User:Egm6321.f11.team1/report6

= R*6.1 Special IFM to solve a nonhomogeneous L2-ODE-CC with general f(t)= == Given == A nonhomogeneous L2-ODE-CC is
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle a_2y''+a_1y'+a_0y=f(t)$"
 * $$\displaystyle (Equation\;6.1.1)
 * $$\displaystyle (Equation\;6.1.1)
 * }
 * }

Find
$$\mathbf{(1)}$$ Find the PDEs that govern the integrating factor h(x,y) for Equation(6.1.1).

$$\mathbf{(2)}$$ Trial solution for the integrating factor $$\displaystyle h(t)=e^{\alpha t}$$ where $$\displaystyle \alpha$$ is unknown to be determined.
 * {| style="width:100%" border="0" align="left"

$$ Because of the integrating factor in exponential form, assume the LHS of Equation(6.1.2) take the form :
 * "$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= \int \, e^{\alpha t}f(t)dt$"
 * $$\displaystyle (Equation\;6.1.2)
 * $$\displaystyle (Equation\;6.1.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= e^{\alpha t}[\bar a_1y'+\bar a_0y]$"
 * $$\displaystyle (Equation\;6.1.3)
 * $$\displaystyle (Equation\;6.1.3)
 * }
 * }

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

$$\mathbf{(2.2)}$$ Find the quadratic equation for $$\displaystyle \alpha$$
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle a_2\alpha ^2-a_1\alpha+a_0=0$"
 * $$\displaystyle (Equation\;6.1.4)
 * $$\displaystyle (Equation\;6.1.4)
 * }
 * }

$$\mathbf{(2.3)}$$ Reduced-order equation: Equation(6.1.3)and Equation(6.1.4) lead to
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\int d^{\alpha t}f(t)dt$"
 * $$\displaystyle (Equation\;6.1.5)
 * $$\displaystyle (Equation\;6.1.5)
 * }
 * }

$$\mathbf{(2.4)}$$ Use the IFM to solve Equation(6.1.5)
 * {| style="width:100%" border="0" align="left"

$$ Find the solution y(t) for general excitation f(t).
 * "$\displaystyle \bar h(t)=exp[\int \, \frac{\bar a_0}{\bar a_1}dt]=e^{\beta t}$"
 * $$\displaystyle (Equation\;6.1.6)
 * $$\displaystyle (Equation\;6.1.6)
 * }
 * }

$$\mathbf{(2.5)}$$ Show that
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle \alpha\beta=\frac{a_0}{a_2}$"
 * $$\displaystyle (Equation\;6.1.7)
 * $$\displaystyle (Equation\;6.1.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ thus $$\displaystyle (\alpha,\beta)$$ are roots of the quadratic equation:
 * "$\displaystyle \alpha+\beta=\frac{a_1}{a_2}$"
 * $$\displaystyle (Equation\;6.1.8)
 * $$\displaystyle (Equation\;6.1.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}=0$|undefined"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.9)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.9)
 * }
 * }

$$\mathbf{(2.6)}$$ Deduce the particular solution $$y_p(t)\,$$ for general excitationf(t).

$$\mathbf{(2.7)}$$ Verify result with table of particular solution for:
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle f(t)=texp(bt)$"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.10)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.10)
 * }
 * }

$$\mathbf{(2.8)}$$ Solve the nonhomogeneous L2-ODE-CC Equation(6.1.1) with the Hyperbolic function:
 * {| style="width:100%" border="0" align="left"

$$ For the coefficients $$(a_0,a_1,a_2)$$, consider two different characteristic equations:
 * "$\displaystyle f(t)=tanht$"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.11)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.11)
 * }
 * }

$$\mathbf{(2.8.1)}$$
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle (r+1)(r-2)=0$"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.12)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.12)
 * }
 * }

$$\mathbf{(2.8.2)}$$
 * {| style="width:100%" border="0" align="left"

$$
 * "$\displaystyle (r-4)^2=0$"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.13)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.13)
 * }
 * }

