaystyle W := u_1 u_2' - u_2 u_1' = \begin{vmatrix} \ u_1 & u_2 \\ u_1' & u_2' \end{vmatrix}$.

$\displaystyle \bold {HINT} \left( \frac{u_2}{u_1} \right)' = \frac{1}{h}$

$\displaystyle \bold {B)}$ Also, discuss the feasibility of the following choices for variation of parameters:

$\displaystyle \bold {B.1)} y(x) = U(x) \pm u_1(x)$.

$\displaystyle \bold {B.2)} y(x) = U(x) / u_1(x)$.

$\displaystyle \bold {B.3)} y(x) = u_1(x) / U(x)$.

## Solution

$\displaystyle \bold {A)}$ Starting with the HINT, we'll apply the following basic differentiation rule:

$\displaystyle \frac{d}{dx}\left[\frac{u}{v}\right]=\frac{vu'-uv'}{v^2}$ to find

$\displaystyle \frac{d}{dx}\left[\frac{u_2}{u_1}\right] = \frac{u_1u_2'-u_2u_1'}{u_1^2}$.

So, $\displaystyle h = \frac{u_1^2}{u_1u_2'-u_2u_1'}$.

Next, we'll use integration by parts, $\displaystyle \int u dv = uv-\int v du$ with the following:

$\displaystyle u = \int h(x) \frac{f(x)}{u_1(x)} dx$.

$\displaystyle dv = \frac{1}{h} = \frac{u_1u_2'-u_2u_1'}{u_1^2} dx$.

$\displaystyle v = \frac{u_2}{u_1}$.

$\displaystyle du = \frac{h(x)f(x)}{u_1(x)} dx$.

$\displaystyle y_P(x) = u_1(x)\int\frac{1}{h(x)}\left[\int h(x)\frac{f(x)}{u_1(x)}\,dx\right]\,dx$

$\displaystyle y_P(x) = u_1(x) \int u dv$

$\displaystyle y_P(x) = u_1(x) [uv - \int v du]$

$\displaystyle y_P(x) = u_1(x) \left[ \frac{u_2(x)}{u_1(x)} \int h(s) \frac{f(s)}{u_1(s)}ds - \int \frac{u_2(s)}{u_1(s)} \frac{h(s)f(s)}{u_1(s)} ds \right]$

$\displaystyle y_P(x) = \cancel{u_1(x)} \frac{u_2(x)}{\cancel{u_1(x)}} \int \frac{u_1(s)^\cancel{2}}{u_1(s)u_2(s)'-u_2(s)u_1(s)'} \frac{f(s)}{\cancel{u_1(s)}}ds - u_1(x) \int \frac{u_2(s)}{\cancel{u_1(s)}} \frac{\cancel{u_1(s)^2}}{u_1(s)u_2(s)'-u_2(s)u_1(s)'} \frac{f(s)}{\cancel{u_1(s)}} ds$

$\displaystyle y_P(x) = \int^x f(s) \frac{u_1(s)u_2(x)}{u_1(s)u_2(s)'-u_2(s)u_1(s)'} ds - \int^x f(s) \frac{u_1(x)u_2(s)}{u_1(s)u_2(s)'-u_2(s)u_1(s)'} ds$

$\displaystyle y_P(x) = \int^x f(s) \frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{u_1(s)u_2(s)'-u_2(s)u_1(s)'} ds$

$\displaystyle y_P(x)=\int^x f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds$

with

$\displaystyle W := u_1 u_2' - u_2 u_1' = \begin{vmatrix} \ u_1 & u_2 \\ u_1' & u_2' \end{vmatrix}$.

$\displaystyle \bold {B.1)} y(x) = U(x) \pm u_1(x)$.

$\displaystyle [y = U(x) \pm u_1(x)] * a_0(x)$.

$\displaystyle [y' = U'(x) \pm u_1'(x)] * a_1(x)$.

$\displaystyle [y'' = U''(x) \pm u_1''(x)] * a_2(x)$.

$\displaystyle a_0y+a_1y'+y'' = U[a_0]+U'[a_1]+U'' +$ left over terms without U and its derivatives.

$\displaystyle \bold {B.2)} y(x) = U(x) / u_1(x)$.

$\displaystyle \left[ y = U(x) / u_1(x) \right] * a_0(x)$.

$\displaystyle \left[ y' = \frac{u_1(x)U'(x)-U(x)u_1'(x)}{u_1(x)^2} \right] * a_1(x)$.

$\displaystyle \left[ y'' = \frac{2U(x)u_1'(x)^2-u_1(x)[U(x)u_1''(x)+2u_1'(x)U'(x)]+u_1(x)^2U''(x)}{u_1(x)^3} \right] * a_2(x)$.

$\displaystyle \bold {B.3)} y(x) = u_1(x) / U(x)$.

$\displaystyle \left[ y = u_1(x) / U(x) \right] * a_0(x)$.

$\displaystyle \left[ y' = \frac{U(x)u_1'(x)-u_1(x)U'(x)}{U(x)^2} \right] * a_1(x)$.

$\displaystyle \left[ y'' = \frac{2u_1(x)U'(x)^2-U(x)[u_1(x)U''(x)+2U'(x)u_1'(x)]+U(x)^2u_1''(x)}{U(x)^3} \right] * a_2(x)$.

