User:Egm6321.f09.Team1.sallstrom/HW4

= Problem 1 = From lecture notes slide 19-1

Given
The Legendre polynomial equation (Eq. 1, slide 14-2)
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ with $$n=0$$, i.e.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' = 0 $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

u_1(x) = 1 $$ $$ is a homogeneous solution.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Use reduction of order method 2 (undetermined factor) to find the second homogeneous solution, u2(x)

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

y(x) = U(x) u_1(x) $$ $$ Given Eq. 3, this simply becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y(x) = U(x) $$ $$ And Eq. 2 becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) U'' - 2 x U' = 0 $$ $$ Now, let
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z = U' $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) Z' - 2 x Z = 0 $$ $$ Rearrange so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{Z'}{Z}= \frac{2 x}{1 - x^2} $$ $$ Integrating both sides yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }


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

\ln Z = - \ln (1 - x^2) + \tilde c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }


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

Z = \frac{1}{1 - x^2} e^{\tilde c_2} = = \frac{1}{1 - x^2} c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Now, we assume that


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

\frac{1}{1 - x^2} = \frac{A}{x+1} + \frac{B}{x-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 12)
 * }
 * }


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

1 = -A(x-1) - B(x+1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 13)
 * }
 * }


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

\begin{cases} A - B = 1 \\ A + B = 0 \end{cases} \Rightarrow \begin{cases} A = \frac{1}{2} \\ B = - \frac{1}{2} \end{cases} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }


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

\frac{1}{1 - x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 15)
 * }
 * }

Using Eqs. 7, 11, and 15, we get
 * {| style="width:100%" border="0" align="left"

U' = \left[\frac{1}{2(x+1)} - \frac{1}{2(x-1)} \right] c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 16)
 * }
 * }

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

U(x) = c_1 + \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] c_2 = c_1 u_1(x) + c_2 u_2(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }

where


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

$$\displaystyle u_2 = \frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) = \frac{1}{2}\ln \left(\frac{x+1}{x-1} \right) = \ln \sqrt{\frac{x+1}{x-1}} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }

= Problem 2 = From lecture notes slide 19-1, 23-1 / King et. al, p. 28, problem 1.1b.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

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

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

u_1 = x^{-1} \sin x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. Slightly helped by the fact that we know the answer, we start with the trial solution


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

\tilde u_1(x) = a x^b \sin(rx) $$ $$ (Note that it would be better to choose $$e^{rx}$$ rather than $$\sin(rx)$$, as it would give us both solutions. However, we don't want to spoil the second part of the problem, so we only solve half of it by using a non-optimal trial solution)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

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

\tilde u'_1(x) = a b x^{b-1} \sin(rx) + a r x^b \cos(rx) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\tilde u''_1(x) = a b (b-1) x^{b-2} \sin(rx) + 2 a b r x^{b-1} \cos(rx) - a r^2 x^b \sin(rx) $$ $$ Plugging $$\tilde u_19x)$$ into Eq. 1 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a x^b \left[ b (b-1) x^{-1} \sin(rx) + 2 b r \cos(rx) - r^2 x \sin(rx) + 2 b x^{-1} \sin(rx) + 2 r \cos(rx) + x \sin(rx) \right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Regrouping the terms yields


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

a x^b \left[ \left(b^2 + b\right) x^{-1} \sin(rx) + 2 r \cos(rx) \left( b + 1 \right) + x \sin(rx) (1 - r^2) \right] = 0 $$ $$ For the above to be satisfied, we need to satisfy the following three equations
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{cases} b^2 + b = 0 \\ b + 1 = 0 \\ 1 - r^2 = 0 \end{cases} $$ $$ This is satisfied only for $$b\displaystyle = -1$$ and $$r = \pm 1$$. Note that changing the sign of $$r$$ has the same effect as changing the sign of the constant $$a$$. We can therefore choose only $$\displaystyle r=1$$ without loosing a solution. Using these constants, we get
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\tilde u_1(x) = a \underbrace{x^{-1} \sin(x)}_{u_1(x)} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) = x^{-1} \sin(x) $$ $$ is a homogeneous solution to Eq. 1.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{2}{x} y' + y = y + \underbrace{\frac{2}{x}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$a_1(x) = 2/x$$ into Eq. 14 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{\tilde c_1}{[u_1(x)]^2} \exp\left( -2 \ln x + \tilde c_2 \right) = \frac{\tilde c_1}{[x u_1(x)]^2} \exp\left( \tilde c_2 \right) = \frac{c}{[x u_1(x)]^2} = \frac{c}{\sin^2 x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \frac{c}{\sin^2 x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \int_0^x \frac{\tilde c_2}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \frac{\sin^2 x + \cos^2 x}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \left[- \frac{(\cos x)'}{\sin x} - \cos x \left(\frac{1}{\sin x}\right)' \right]\, \textrm{d}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }


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

U(x) = c_1 + \tilde c_2 \int_0^x \left(- \frac{\cos x}{\sin x}\right)'\, \textrm{d}x = c_1 + c_2 \frac{\cos x}{\sin x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

Then


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

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{\frac{\cos x}{x}}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * }

So


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

$$\displaystyle u_2(x) = x^{-1} \cos(x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }

3. Assume the full solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c_1(x) u_1(x) + c_2(x) u_2(x) $$ $$ Differentiate
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c'_1(x) u_1(x) + c_1(x) u'_1(x) + c'_2(x) u_2(x) + c_2(x) u'_2(x) $$ $$ Now assume that
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u_1(x) + c'_2(x) u_2(x) = 0 $$ $$ Then
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c_1(x) u'_1(x) + c_2(x) u'_2(x) $$ $$ Differentiate again
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u''_2(x) $$ $$ Now consider the form of a L2-ODE-VC
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y''(x) + a_1(x) y'(x) + a_0(x) y(x) = f(x) $$ $$ Substitute Eqs. 22, 25, 26 into Eq. 27
 * <p style="text-align:right;">$$\displaystyle (Eq. 27)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u_2(x) + $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_1(x) \left[c_1(x) u'_1(x) + c_2(x) u'_2(x)\right] + a_0(x) \left[c_1(x) u_1(x) + c_2(x) u_2(x)\right] = f(x) $$ $$ Rearrange the terms
 * <p style="text-align:right;">$$\displaystyle (Eq. 28)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) \underbrace{\left[ u''_1(x) + a_1(x) u'_1(x) + a_0(x) u_1(x) \right]}_{=0} + c'_1(x) u'_1(x) + $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) \underbrace{\left[ u''_2(x) + a_1(x) u'_2(x) + a_0(x) u_2(x) \right]}_{=0} + c'_2(x) u'_2(x) = f(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c'_2(x) u'_2(x) = f(x) = 1 $$ $$ Combining Eqs. 24 and 30 yields
 * <p style="text-align:right;">$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \underbrace{\left[\begin{matrix} u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x) \end{matrix}\right]}_{\mathbf{W}} \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ Where $$\mathbf{W}$$ is the Wronskian matrix. We can then solve for $$c'_1(x)$$ and $$c'_2(x)$$ be inverting $$\mathbf{W}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 31)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ We need the derivatives of the homogeneous solutions:
 * <p style="text-align:right;">$$\displaystyle (Eq. 32)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_1(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_2(x) = - \frac{\sin x}{x} - \frac{\cos x}{x^2} $$ $$ Next, we need the Wronskian, i.e. the determinant of the Wronskian matrix
 * <p style="text-align:right;">$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \det \mathbf{W} = u_1(x) u'_2(x) - u'_1(x) u_2(x) = - \frac{\sin^2 x}{x^2} - \frac{\sin x\cos x}{x^3} - \frac{\cos^2 x}{x^2} + \frac{\sin x \cos x}{x^3} = - \frac{1}{x^2} $$ $$ The inverse of the Wronskian is then
 * <p style="text-align:right;">$$\displaystyle (Eq. 35)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Now, equation 32 can be evaluated
 * <p style="text-align:right;">$$\displaystyle (Eq. 36)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} = x^2 \left\{\begin{matrix} -u_2(x) f(x) \\ u_1(x) f(x) \end{matrix}\right\} = -x \left\{\begin{matrix} -\cos x \\ \sin x \end{matrix}\right\} $$ $$ The coefficients can then be integrated
 * <p style="text-align:right;">$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) = \int^x \xi \cos \xi \, \textrm{d}\xi = \cos(x) + x \sin(x) + C_1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 38)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \xi \sin \xi \, \textrm{d}\xi = -\sin(x) + x \cos(x) + C_2 $$ $$ Now, we can substitute into Eq. 22
 * <p style="text-align:right;">$$\displaystyle (Eq. 39)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \left[\cos(x) + x \sin(x) + C_1\right] \frac{\sin(x)}{x} + \left[-\sin(x) + x \cos(x) + C_2\right] \frac{\cos(x)}{x} = $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{\sin(x)\cos(x)}{x} + \sin^2(x) + C_1 u_1(x) -\frac{\sin(x)\cos(x)}{x} + \cos^2(x) + C_2 u_2(x) = C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 40)
 * }
 * }