$$\mathbf{(2.9)}$$ For each case in Equation(6.1.12)and Equation(6.1.13), determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Solution
$$\mathbf{(1)}$$ Multiply Equation(6.1.1) by the integrating factor h(t,y): "$\underbrace{h(t,y)a_2}_{\displaystyle f}y''+\underbrace{h(t,y)a_1y'+h(t,y)a_0y-h(t,y)f(t)}_{\displaystyle g}=0 $" Recall the 2nd exactness condition for N2-ODEs: "$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$" "$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $" Calculate the following terms in the exactness condition: "$\displaystyle f_t=a_2h_t\,;\,f_{tt}=a_2h_{tt}\,;\,f_{ty}=a_2h_{ty}\,;\,f_{tp}=0$" "$\displaystyle f_y=a_2h_y\,;\,f_{yy}=a_2h_{yy}\,;\,f_{yp}=0$" "$\displaystyle g_t=a_1ph_t+a_0yh_t-f(t)\cdot h_t-h\cdot f'(t)$" "$\displaystyle g_{tp}=a_1h_t$" "$\displaystyle g_y=a_1ph_y+a_0yh_y+a_0h-f(t)h_y$" "$\displaystyle g_{yp}=a_1h_y\,;\,g_p=a_1h\,;\,g_{pp}=0$" Substituting those calculated terms into 2nd exactness condition: "$\displaystyle a_2h_{tt}+2p\cdot a_2 h_{ty}+p^2\cdot a_2h_{yy}=a_1h_t+\cancel{p\cdot a_1h_y}-\cancel{a_1ph_y}-a_0yh_y-a_0h+f(t)h_y$" "$\displaystyle 0+p\cdot 0+2\cdot a_2h_y=0$" From the second equation, we can obtain that "$\displaystyle h_y=0$" which means that integrating factor $$\displaystyle h(t,y)=h(t)$$. Substituting h(t) into the first equation of 2nd exactness condition, we can obtain an L2-ODE-CC, "$\displaystyle a_2h_{tt}-a_1h_t+a_0h=0$" Therefore, that PDE reduces into an ODE, which can be solved.

$$\mathbf{(2.1)}$$ Differentiate the RHS of Equation(6.1.3), "$\displaystyle \frac{d}{dt}\left[e^{\alpha t}(\bar a_1y'+\bar a_0y)\right]=\alpha e^{\alpha t}(\bar a_1y'+\bar a_0y)+e^{\alpha t}(\bar a_1 y''+\bar a_0 y')$" "$\displaystyle =e^{\alpha t}[\bar a_1y''+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]$" Compare the coefficient with the integrand on LHS of Equation(6.1.3), "$\displaystyle e^{\alpha t}[\bar a_1y+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]=e^{\alpha t}[a_2y+a_1y'+a_0y]$" "$\displaystyle \bar a_1=a_2$" "$\displaystyle \bar a_0+\alpha \bar a_1=a_1$" "$\displaystyle \alpha \bar a_0=a_0$"

We can find $$\displaystyle (\bar a_1,\ \bar a_0)$$ in terms of $$\displaystyle (a_0,\ a_1,\ a_2)$$ that "$\displaystyle \bar a_1=a_2$" "$\displaystyle \bar a_0=\frac{a_0}{\alpha}$"

$$\mathbf{(2.2)}$$ Substituting $$\displaystyle \bar a_0$$ and $$\displaystyle \bar a_1$$ into $$\displaystyle \bar a_0+\alpha \bar a_1=a_1$$, we can get that <\br> "$\displaystyle \frac{a_0}{\alpha}+\alpha\cdot a_2=a_1$" Rearrange it, we obtain the quadratic equation for $$\displaystyle \alpha$$: "$\displaystyle a_2 \alpha^2-a_1\alpha+a_0=0$"

$$\mathbf{(2.3)}$$ Therefore, the reduced-order equation is obtained "$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$" Rewrite it,
 * {| style="width:100%" border="0" align="left"

$$ Equation(6.1.14) is a L1-ODE-CC, which can be solved by the IFM.
 * "$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\,\int e^{\alpha t}f(t)dt$"
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.14)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.14)
 * }
 * }