None of these alternative trial solutions are feasible because they don't produce the desired homogeneous equation that cancels out to simplify the equation. Also, there are several terms that do not contain $\displaystyle U(x)$ which were a requirement as well.

## Author

Egm6341.f11.team4.allen 06:00, 16 November 2011 (UTC)

# Problem R*6.3 - Invalid solution

From Mtg 35-4

## Given

 $\displaystyle (x-1)y^{''}-xy^'+y=0$ (3.1)

The first homogeneous solution:

 $\displaystyle \displaystyle u_1(x)=e^x$ (3.2)

Trial solution:

 $\displaystyle y=e^{rx}, r=\mbox{constant}$ (3.3)

Characteristic equation:

 $\displaystyle (x-1)r^2-xr+1 = 0$ (3.4)
 \displaystyle \begin{align} &r_1 = 1 \\ &r_2(x) = \frac{1}{x-1} \end{align} (3.5)

## Find

Explain why $\displaystyle r_2(x)$ is not a valid root, i.e., $\displaystyle u_2(x) = e^{xr_{2}(x)}$ is not a valid solution.

## Solution

- Solved on our own


Validity of $\displaystyle u_2(x) = e^{xr_2(x)}$ can be shown by substituting $\displaystyle y= u_2(x)$ into the Eq. (4.1). Before the substitution, compute the first and the second derivatives of $\displaystyle y$.

 \displaystyle \begin{align} y &= e^{\frac{x}{x-1}} \\ y^{'} &= ( \frac{x}{x-1})^{'}e^{\frac{x}{x-1}} \\ &= \frac{(x-1)-x}{(x-1)^2}e^{\frac{x}{x-1}} \\ &= \frac{-1}{(x-1)^2}e^{\frac{x}{x-1}} \\ y^{''} &= \left[\frac{-1}{(x-1)^2}\right]^{'}e^{\frac{x}{x-1}} + \frac{-1}{(x-1)^2}\left[e^{\frac{x}{x-1}}\right]^{'} \\ &= \frac{-(-1) \cdot 2(x-1)}{(x-1)^4}e^{\frac{x}{x-1}} + \frac{-1}{(x-1)^2}\left[ \frac{-1}{(x-1)^2}e^{\frac{x}{x-1}} \right] \\ &= \frac{2(x-1)}{(x-1)^4} e^{\frac{x}{x-1}} + \frac{1}{(x-1)^4}e^{\frac{x}{x-1}} \\ &= \frac{2x-1}{(x-1)^4}e^{\frac{x}{x-1}} \end{align} (3.6)

Substitute $\displaystyle y, y^{'}, y^{''}$ in the Eq. (3.6) into the Eq. (3.1),

 \displaystyle \begin{align} (x-1)y^{''}-xy^'+y &= (x-1) \cdot \frac{2x-1}{(x-1)^4}e^{\frac{x}{x-1}} - x \cdot \frac{-1}{(x-1)^2}e^{\frac{x}{x-1}} + e^{\frac{x}{x-1}} \\ &= \left[ (x-1) \cdot \frac{2x-1}{(x-1)^4} - x \cdot \frac{-1}{(x-1)^2}+1 \right] e^{\frac{x}{x-1}} \\ &= \left[ \frac{2x-1}{(x-1)^3} + \frac{x}{(x-1)^2}+1 \right] e^{\frac{x}{x-1}} \\ &= \left[ \frac{2x-1}{(x-1)^3} + \frac{x(x-1)}{(x-1)^3}+ \frac{(x-1)^3}{(x-1)^3} \right] e^{\frac{x}{x-1}} \\ &= \left[ \frac{2x-1+x^2-x+x^3-3x^2+3x-1}{(x-1)^3} \right] e^{\frac{x}{x-1}} \\ &= \left[ \frac{x^3-2x^2+4x-2}{(x-1)^3} \right] e^{\frac{x}{x-1}} \\ &\neq 0 \end{align} (3.7)
- As shown in the Eq. (3.7), $\displaystyle u_2(x)$ is not a valid homogeneous solution.


# Problem R*6.4 - 2nd homogeneous solution by variation of parameters

From Mtg 35-4

## Given

Application: King 2003 p.28

 $\displaystyle (x-1)y^{''}-xy^{'}+y=0$ (4.1)

## Find

For the L2-ODE-VC Eq. (4.1), select a valid homogeneous solution, and call it $\displaystyle u_1$. Find the 2nd homegeneous solution $\displaystyle u_2(x)$ by variation of parameters, and compare to $\displaystyle e^{xr_2(x)}$.

## Solution

- Solved on our own


Let trial solution be:

 $\displaystyle y=e^{rx}, \mbox{ }r=\mbox{constant}$ (4.2)

Then, characteristic equation is:

 $\displaystyle (x-1)r^2-xr+1 = 0$ (4.3)

After factoring, the Eq. (4.3) becomes:

 $\displaystyle (r-1)\left[(x-1)r-1 \right] = 0$ (4.4)
 \displaystyle \begin{align} r_1 &= 1 \\ r_2(x) &= \frac{1}{x-1} \end{align} (4.5)

Therefore, the first homogeneous solution is:

 \displaystyle \begin{align} u_1(x) = e^x \end{align} (4.6)

Since it was proven that $\displaystyle r_2(x)$ is not a valid root in the problem R*6.3, $\displaystyle r_2(x)$ is ignored.
The second homogeneous solution can be obtained using the equation (4) in the class note 34-5. The Eq. (4.7) depicts the equation (4) in the class note 34-5.