4. Assume we have the homogeneous solution $$u_1(x)$$ and the full solution has the form


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

$$\displaystyle y(x) = U(x) u_1(x) $$ $$ Then (see slide 21-3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 41)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) Z'(x) + \left[a_1(x) u_1(x) + 2 u'_1(x) \right]Z(x) = f(x) $$ $$ where $$Z(x)=U'(x)$$ and $$a_1(x) = \frac{2}{x}$$. We can rewrite Eq. 42
 * <p style="text-align:right;">$$\displaystyle (Eq. 42)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$ Then
 * <p style="text-align:right;">$$\displaystyle (Eq. 43)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left(Z(x) e^{g(x)}\right)' = e^{g(x)} \tilde f(x) $$ $$ We can integrate
 * <p style="text-align:right;">$$\displaystyle (Eq. 44)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ where
 * <p style="text-align:right;">$$\displaystyle (Eq. 45)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x}{\sin x} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 46)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = 2 \cot x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 47)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = 2 \ln\left(\sin(x)\right) = \ln\left(\sin^2(x)\right) $$ $$ Substitute into Eq. 45
 * <p style="text-align:right;">$$\displaystyle (Eq. 48)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = \frac{1}{\sin^2(x)} \int^x \xi \sin(\xi) \, \textrm{d}\xi = \frac{1}{\sin^2(x)} \left\{\left[-\xi \cos(\xi)\right]^x + \int^x \cos(\xi) \, \textrm{d}\xi \right\} $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle =\frac{1}{\sin^2(x)} \left\{-x \cos(x) + \left[\sin(\xi)\right]^x \right\} = -x \frac{\cos(x)}{\sin^2(x)} + \frac{1}{\sin(x)} + \frac{C}{\sin^2(x)} = U'(x) $$ $$ Integrating $$Z$$ to get $$U(x)$$ yields
 * <p style="text-align:right;">$$\displaystyle (Eq. 49)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \frac{x}{\sin x} - C \frac{\cos(x)}{\sin(x)} + C_1 = C_1 + C_2 \frac{\cos(x)}{\sin(x)} + \frac{x}{\sin x} $$ $$ Finally, we have that
 * <p style="text-align:right;">$$\displaystyle (Eq. 50)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = C_1 u_1(x) + C_2 \frac{\cos(x)}{x} + 1 = C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 51)
 * }
 * }

= Problem 3 = From lecture notes slides 21-3, 21-4, 23-1 / King et. al, p. 28, problem 1.1a.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

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

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

u_1 = e^x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. We start with the trial solution


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

$$ \displaystyle u_1(x) = e^{rx} $$ $$ We differentiate
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u'_1(x) = r e^{rx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }


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

$$ \displaystyle u''_1(x) = r^2 e^{rx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Plugging $$u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Rearranging the above terms yields

Plugging $$u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

For the above equation to be satisfied for arbitrary $$x$$, we need the following equations to be satisfied


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

$$ \displaystyle \begin{cases} r^2 - r = r (r - 1) = 0 \\ 1 - r^2 = -(r + 1)(r - 1) = 0 \end{cases} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