$$\mathbf{(2.4)}$$ Recall the IFM of general non-homogeneous L1-ODE-VC, "$\displaystyle 1\cdot y'+\underbrace{\frac{Q(x)}{P(x)}}_{a_0(x)}y=\underbrace{\frac{R(x)}{P(x)}}_{b(x)}$|undefined" The integrating factor is "$\displaystyle h(x)=exp[\int^x a_0(s)ds]$" The solution is "$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k\right]$" Divided Equation(6.1.14) by $$\displaystyle \bar a_1$$, "$\displaystyle 1\cdot y'+\underbrace{\frac{\bar a_0}{\bar a_1}}_{\displaystyle \beta}y=\underbrace{\frac{e^{-\alpha t}}{\bar a_1}\, \int e^{\alpha t} f(t)dt}_{\displaystyle b(t)}$|undefined" Therefore, we can obtain the integrating factor "$\displaystyle \bar h(t)=exp[\int\,\frac{\bar a_0}{\bar a_1} dt]$" "$\displaystyle =exp[\int\,\beta dt]$" "$\displaystyle =e^{\beta t}$" Therefore, the solution of Equation(6.1.14) is "$\displaystyle y(t)=\frac{1}{\bar h(t)}\left[ \int^t h(s)b(s)ds+k\right]$" "$\displaystyle =\frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$|undefined"

$$\mathbf{(2.5)}$$ Substitute $$\displaystyle \bar a_1=a_2\,;\,\bar a_0+\alpha \bar a_1=a_1\,;\,\alpha \bar a_0=a_0$$ into $$\displaystyle \alpha$$ and$$\displaystyle \beta$$, we can obtain that "$\displaystyle \alpha\beta=\alpha \frac{\bar a_0}{\bar a_1}=\frac{a_0}{\bar a_1}=\frac{a_0}{a_2}$" "$\displaystyle \alpha +\beta= \frac{a_1-\bar a_0}{\bar a_1}+\frac{\bar a_0}{\bar a_1}=\frac{a_1}{\bar a_1}=\frac{a_1}{a_2}$" Thus, $$\displaystyle (\alpha,\beta)$$ are the roots of the following quadratic equtation: "$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}$|undefined" Obviously, it is the same as the quadratic equation obtained by $$\mathbf{(2.2)}$$.

$$\mathbf{(2.6)}$$ Rearrange the solution of Equation(6.1.14) "$\displaystyle \frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$|undefined" "$\displaystyle =e^{-\beta t}[\int^t\frac{e^{(\beta-\alpha)s}}{\bar a_1}\int^s e^{\alpha \tau}f(\tau)d\tau]ds$|undefined" "$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int^t e^{(\beta-\alpha)s}[\int e^{\alpha s}f(s)ds+k_1]ds]$|undefined" "$\displaystyle =\frac{e^{-beta t}}{\bar a_1}[\int^te^{\beta-\alpha)s}[\int e^{\alpha s}f(s)ds]ds+\int^t e^{(\beta-\alpha)s}k_1ds]$|undefined" "$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt+k_2+\frac{k_1}{\beta-\alpha}e^{(\beta-\alpha)t}]$|undefined" "$\displaystyle =\underbrace{\frac{k_1}{\bar a_1(\beta-\alpha)}e^{-\alpha t}}_{\displaystyle C_1 y_H^1}+\underbrace{\frac{k_2}{\bar a_1}e^{-\beta t}}_{\displaystyle C_2 y_H^2}+\underbrace{\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt}_{\displaystyle y_P}$|undefined" The particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$ is "$\displaystyle y_P(t)=\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt$|undefined" $$\mathbf{(2.7)}$$

The particular solution for an arbitrary $$\displaystyle f(x)$$ is given directly above in section 2.6. The solutions for $$\displaystyle f(x)=t^2$$, as evaluated by Wolfram-Alpha is:

$$y_P(t)=\frac{2}{\alpha^3 a_1 \beta}+\frac{2}{\alpha^2 a_1 \beta^2}+\frac{2}{\alpha a_1 \beta^3}+\frac{t^2}{\alpha a_1 \beta}-\frac{2t}{\alpha^2 a_1 \beta}-\frac{2t}{\alpha a_1 \beta^2}$$

The solution as found in Table 3.6.1 on page 1 of the following link (note use $$\displaystyle s=0$$ accordingly):

$$\displaystyle y_P(t)=A_0t^2+A_1t+A_0$$

Choosing the coefficients $$\displaystyle A$$ wisely yields a perfect match. Now check solutions for $$\displaystyle f(x)=te^{bt}$$, again use Wolfram-Alpha:

$$y_P(t)=\frac{e^{bt}(-\alpha+2\beta b t-b-2\beta+2\alpha\beta t)}{4a_1\beta^2(\alpha+\beta)^2}$$

The solution for this case also is found in the same link above:

$$\displaystyle y_P(t)=(A_1t+A_0)e^{bt}$$

Similarly, a careful choice of coefficients proves equivalency.