 $\displaystyle u_2(x) = u_1(x) \int \frac{1}{h(x)} dx$ (4.7)
 $\displaystyle h(x) = u_1(x)^2 \cdot \exp{\left[ \int a_1(x) dx \right]}$ (4.8)

The term $\displaystyle a_1(x)$ is determined by changing the form of the Eq. (4.1) as follows.

 $\displaystyle y^{''}\underbrace{-\frac{x}{x-1}}_{a_1(x)}y^'+\underbrace{\frac{1}{x-1}}_{a_0(x)}y=0$ (4.9)

First, the term $\displaystyle \frac{1}{h(x)}$ is first computed in order to solve the Eq. (4.7). Substitute $\displaystyle a_1(x)$ in the Eq. (4.9) into the Eq. (4.8), then compute $\displaystyle \frac{1}{h(x)}$,

 \displaystyle \begin{align} \frac{1}{h(x)} &= \frac{1}{u_1(x)^2} \cdot \exp{\left[ - \int^x \frac{-s}{s-1} ds \right]} \\ &= \frac{1}{u_1(x)^2} \cdot \exp{\left[ - \int^x \frac{-s+1-1}{s-1} ds \right]} \\ &= \frac{1}{u_1(x)^2} \cdot \exp{\left[ - \int^x -1-\frac{1}{s-1} ds \right]} \\ &= \frac{1}{u_1(x)^2} \cdot \exp{\left[ x + \log{(x-1)} \right]} \\ &= \frac{1}{u_1(x)^2} \cdot \exp{(x)} \cdot \exp{\left[ \log{(x-1)} \right]} \\ &= \frac{1}{e^{2x}} \cdot e^x \cdot e^{\log{(x-1)}} \leftarrow (u_1(x)=e^x)\\ &= (x-1)e^{-x} \end{align} (4.10)

Then, substitute $\displaystyle \frac{1}{h(x)}$ from the Eq. (4.10) into the Eq. (4.7),

 \displaystyle \begin{align} u_2(x) &= u_1(x) \int \frac{1}{h(x)} dx \\ &= u_1(x) \int^x (s-1)e^{-s} ds \\ &= u_1(x) \left[ -(x-1)e^{-x} + \int^x e^{-s} ds \right] \mbox{ : integration by parts} \\ &= u_1(x) \left[ -(x-1)e^{-x}-e^{-x} \right] \\ &= u_1(x) \left[ -xe^{-x} \right] \\ &= e^x \left[ -xe^{-x} \right] \\ &= -x \end{align} (4.11)
- Since the second homogeneous solution has the form of $\displaystyle u_2(x) = -x$, $\displaystyle u_2(x) = e^{x r_2(x)}$ is not valid solution.


# Problem R*6.5 - Homogeneous solutions of L2-ODE-VC

From Mtg 36-3

## Given

A L2-ODE-VC

 $\displaystyle (x-1)y^{''}-(2x+3)y^{'}+2y=0$ (5.1)

## Find

For the L2-ODE-VC Eq. (5.1), find 2 homogeneous solutions $\displaystyle u_1(x)$ and $\displaystyle u_2(x)$ using the trial solution $\displaystyle y(x) = e^{r(x)}$.

## Solution

- Solved on our own


Let trial solution be:

 $\displaystyle y=e^{rx}, \mbox{ }r=\mbox{constant}$ (5.2)

Then, characteristic equation is:

 $\displaystyle (x+1)r^2-(2x+3)r+2 = 0$ (5.3)

Upon factoring we get,

 $\displaystyle (r-2)\left[(x+1)r-1 \right] = 0$ (5.4)
 \displaystyle \begin{align} r_1 &= 2 \\ r_2(x) &= \frac{1}{x+1} \end{align} (5.5)

First homogeneous solution is thus:

 \displaystyle \begin{align} u_1(x) = e^{2x} \end{align} (5.6)

Since $\displaystyle r_2(x)$ is not a constant, therefore it's not a valid root and we have only one valid root, i.e. $\displaystyle r_1 = 2$.
$\displaystyle u_2(x)$ can be obtained using procedure, Variation of parameters, as given in equation (4) in class notes 34-5. Therefore we have,

 $\displaystyle u_2(x) = e^{2x} \int \frac{1}{h(x)} dx$ (5.7)

h(x) is given by,

 $\displaystyle h(x) = u_1(x)^2 \cdot \exp{\left[ \int a_1(x) dx \right]}$ (5.8)

$\displaystyle a_1(x)$ is determined from Eq. (5.1) written in following form,

 $\displaystyle y^{''}\underbrace{-\frac{2x+3}{x+1}}_{a_1(x)}y^'+\underbrace{\frac{2}{x+1}}_{a_0(x)}y=0$ (5.9)

Therefore $\displaystyle h(x)$ is given by,

 $\displaystyle h(x) = e^{2x} \cdot \exp{\left[ \int -\frac{(2x+3)}{(x+1)} dx \right]}$ $\displaystyle h(x) = e^{2x} \cdot \exp{\left[-2x - ln(x+1) \right]}$ $\displaystyle h(x) = \frac {1}{x+1}$ (5.10)

Now substituting $\displaystyle \frac{1}{h(x)}$ from the Eq. (5.10) into the Eq. (5.7),

 \displaystyle \begin{align} u_2(x) &= u_1(x) \int \frac{1}{h(x)} dx \\ &= e^{2x} \int^x (s+1) ds \\ &= e^{2x} \frac{(x+1)^2}{2} \end{align} (5.11)
- The homogenous solutions for Eq. (5.1) are, $\displaystyle u_1(x) = e^{2x}$ and $\displaystyle u_2(x) = e^{2x} \frac {(x+1)^2}{2}$.