$$\mathbf{(2.8.1)}$$ $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ is the roots of quadratic equation: "$\displaystyle (r+1)(r-2)=0.$" Therefore let $$\displaystyle \alpha=-1\,;\,\beta=2$$ As for the excitation f(t) is Hyperbolic function: "$\displaystyle f(t)=tanh(t).$" For the general solution Equation(6.1.15) "$\displaystyle y(t)=\frac{1}{6}\left[1+2e^{-2t}log(e^{2t}+1)-4e^ttan^{-1}(e^{-t})\right]$" This is calculated by www.wolframalpha.com.

$$\mathbf{(2.8.2)}$$ The quadratic equation is: "$\displaystyle (r-4)^2=0.$" Therefore let $$\displaystyle \alpha=\beta=4$$ As for the excitation f(t) is Hyperbolic function: "$\displaystyle f(t)=tanh(t).$" For the general solution Equation(6.1.15) "$\displaystyle y(t)=-\frac{Li_2(-e^{2t}\cdot e^{-4t})}{2}-\frac{e^{-2t}}{2}+\frac{1}{16}$|undefined" where $$\displaystyle Li_2(t)\,$$ is the polylogarithm function. This is calculated by www.wolframalpha.com.

$$\mathbf{(2.9)}$$ Work is in progress.

=R*6.2 Showing the form of particular solution in notes agree with the one in book by King and others =

Given
And

Find
with

Solution
- Solved on our own $$\frac{1}{h(x)}=\frac{u_1(x)u_2'(x)-u_2(x)u_1'(x)}{{u_1}^2(x)}=\frac{W(x)}{{u_1}^2(x)}$$ <p style="text-align:right"> (6.2.5)

Substituting the value of 'h(x)' from Equation 6.2.5 into Equation 6.2.1

$$y_P(x)=u_1(x)\int{\frac{W(x)}{{u_1}^2(x)}}\left[\int \frac{{f(x)}u_1(x)}{W(x)}\,dx \right]\,dx =u_1(x)\left[\left(\int\frac{W(x)}dx \right)\frac{u_2(x)}{u_1(x)}-\int\frac{W(x)}dx \right]$$

Applying dummy integrals:

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

Taking $$\displaystyle u_1(x) \,\, u_2(x)$$ inside integral.

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

This shows that the form of particular solution in notes agree with the one in book by King and others

=R*6.3 Invaid Solution=

Given
From [[media:Pea1.f11.mtg35.djvu|Mtg 35-4]]

The first homogeneous solution: Trial solution: Characteristic equation:

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 As $$\displaystyle u_2(x) = e^{xr_2(x)}$$

Validity of $$\displaystyle y= u_2(x) $$ can be shown by substituting it into the Eq. (6.3.1).

Substitute $$\displaystyle y, y^{'}, y^{''}$$ in the Eq. (6.3.6) into the Eq. (6.3.1), As shown in the Eq. (6.3.7), $$\displaystyle u_2(x) $$ is not a valid homogeneous solution.

= Problem R*6.4 For a L2-ODE-VC using variation of parameters find the 2nd homogeneous solution and compare =

Reference Lecture 35-4 found here:[[media:pea1.f11.mtg35.djvu|Mtg 35]]

Given
The following liner, second order, ordinary differential equation, with varying coefficients (L2-ODE-VC)

"$\displaystyle (x - 1) y'' - x y' + y = x$" <p style="text-align:right;">$$\displaystyle (6.4.1) $$

Find
A valid homogeneous solution and call it $$\displaystyle u_1 $$

Find the particular solution using variation of parameters method and compare it to $$\displaystyle e^{xr_2(x)} $$

Solution
Lets start by validating that the homogeneous solution selected below

$$\displaystyle u_1 = e^x $$ <p style="text-align:right;">$$\displaystyle (6.4.2) $$

Use the trial solution:

"$ \displaystyle u_1(x) = e^{rx}$" <p style="text-align:right;">$$\displaystyle (6.4.3) $$

Differentiate the above to get "$ \displaystyle u'_1(x) = r e^{rx}$" <p style="text-align:right;">$$\displaystyle (6.4.4)$$

"$ \displaystyle u''_1(x) = r^2 e^{rx} $" <p style="text-align:right;">$$\displaystyle (6.4.5)$$

Plugging these derivatives and the trial solution into 6.4.1 yields "$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0$" <p style="text-align:right;">$$\displaystyle (6.4.6)$$