# Problem R*6.6 - Find the homogeneous L2-ODE-VC

From Mtg 36-10

## Given

Trial solution

 \begin{align} y(x)=\frac{e^{rx}}{{sinx}} \end{align} (6.1)

Characteristic equation

 \begin{align}{r_1=2}\end{align} (6.2)
 \begin{align}{r_2=\frac{1}{x+1}}\end{align} (6.3)
 \begin{align}{(r-2)(r-\frac{1}{x+1})=0}\end{align} (6.4)

## Find

The homogeneous L2-ODE-VC

## Solution

 Solved on our own

According to Eq.6.1, we can get

 \begin{align} & y=\frac{e^{xr}}{sinx} \\ & y'=\frac{e^{xr}}{sinx}\cdot (r-\frac{cosx}{sinx}) \\ & y''=\frac{e^{xr}}{sinx}\cdot (\frac{r^2+1-2cos^2x}{sin^2x}+\frac{2rcosx}{sinx}) \\ \end{align} (6.5)

So we have

 \begin{align} {a_2}\cdot \frac{r^2+1-2cos^2x+2rsinxcosx}{sin^2x}+{a_1}\cdot (r-\frac{cosx}{sinx})+{a_0}=(r-2)(r-\frac{1}{x+1}) \end{align} (6.6)

Rearrange the Eq.6.6, we obtain

 \begin{align} {r^2}\cdot \frac{a_2}{sin^2x}+{r}\cdot (\frac{2a_2cosx}{sinx}-a_1)+(\frac{a_2(1-2cos^2x)}{sin^2x}-\frac{a_1cosx}{sinx}+a_0)=r^2-r\cdot \frac{2x+3}{x+1}+\frac{2}{x+1} \end{align} (6.7)

Hence, we get the following equations

 \begin{align} \frac{a_2}{sin^2x}=1 \end{align} (6.8)
 \begin{align} \frac{2a_2cosx}{sinx}-a_1=-\frac{2x+3}{x+1} \end{align} (6.9)
 \begin{align} \frac{a_2(1-2cos^2x)}{sin^2x}-\frac{a_1cosx}{sinx}+a_0=\frac{2}{x+1} \end{align} (6.10)

Solving the equations above, we have

 \begin{align} a_2=sin^2x \end{align} (6.11)
 \begin{align} a_1=\frac{2x+3}{x+1}+2sinxcosx \end{align} (6.12)
 \begin{align} a_0=\frac{2sinx+(2x+3)cosx}{(x+1)sinx}+4cos^2x-1 \end{align} (6.13)

Therefore, the L2-ODE-VC is

 \begin{align} sin^2x\cdot y''+(\frac{2x+3}{x+1}+2sinxcosx)\cdot y'+[\frac{2sinx+(2x+3)cosx}{(x+1)sinx}+4cos^2x-1]\cdot y=0 \end{align} (6.14)

## Author

Contributed by Kexin Ren

# Problem R*6.7 - Show 37-4 (2) using variation of parameters

From Mtg 37-4

## Given

 \displaystyle \begin{align} P_2(x)=\frac{1}{2}(2x^2-1) \end{align} (7.1)

## Find

Show that 2nd homogeneous solutions of the Legendre equation can be obtained from the 1st homogeneous solutions by variation os parameters.

 \displaystyle \begin{align} Q_2(x)=\frac{1}{4}(3x^2-1)log(\frac{1+x}{1-x})-\frac{3}{2}x \end{align} (7.2)

## Solution

- Solved on our own


Given

 \displaystyle \begin{align} P_2(x)=\frac{1}{2}(2x^2-1) \end{align}

should be

 \displaystyle \begin{align} P_2(x)=\frac{1}{2}(3x^2-1) \end{align} (7.1)

The Legendre function is

 \displaystyle \begin{align} (1-x^2)y''-2xy'+n(n+1)y=0 \end{align} (7.3)

For a second order equation, n=2.

 \displaystyle \begin{align} (1-x^2)y''-2xy'+6y=0 \end{align} (7.4)

The general from of a homogeneous L2-ODE-VC is

 \displaystyle \begin{align} a_2y''+a_1y'+a_0y=0 \end{align} (7.5)

The general solution form is

 \displaystyle \begin{align} y(x)=U(x)u_1(x) \end{align} (7.6)

U(x) is unknown. $u_1(x)=P_2(x)$. So solve $u_2(x)$ with $u_1(x)$.

 \displaystyle \begin{align} y(x)=U(x)u_1(x) \end{align} (7.7)
 \displaystyle \begin{align} y'=U'(x)u_1(x)+U(x)u_1'(x) \end{align} (7.8)
 \displaystyle \begin{align} y''(x)&=U''(x)u_1(x)+U'(x)u_1'(x) + U'(x)u_1'(x)+ U(x)u_1''(x) \\ &=U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x) \end{align} (7.9)

Substitute (7.7),(7.8),(7.9) into (7.5)

 \displaystyle \begin{align} a_2(U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x)) + a_1(U'(x)u_1(x)+U(x)u_1'(x))+a_0U(x)u_1(x)=0 \end{align} (7.10)
 \displaystyle \begin{align} U(x)\cancelto{0} {(a_2u_1''(x)+a_1u_1'(x)+a_0u_1(x))} + U'(x)(2a_2u_1'(x)+a_1u_1(x))+U''(x)a_2u_1(x)=0 \end{align} (7.11)
 \displaystyle \begin{align} U'(x)(2a_2u_1'(x)+a_1u_1(x))+U''(x)a_2u_1(x)=0 \end{align} (7.12)

To simplify, set $a_2=1$.

 \displaystyle \begin{align} U'(x)(2u_1'(x)+a_1u_1(x))+U''(x)u_1(x)=0 \end{align} (7.13)

Set Z=U' to reduce the order,

 \displaystyle \begin{align} Z(x)(2u_1'(x)+a_1u_1(x))+Z'(x)u_1(x)=0 \end{align} (7.14)

Divide $u_1(x)$.

 \displaystyle \begin{align} Z(x)(2\frac{u_1'(x)}{u_1(x)}+a_1)+Z'(x)=0 \end{align} (7.15)

To solve (7.15), use IFM to make the equation exact. Then it will be easier to solve the equation.

 \displaystyle \begin{align} h(x)&=\exp{\left[ \int a_0(s)ds \right] } \\ &=\exp{\left[ \int ( \frac{2u_1'(x)}{u_1(x)}+a_1) ds \right] } \\ &=\exp{\left[ 2ln(u_1(x))+ \int (a_1) ds \right] } \\ &= u_1(x)^2 \exp{\left[ \int (a_1) ds \right] } \end{align} (7.16)

Multipliy h to the equation (7.15)

 \displaystyle \begin{align} & h(x)Z(x)(2\frac{u_1'(x)}{u_1(x)}+a_1)+h(x)Z'(x) \\ &=u_1(x)^2 \exp{\left[ \int (a_1) ds \right] } (2\frac{u_1'(x)}{u_1(x)}+a_1)Z(x)+hZ'(x) \\ &=h'(x)Z(x)+h(x)Z(x) \\ &=(h(x)Z(x))' \\ &= 0 \end{align} (7.17)
 \displaystyle \begin{align} Z(x)&=\frac{1}{h(x)} \int0dx \\ &=\frac{C_1}{h(x)} \end{align} (7.18)
 \displaystyle \begin{align} U(x) &= \int Z(x) dx \\ &= \int \frac{C_1}{h(x)} \\ &= \int \frac{C_1}{h(x)} + C_2 \end{align} (7.19)
 \displaystyle \begin{align} y(x)&=\left[ \int \frac{C_1}{h(x)} + C_2 \right]u_1(x) \\ &=\int \frac{C_1}{h(x)}u_1(x) + C_2u_1(x) \end{align} (7.20)
 \displaystyle \begin{align} \therefore u_2=u_1(x)\int \frac{C_1}{h(x)} \end{align} (7.21)

At (7.13), we set $a_2=1$ to simplify the process. So,

 \displaystyle \begin{align} y''-\frac{2x}{1-x^2}y'+\frac{6}{1-x^2}y=0 \end{align} (7.22)
 \displaystyle \begin{align} a_1=\frac{-2x}{1-x^2} \end{align} (7.23)
 \displaystyle \begin{align} a_0=\frac{6}{1-x^2} \end{align} (7.24)

From (7.16), (7.23) and (7.1),

 \displaystyle \begin{align} h(x)&=\left[ \frac{1}{2}(3x^2-1) \right]^2 \exp \left[ \int\frac{-2x}{1-x^2} \right] \\ &=\frac{1}{4}(3x^2-1)^2(1-x^2) \end{align} (7.25)

From (7.21),

 \displaystyle \begin{align} u_2(x)&=u_1(x) \int \frac{C_1}{h(x)} dx\\ &=\frac{1}{2}(3x^2-1)\int \frac{1}{\frac{1}{4}(3x^2-1)^2(1-x^2)} dx \\ &=\frac{1}{2}(3x^2-1)\frac{1}{2}\left[ \frac{-6x}{3x^2-1}-\log(1-x)+\log(1+x) \right]\\ &=\frac{1}{4}(3x^2-1)\log(\frac{1+x}{1-x})-\frac{3}{2}x \end{align} (7.26)

 \displaystyle \begin{align} \therefore Q_2(x) = u_2(x)&= \frac{1}{4}(3x^2-1)\log(\frac{1+x}{1-x})-\frac{3}{2}x \end{align} (7.27)

# Problem R*6.8 - Solve Non-homogeneous L2-ODE-VCs

From Mtg 37-4

## Given

Non-homogeneous L2-ODE-VCs:

 $\displaystyle (x-1)y''-xy'+y=f(x)$ (8.1)
 $\displaystyle xy''+2y'+xy=f(x)$ (8.2)

## Find

a) Find the 1st homogeneous solution by trial solution.
b) Find the complete solution by variation of parameters with $\displaystyle f(x)=e^{\alpha x}$

## Solution

 - Solved on our own


### Solution of (8.1)

Assume a trial solution: $\displaystyle u_1(x)=e^{rx}$. So the characteristic equation is:

 $\displaystyle (x-1)r^2-xr+1=0$ (8.3)

From Problem R*6.3 we know that $\displaystyle r(x)= \frac{1}{x-1}$ is not a valid root. So $\displaystyle r=1$ and thus the 1st homogeneous solution is:

 $\displaystyle u_1=e^x$ (8.4)

Assume the full solution of the form:

 $\displaystyle y(x)=U(x)u_1(x)$ (8.5)

then:

 \displaystyle \begin{align} y'=U'(x)u_1(x)+U(x)u_1'(x) \end{align} (8.6)
 \displaystyle \begin{align} y''(x)&=U''(x)u_1(x)+U'(x)u_1'(x) + U'(x)u_1'(x)+ U(x)u_1''(x) \\ &=U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x) \end{align} (8.7)

Change (8.1) into the form:

 $\displaystyle a_2 y''+ a_1 y' + a_0 = f(x)$ (8.8)

where: $\displaystyle a_2=x-1$, $\displaystyle a_1= -x$ and $\displaystyle a_0= 1$. So:

 \displaystyle \begin{align} a_2(U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x)) + a_1(U'(x)u_1(x)+U(x)u_1'(x))+a_0U(x)u_1(x)=f(x) \end{align} (8.9)
 \displaystyle \begin{align} U(x)\cancelto{0} {(a_2u_1''(x)+a_1u_1'(x)+a_0u_1(x))} + U'(x)(2a_2u_1'(x)+a_1u_1(x))+U''(x)a_2u_1(x)=f(x) \end{align} (8.10)

We get:

 \displaystyle \begin{align} U''(x)+\frac{2a_2u_1'(x)+a_1u_1(x)}{a_2u_1(x)}U'(x)=\frac{f(x)}{a_2u_1(x)} \end{align} (8.11)

Now in order to solve $\displaystyle U$, assume $\displaystyle Z:=U'$. So:

 $\displaystyle Z'+ \frac{2a_2u_1'(x)+a_1u_1(x)}{a_2u_1(x)}Z=\frac{f(x)}{a_2u_1(x)}$ (8.12)

Use IFM to make the equation exact:

 \displaystyle \begin{align} h(x)&=\exp{\left[ \int \frac{a_1u_1+2a_2u_1'}{a_2u_1}dx \right] } \\ &=\exp{\left[ \int ( \frac{a_1}{a_2}+\frac{2u_1'}{u_1}) dx \right] } \\ &=u_1^2(x) \exp{\left[ \int \frac{a_1}{a_2} dx \right] } \end{align} (8.13)

and:

 \displaystyle \begin{align} Z(x)&=\frac{1}{h(x)} \left[ k+\int h(x)\frac{f(x)}{u_1(x)} \right] \end{align} (8.14)
 \displaystyle \begin{align} U(x) &= \int Z(x) dx + k\\ \end{align} (8.15)

Then we get:

 \displaystyle \begin{align} y(x)&=U(x)u_1(x) \\ &=k_1 u_1(x)+u_1(x) \int \frac{1}{h(x)} \left[ k_2 + \int h(x) \frac{f(x)}{a_2(x) u_1(x)}dx \right] dx \\ &=k_1 u_1(x)+k_2 u_1(x) \int \frac{1}{h(x)} dx + u_1(x) \int \frac{1}{h(x)} \left[ \int h(x) \frac{f(x)}{a_2(x)u_1(x)} dx \right] dx \\ &=k_1 e^x + k_2 e^x \int \left[ \frac{1}{e^{2x} exp(\int \frac{-x}{x-1}dx)} dx \right] + e^x \int \left[ \frac{1}{e^{2x} exp(\int \frac{-x}{x-1}dx)} \int \left[ e^{2x} exp(\int \frac{-x}{x-1}dx) \frac{e^{\alpha x}}{e^x(x-1)} \right] dx \right] dx \\ &=k_1 e^x + k_2 e^x \int\left[ \frac{x-1}{e^x} \right]dx + e^x \int \left[ \frac{x-1}{e^x} \int \frac{e^{\alpha x}}{(x-1)^2} dx \right]dx \\ &=k_1 e^x - k_2 x + e^x \int \left[ \frac{x-1}{e^x} \int \frac{e^{\alpha x}}{(x-1)^2} dx \right]dx \\ \end{align} (8.16)

So the complete solution is:

 \displaystyle \begin{align} y(x)&=U(x)u_1(x)\\ &=k_1 x + k_2 e^x + e^x \int \left( \frac{x-1}{e^x} \int \frac{e^{\alpha x}}{(x-1)^2} dx \right) dx \\ \end{align} (8.17)

And the 2nd homogeneous solution is:

 \displaystyle \begin{align} u_2(x)&= x \\ \end{align} (8.18)