Rearranging by grouping the terms with x from above yields

"$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0$" <p style="text-align:right;">$$\displaystyle (6.4.7)$$

In order for the above equation to be equal to zero, the following equations must be satisfied

"$ \displaystyle r^2 - r = r (r - 1) = 0 $"

"$ \displaystyle 1 - r^2 = -(r + 1)(r - 1) = 0 $" <p style="text-align:right;">$$\displaystyle (6.4.8)$$

The solution of the above equations is $$\displaystyle r=1$$. So 6.4.3 can be written as

"$ \displaystyle u_1(x) = e^x $" <p style="text-align:right;">$$\displaystyle (6.4.9)$$

Which is a valid homogeneous solution

So now lets find the 2nd homogeneous solution using variation of parameters

We can rewrite 6.4.1 as

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y  = \underbrace{\frac{x}{x-1}}_{f(x)} $$ <p style="text-align:right;">$$\displaystyle (6.4.10) $$

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(x)}{x-1}} \ dx] \ dx $$ <p style="text-align:right;">$$\displaystyle (6.4.11) $$

Using Wolfram Alpha to evaluate both integrals and simplify it can be shown that

$$\displaystyle {u}_{2}(x)= x $$ <p style="text-align:right;">$$\displaystyle (6.4.12)$$

Which is the 2nd homogeneous solution

This can can then be compared to $$\displaystyle e^{xr_2(x)} $$ as requested

if $$\displaystyle r_2(x) = ln(x)/x $$ the expressions for $$\displaystyle{u}_{2}$$ are equivalent

= R*6.5: Find Two Homogeneous Solutions using a Trial Solution =

Reference lecture notes page 36-3 found here [[media:pea1.f11.mtg36.djvu|Mtg 36]]

Given
Find u1 and u2 of Equation 6.5.1 below using the trial solution given in equation 6.5.2:

"$\displaystyle (x +1) y'' - (2x+3)y' + 2y = 0 $" <p style="text-align:right">(6.5.1)

"$\displaystyle y(x)=e^{rx}$" <p style="text-align:right">(6.5.2)

Solution
Lets start by differentiating the trial solution as follows:

"$ \displaystyle y'(x) = r e^{rx}$" <p style="text-align:right">(6.5.3)

"$ \displaystyle y''(x) = r^2 e^{rx}$" <p style="text-align:right">(6.5.4)

The next step is to plug the derivatives and the original trail function into equation 6.5.1 as follows:

"$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $" <p style="text-align:right">(6.5.5)

Divide $$ \displaystyle e^{rx} $$ by both side and group all the $$ \displaystyle x $$ terms to obtain:

"$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0$" <p style="text-align:right">(6.5.6)

The following conditions must be satisfy for equation 6.5.6 to be true:

"$ \displaystyle r(r-2)=(r-2)(r-1)=0 $" <p style="text-align:right"> (6.5.7)

In order for 6.5.6 to be true it can be seen that "$ \displaystyle r=2 $" <p style="text-align:right">(6.5.8)

The first homogeneous solution is obtained by plugging equation 6.5.8 into the trial solution given in equation 6.5.2:

"$ \displaystyle u_1(x)=e^{2x} $" <p style="text-align:right">(6.5.9)

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

"$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$|undefined" <p style="text-align:right">(6.5.10)

Equation 6.5.1 can be rearranged into the desired form shown:

"$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0$|undefined" <p style="text-align:right"> (6.5.11)

The value for $$ \displaystyle a_1 $$ from equation 6.5.11 is then substituted into equation 6.5.10, additionally the first homogeneous solution from equation 6.5.9 is substituted to solve for the second homogeneous solution as shown:

"$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$|undefined" <p style="text-align:right"> (6.5.12)

Evaluating the inner most integral yields:

"$ \displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$|undefined" <p style="text-align:right">(6.5.13)

Finally evaluating the last integral and simplifying yields:

"$ \displaystyle u_2(x)=\frac{-1}{4}(2x+3)$" <p style="text-align:right">(6.5.14)

This is the 2nd homogeneous solution.