### Solution of (8.2)

Assume a trial solution: $\displaystyle u_1(x)=x^c \sin x$. So the characteristic equation is:

 $\displaystyle (c^2+c)x^{c-1} \sin x+2(c+1)x^c \cos x=0$ (8.19)

So $\displaystyle c=-1$ and thus the 1st homogeneous solution is:

 $\displaystyle u_1= \frac{ \sin x}{x}$ (8.20)

Just like what we did above, assume the full solution of the form:

 $\displaystyle y(x)=U(x)u_1(x)$ (8.21)

So we get:

 \displaystyle \begin{align} U''(x)+\frac{2a_2u_1'(x)+a_1u_1(x)}{a_2u_1(x)}U'(x)=\frac{f(x)}{a_2u_1(x)} \end{align} (8.22)

where: $\displaystyle a_2=x$, $\displaystyle a_1= 2$ and $\displaystyle a_0= x$.

Again, assume $\displaystyle Z:=U'$ and use IFM, then we can get the full solution:

 \displaystyle \begin{align} y(x)&=U(x)u_1(x) \\ &=k_1 u_1(x)+u_1(x) \int \frac{1}{h(x)} \left[ k_2 + \int h(x) \frac{f(x)}{a_2(x) u_1(x)}dx \right] dx \\ &=k_1 u_1(x)+k_2 u_1(x) \int \frac{1}{h(x)} dx + u_1(x) \int \frac{1}{h(x)} \left[ \int h(x) \frac{f(x)}{a_2(x)u_1(x)} dx \right] dx \\ &=k_1 \frac{\sin x}{x} + k_2 \frac{\sin x}{x} \int \left[ \frac{1}{ \left( \frac{\sin x}{x} \right)^2 exp(\int \frac{2}{x}dx)} dx \right] + \frac{\sin x}{x} \int \left[ \frac{1}{ \left( \frac{\sin x}{x} \right)^2 exp(\int \frac{2}{x}dx)} \int \left[ \left( \frac{\sin x}{x} \right)^2 exp(\int \frac{2}{x}dx) \frac{e^{\alpha x}}{x \frac{\sin x}{x}} \right] dx \right] dx \\ &=k_1 \frac{\sin x}{x} + k_2 \frac{\sin x}{x} \int\left[ \frac{1}{(\sin x)^2} \right]dx + \frac{\sin x}{x} \int \left[ \frac{1}{(\sin x)^2} \int (\sin x)^2 \frac{e^{\alpha x}}{\sin x} dx \right]dx \\ &=k_1 \frac{\sin x}{x} - k_2 \frac{\cos x}{x} + \frac{e^{\alpha x}}{({\alpha}^2+1)x} \\ \end{align} (8.23)

So the complete solution is:

 \displaystyle \begin{align} y(x)&=U(x)u_1(x)\\ &=k_1 \frac{\sin x}{x} + k_2 \frac{\cos x}{x} + \frac{e^{\alpha x}}{({\alpha}^2+1)x} \\ \end{align} (8.24)

And the 2nd homogeneous solution is:

 \displaystyle \begin{align} u_2(x)&= \frac{\cos x}{x} \\ \end{align} (8.25)

YuChen

# Problem R6.9 - Non-homogeneous Legendre equation

From Mtg 37-5

## Given

 \displaystyle \begin{align} (1-x^2)y''-2xy'+2y=f(x) \end{align} (9.1)
 \displaystyle \begin{align} f(x)=1 \end{align} (9.2)

## Find

Find the final solution y(x) by variation of parameters.

## Solution

- Solved on our own


The given function is

 \displaystyle \begin{align} (1-x^2)y''-2xy'+2y=f(x) \end{align} (9.3)

The general solution form is

 \displaystyle \begin{align} y(x)=U(x)u_1(x) \end{align} (9.4)

U(x) is unknown. $u_1(x)=P_2(x)$.

 \displaystyle \begin{align} y(x)=U(x)u_1(x) \end{align} (9.5)
 \displaystyle \begin{align} y'=U'(x)u_1(x)+U(x)u_1'(x) \end{align} (9.6)
 \displaystyle \begin{align} y''(x)&=U''(x)u_1(x)+U'(x)u_1'(x) + U'(x)u_1'(x)+ U(x)u_1''(x) \\ &=U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x) \end{align} (9.7)

$u_1(x)$ is a homogeneous solution,

 \displaystyle \begin{align} a_2u_1''(x)+a_1u_1'(x)+a_0u_1(x)=0 \end{align} (9.8)

From (7.23), (7.24), (7.25)

 \displaystyle \begin{align} a_2&=1 \\ a_1&=\frac{-2x}{1-x^2} \end{align} (9.9)

From (9.5),(9.6),(9.7)

 \displaystyle \begin{align} a_2(U''(x)u_1(x)+2U'(x)u_1'(x)+ U(x)u_1''(x)) + a_1(U'(x)u_1(x)+U(x)u_1'(x))+a_0U(x)u_1(x)=f(x) \end{align} (9.10)
 \displaystyle \begin{align} U(x)\cancelto{0} {(a_2u_1''(x)+a_1u_1'(x)+a_0u_1(x))} + U'(x)(2a_2u_1'(x)+a_1u_1(x))+U''(x)a_2u_1(x)=f(x) \end{align} (9.11)
 \displaystyle \begin{align} U'(x)(2a_2u_1'(x)+a_1u_1(x))+U''(x)a_2u_1(x)=f(x) \end{align} (9.12)

To simplify, set $a_2=1$.

 \displaystyle \begin{align} U''(x)+\frac{2a_2u_1'(x)+a_1u_1(x)}{a_2u_1(x)}U'(x)=\frac{f(x)}{a_2u_1(x)} \end{align} (9.13)

Set Z=U' to reduce the order,

 \displaystyle \begin{align} Z'(x)+\frac{2a_2u_1'(x)+a_1u_1(x)}{a_2u_1(x)}Z(x)=\frac{f(x)}{a_2u_1(x)} \end{align} (9.14)

To solve (.15), use IFM to make the equation exact.

 \displaystyle \begin{align} h(x)&=\exp{\left[ \int \frac{a_1u_1+2a_2u_1'}{a_2u_1}dx \right] } \\ &=\exp{\left[ \int ( \frac{a_1}{a_2}+\frac{2u_1'}{u_1}) dx \right] } \\ &=u_1^2(x) \exp{\left[ \int \frac{a_1}{a_2} dx \right] } \end{align} (9.15)

 \displaystyle \begin{align} Z(x)&=\frac{1}{h(x)} \left[ k+\int h(x)\frac{f(x)}{u_1(x)} \right] \end{align} (9.16)
 \displaystyle \begin{align} U(x) &= \int Z(x) dx + k\\ \end{align} (9.17)

From (9.6), (9.9), (9.17),

 \displaystyle \begin{align} y(x)&=U(x)u_1(x) \\ &=k_1u_1(x)+u_1(x) \int \frac{1}{h(x)}\left[ k+ \int h(x)\frac{f(x)}{a_2(x)u_1(x)}dx \right] dx \\ &=k_1u_1(x)+ku_1(x) \int \frac{1}{h(x)} dx + u_1(x) \int \frac{1}{h(x)} \left[ \int h(x)\frac{f(x)}{a_2(x)u_1(x)} \right] dx \\ &=k_1x + k_2x \int \left[ \frac{1}{x^2exp(\int \frac{-2x}{1-x^2}dx)} dx\right] + x\int \left[ \frac{1}{x^2exp(\int \frac{-2x}{1-x^2}dx) } \int \left[ xexp( \int \frac{-2x}{1-x^2}dx) \right]dx \right] dx\\ &=k_1x + k_2x \int\left[ \frac{1}{x^2(1-x^2)} \right]dx + x \int \left[ \frac{1}{x^2(1-x^2)}\int x(1-x^2) dx \right]dx \\ &=k_1x + k_2x\left[ \frac{1}{2}\log(\frac{1+x}{1-x})-\frac{1}{x}\right] + x \int \left[ \frac{1}{x^2(1-x^2)}(\frac{1}{2}x^2-\frac{1}{4}x^4) \right] dx \\ &=k_1x + k_2x\left[ \frac{1}{2}\log(\frac{1+x}{1-x})-\frac{1}{x}\right] + x \int \left[ \frac{2-x^2}{4(1-x^2)}\right] dx \\ &=k_1x + k_2x\left[ \frac{1}{2}\log(\frac{1+x}{1-x})-\frac{1}{x}\right] + \frac{x}{8}\left[ 2x +\log(\frac{x+1}{1-x}) \right] \\ \end{align} (9.18)
 \displaystyle \begin{align} \therefore y(x)=k_1x + k_2x\left[ \frac{1}{2}\log(\frac{1+x}{1-x})-\frac{1}{x}\right] + \frac{x}{8}\left[ 2x +\log(\frac{x+1}{1-x}) \right] \\ \end{align} (9.19)

As a result, the second homogeneous solution is

 \displaystyle \begin{align} y(x)&=x\left[ \frac{1}{2}\log(\frac{1+x}{1-x})-\frac{1}{x} \right] \\ &=\frac{x}{2}\log(\frac{1+x}{1-x})-1 \end{align} (9.20)

# Contributing Members

 Team Contribution Table Problem Number Lecture (Mtg) Assigned To Solved By Typed By Signature (Author) R*6.1 Mtg 33-1 Allen Allen Allen Allen 13 Nov. 2011 at 21:09 (UTC) R*6.2 Mtg 35-1 Allen Allen Allen Allen 13 Nov. 2011 at 21:09 (UTC) R*6.3 Mtg 35-4 Shin Shin Shin Shin 13 Nov. 2011 at 21:09 (UTC) R*6.4 Mtg 35-4 Shin Shin Shin Shin 13 Nov. 2011 at 21:09 (UTC) R*6.5 Mtg 36-3 Ankush Ankush Ankush Ankush 16 Nov. 2011 at 10:09 (UTC) R*6.6 Mtg 36-10 Ren Ren Ren Ren 16 Nov. 2011 at 1:57 (UTC) R*6.7 Mtg 37-4 Chung Chung Chung Chung 15 Nov. 2011 at 20:09 (UTC) R*6.8 Mtg 37-4 YuChen YuChen YuChen YuChen16 Nov. 2011 at 10:34(UTC) R6.9 Mtg 37-5 Chung Chung Chung Chung15 Nov. 2011 at 20:09 (UTC)