= R*6.6 Reverse Engineering ODE Equations =

Given
Trial solutions:


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

$$
 * $$\frac{e^{rx}}{x^2}, r=\text{constant}$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.1)
 * }
 * }


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

$$
 * $$\frac{e^{rx}}{\sin x}, r=\text{constant}$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.2)
 * }
 * }

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

$$
 * $$\displaystyle r^2+3=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.3)
 * }
 * }


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

$$
 * $$\displaystyle r_1=2$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.4)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$r_2(x)=\frac{1}{x+1}$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.5)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$.
 * $$\displaystyle (r-r_1)[r-r_2(x)]=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.6)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.6.6)
 * }
 * }

== Find ==

Find a homogeneous L2-ODE-VC that accepts the trial solutions Equation 6.6.1 and Equation 6.6.2, and their respective characteristic equations, Equation 6.6.3, and Equation 6.6.4 thru Equation 6.6.6.

Solution
I will mirror the approach laid out in Meeting 36.

Multiply the trial solution of Equation 6.6.1 (and its derivatives) by unknown coefficients $$\displaystyle a(x)$$.

$$\left[\frac{e^{rx}}{x^2}\right]\times a_0(x)$$

$$\left[\frac{d}{dx}\left(\frac{e^{rx}}{x^2}\right)=\left(r-\frac{2}{x}\right)\frac{e^{rx}}{x^2}\right]\times a_1(x)$$

$$\left[\frac{d^2}{dx^2}\left(\frac{e^{rx}}{x^2}\right)=\left(r^2-\frac{4r}{x}+\frac{6}{x^2}\right)\frac{e^{rx}}{x^2}\right]\times a_2(x)$$

Now equate the sum of the above three results to the product of the trial solution (Equation 6.6.1) and the characteristic equation (Equation 6.6.3):

$$\frac{e^{rx}}{x^2}\left[a_2\left(r^2-\frac{4r}{x}+\frac{6}{x^2}\right)+a_1\left(r-\frac{2}{x}\right)+a_0\right]=\frac{e^{rx}}{x^2}(r^2+3)=0$$

$$a_2r^2+\left(\frac{4a_2}{x}+a_1\right)r+\left(\frac{6a_2}{x^2}-\frac{2a_1}{x}+a_0\right)=r^2+3$$

Comparing the polynomial coefficients of $$\displaystyle r$$ yields the following:

$$a_2=1, \quad a_1=-\frac{4}{x}, \quad a_0=3-\frac{48}{x^2}$$

If we define $$f:=\frac{e^{rx}}{x^2}$$, the following L2-ODE-VC will be satisfied:

Now the same technique will be applied to Equation 6.6.2 and its characteristic equation. Even though one root is a function of x, the technique is still valid as long as one root is constant. See page (36-2) in Meeting 36.

Multiply the trial solution of Equation 6.6.2 (and its derivatives) by unknown coefficients $$\displaystyle a(x)$$.

$$\left[\frac{e^{rx}}{\sin x}\right]\times a_0(x)$$

$$\left[\frac{d}{dx}\left(\frac{e^{rx}}{\sin x}\right)=\left(r-\cot x\right)\frac{e^{rx}}{\sin x}\right]\times a_1(x)$$

$$\left[\frac{d^2}{dx^2}\left(\frac{e^{rx}}{\sin x}\right)=\left(r^2-2r\cot x+\cot^2x+\csc^2x\right)\frac{e^{rx}}{\sin x}\right]\times a_2(x)$$

Now equate the sum of the above three results to the product of the trial solution (Equation 6.6.2) and the characteristic equation (Equation 6.6.6):

$$\frac{e^{rx}}{\sin x}\left[a_2\left(r^2-2r\cot x+\cot^2x+\csc^2x\right)+a_1\left(r-\cot x\right)+a_0\right]=\frac{e^{rx}}{\sin x}(r-r_1)[r-r_2(x)]=0$$

$$a_2r^2+\left(a_1-2a_2\cot x\right)r+\left(a_0-a_1\cot x +\cot^2x+\csc^2x\right)=r^2-2r-\frac{r}{x+1}+\frac{2}{x+1}$$

Comparing the polynomial coefficients of $$\displaystyle r$$ yields the following:

$$a_2=1, \quad a_1=2\cot x-\frac{1}{x+1}-2, \quad a_0=\cot x\left(2\cot x-\frac{1}{x+1}-2\right)-\cot^2x-\csc^2x+\frac{2}{x+1}$$

If we define $$f:=\frac{e^{rx}}{\sin x}$$, the following L2-ODE-VC will be satisfied:

= 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 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.1)
 * }
 * }

For n=2
 * {| style="width:100%" border="0" align="left"

\displaystyle (1-x^2)y''-2xy'+6y=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.2)
 * }
 * }

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

\displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.3)
 * }
 * }

== Find == Show the second homogeneous solution:
 * {| style="width:100%" border="0" align="left"

\displaystyle Q_2(x)=\frac{1}{4}(3x^2-1)\log \left ( \frac{1+x}{1-x} \right )-\frac{3}{2}x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.4)
 * }
 * }

Solution
For a Non-homogeneous L2-ODE-VC:
 * {| style="width:100%" border="0" align="left"

\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.5)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x). $$ \displaystyle y(x)=U(x)u_1(x) $$

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


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

\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) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.6)
 * }
 * }

Let $$ \displaystyle Z:=U' $$

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

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


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

\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 * <p style="text-align:right;">$$\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 $$ \displaystyle \begin{align} u_1(x)&=P_2(x)=\frac{1}{2}(3x^2-1) \\ \displaystyle a_1(x)&=\frac{-2x}{1-x^2} \end{align} $$


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

\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} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.9)
 * }
 * }


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

\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} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.7.10)
 * }
 * }

Rearranging:

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

= R*6.8 Find Solutions to L2-ODE-VC with Trial Solution and Variation of Parameters with Excitation=

Given
Two Non-Homogeneous L2-ODE-VC


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

$$
 * $$\displaystyle (x-1)y''-xy'+y=f(x)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.1)
 * }
 * }


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

$$
 * $$\displaystyle xy''+2y'+xy=f(x)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.2)
 * }
 * }

== Find ==

Find the 1st homogeneous solution by trial solution.

Find the complete solution by variation of parameters with the following excitation:


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


 * $$\displaystyle f(x)=\sin x$$


 * $$\displaystyle f(x)=\exp(\alpha x)$$

Part 1
Find the 1st homogeneous solution by trial solution. Since we are looking for homogenous solution, $$\displaystyle f(x)=0$$. For Equation 6.8.1 we will assume the form $$\displaystyle y(x)=e^{rx}$$, where $$\displaystyle r$$ is an undetermined coefficient. Thus Equation 6.8.1 becomes:

$$\displaystyle (x-1)r^2e^{rx}-xre^{rx}+e^{rx}=0$$

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

$$\displaystyle [(x-1)r-1](r-1)=0$$

Thus the two roots are:

$$\displaystyle r_1=1$$

$$\displaystyle r_2=\frac{1}{x-1}$$

The second root is not valid, since it undermines the assumption of form in the trial solution due to it being a function of x. See R*6.3 for additional detail. Therefore the 1st homogeneous solution for Equation 6.8.1 is:

For Equation 6.8.2 we will assume the form $$\displaystyle y(x)=\frac{\sin rx}{x}$$, where $$\displaystyle r$$ is an undetermined coefficient. Thus Equation 6.8.2 becomes:

$$x\left(-\frac{(r^2x^2-2)\sin rx +2rx\cos rx}{x^3}\right)+2\frac{rx\cos rx -\sin rx}{x^2}+x\left(\frac{\sin rx}{x}\right)=0$$

$$-\frac{(r^2x^2-2)\sin rx +2rx\cos rx}{x^2}+2\frac{rx\cos rx -\sin rx}{x^2}+\sin rx=0$$

$$\frac{-r^2x^2\sin rx}{x^2}+\sin rx=0$$

$$\displaystyle -r^2\sin rx+\sin rx=0$$

$$\displaystyle 1-r^2=0$$

Thus the two roots are:

$$\displaystyle r_1=1$$

$$\displaystyle r_2=-1$$

Therefore, taking note that sine is an odd function, the 1st homogeneous solution for Equation 6.8.2 is either:

Section 1
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=x$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{x}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{x}{x-1}$$

We also deduced from Part 1 that

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

Now via (4) on (34-5) of Meeting 34:


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

$$
 * $$u_2(x):=u_1(x)\int\frac{1}{u_1^2(x)}\exp\left[-\int a_1(x)dx\right]dx$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.3)
 * }
 * }

$$u_2(x):=e^x\int\frac{1}{e^{2x}}\exp\left[-\int -\frac{x}{x-1}\right]dx$$

$$\displaystyle u_2(x)=-x$$

Now via (1) on (34-6) of Meeting 34:


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

$$
 * $$y_P(x)=u_1(x)\int\frac{1}{h(x)}\left[\int h(x)\frac{f(x)}{u_1(x)}dx\right]dx$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.4)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.4)
 * }
 * }

Where

$$h(x)=u_1^2(x)\exp\left[\int a_1(x)dx\right]$$

$$\displaystyle h(x)=\frac{e^x}{x-1}$$

So

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{x}{x-1}}{e^x}dx\right]dx$$

To find the complete solution we use (3) on (34-5) from Meeting 34, where constants are determined by initial conditions:


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

$$
 * $$\displaystyle y(x)=k_1u_1(x)+k_2u_2(x)+y_P(x)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.5)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.8.5)
 * }
 * }

Section 2
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=\sin x$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{\sin x}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{\sin x}{x-1}$$

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

The solution is the same as that of Section 1, except now

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{\sin x}{x-1}}{e^x}dx\right]dx$$

Section 3
Find the complete solution by variation of parameters to

$$\displaystyle (x-1)y''-xy'+y=\exp(\alpha x)$$

$$\displaystyle y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=\frac{\exp(\alpha x)}{x-1}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{x}{x-1}$$

$$f(x)=\frac{\exp(\alpha x)}{x-1}$$

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

The solution is the same as that of Section 1, except now

$$y_P(x)=e^x\int\frac{1}{\frac{e^x}{x-1}}\left[\int \frac{e^x}{x-1}\frac{\frac{\exp(\alpha x)}{x-1}}{e^x}dx\right]dx$$

Section 4
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=x$$

$$\displaystyle y''-\frac{2}{x}y'+y=1$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

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

We also deduced from Part 1 that

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

Now via Equation 6.8.3

$$u_2(x):=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2 x}{x^2}}\exp\left[-\int -\frac{2}{x}\right]dx$$

Now use Equation 6.8.4, where

$$h(x)=u_1^2(x)\exp\left[\int a_1(x)dx\right]$$

$$\displaystyle h(x)=\frac{\sin^2x}{x^4}$$

So

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{1}{\frac{\sin x}{x}}dx\right]dx$$

To find the complete solution use Equation 6.8.5

Section 5
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=\sin x$$

$$\displaystyle y''-\frac{2}{x}y'+y=\frac{\sin x}{x}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

$$\displaystyle f(x)=\frac{\sin x}{x}$$

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

The solution is the same as that of Section 4, except now

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{\frac{\sin x}{x}}{\frac{\sin x}{x}}dx\right]dx$$

Section 6
Find the complete solution by variation of parameters to

$$\displaystyle xy''-2y'+xy=\exp(\alpha x)$$

$$\displaystyle y''-\frac{2}{x}y'+y=\frac{\exp(\alpha x)}{x}$$

Thus following the syntax of Meeting 34,

$$a_1(x)=-\frac{2}{x}$$

$$\displaystyle f(x)=\frac{\exp(\alpha x)}{x}$$

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

The solution is the same as that of Section 4, except now

$$y_P(x)=\frac{\sin x}{x}\int\frac{1}{\frac{\sin^2x}{x^4}}\left[\int \frac{\sin^2x}{x^4}\frac{\frac{\exp(\alpha x)}{x}}{\frac{\sin x}{x}}dx\right]dx$$

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

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

\displaystyle (1-x^2)y''-2xy'+2y=f(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.1)
 * }
 * }


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

\displaystyle f(x)=1 $$
 * <p style="text-align:right;">
 * }
 * }

First homogeneous solution:


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

\displaystyle u_1(x)=P_1(x)=x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.2)
 * }
 * }

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

Solution
For a Non-homogeneous L2-ODE-VC:
 * {| style="width:100%" border="0" align="left"

\displaystyle y''+a_1(x)y'+a_0(x)y=f(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.3)
 * }
 * }

Given 1st homogeneous solution u_1(x) Find 2nd homogeneous solution u_2(x). $$ \displaystyle y(x)=U(x)u_1(x) $$

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


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

\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) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.4)
 * }
 * }

Let $$ \displaystyle Z:=U' $$

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

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


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

\displaystyle Z(x)=\frac{1}{h(x)}[k_2+\int h(x)\frac{f(x)}{u_1(x)}dx] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle U(x)=k_1+\int Z(x)dx $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 6.9.6)
 * }
 * }

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


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

\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)
 * }
 * }


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

\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)
 * }
 * }


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

\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 =

= Team Work Distribution =

{| cellpadding="10" cellspacing="5" style="width: 95%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 20%; background-color: wit; border: 1px solid wit; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |


 * style="width: 80%; background-color: wit; border: 1px solid wit; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 10px;" |