The only solution to both the above equations is $$r=1$$. Hence


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

$$ \displaystyle u_1(x) = e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{-x}{x-1} y' + \frac{1}{x-1} y = y + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$a_1(x) = \frac{-x}{x-1}$$ into Eq. 13 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x}{x-1}\right) ds \right\} = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x + 1 - 1}{x-1}\right) ds \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(-1 - \frac{ 1}{x-1}\right) ds \right\} = \frac{c_1}{[u_1(x)]^2} \exp\left(x + \ln(x-1)\right) = \frac{c_1 (x-1)}{[u_1(x)]^2} e^x = \tilde c_2 (x-1) e^{-x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \tilde c_2 (x-1) e^{-x} $$ $$ Integrating yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \int_0^x (s-1) e^{-s}\,\textrm{d}s = c_1 + \tilde c_2 \left\{ \left[ - (s-1) e^{-s}\right]_{s=0}^x + \int_0^x e^{-s} \,\textrm{d}x \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 + \left[ - e^{-s}\right]_{s=0}^x \right\} = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 - e^{-x} + 1\right\} = c_1 + \tilde c_2 \left\{ - x e^{-x} \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 - \tilde c_2 x e^{-x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * }

Then


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

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{x}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }

So


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

$$\displaystyle u_2(x) = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }

3. We can rewrite Eq. 1 as


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

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y = \underbrace{\frac{x}{x-1}}_{f(x)} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * }

Then we can use Eq. 32 and 37 from problem 2
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{ \begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix} \right\} = \mathbf{W}^{-1} \left\{ \begin{matrix} 0 \\ f(x) \end{matrix} \right\} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Eq. 25 evaluates to
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{e^x - x e^x} \left[ \begin{matrix} 1  & -x   \\ -e^x & e^x \end{matrix} \right] $$ $$ So Eq. 24 becomes
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) = \frac{-x f(x)}{e^x - x e^x} = \frac{x^2}{e^x (1 - x)^2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 27)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_2(x) = \frac{e^x f(x)}{e^x - x e^x} = -\frac{x}{(1 - x)^2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 28)
 * }
 * }

Integrating the coefficients, $$c'_1(x)$$ and $$c'_2(x)$$ yields


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

$$\displaystyle c_1(x) = \int^x \frac{\xi^2}{e^\xi (1 - \xi)^2} \, \textrm{d}\xi = \int^x e^{-\xi}\left(1 + \frac{2}{x-1} + \frac{1}{(x-1)^2}\right) \, \textrm{d}\xi = \left[-e^{-\xi} - \frac{e^{-\xi}}{x-1}\right]^x + \int^x \frac{e^{-\xi}}{x-1} \, \textrm{d}\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = \frac{-x e^{-x}}{x-1} + \int^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \frac{\xi}{(1 - \xi)^2} \, \textrm{d}\xi = -\int^x \frac{1}{\xi - 1} + \frac{1}{(\xi - 1)^2} \, \textrm{d}\xi = -\ln(x-1) + \frac{1}{x-1} + C_2 $$ $$ Hence, the solution is
 * <p style="text-align:right;">$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \frac{-x}{x - 1} + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi + C_1 e^x + \frac{x}{x - 1} - x \ln(x - 1) + C_2 x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 31)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = C_1 e^x + C_2 x- x \ln(x - 1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 32)
 * }
 * }

4. We start with equation 43 and 45 from problem 2


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

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ We need the following functions:
 * <p style="text-align:right;">$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x e^{-x}}{x-1} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 35)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = \frac{-x}{x-1} + 2 = 1-\frac{1}{x-1} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 36)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = x - \ln(x - 1) + \tilde C $$ $$ Substituting into Eq. 34 yields
 * <p style="text-align:right;">$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = (x-1) e^{-x} \int^x \frac{x}{(\xi-1)^2} \, \textrm{d}\xi =  (x-1) e^{-x} \left[\ln (x-1) - \frac{1}{x-1} + \bar C \right] $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = (x-1) e^{-x} \ln (x-1) - e^{-x}  + \bar C (x-1) e^{-x} $$ $$ Now we integrate Eq. 38 to get $$U(x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 38)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \int^x \left[ (\xi-1) e^{-\xi} \ln (\xi-1) - e^{-\xi} + \bar C (\xi-1) e^{-\xi} \right] {\rm d}\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-1) e^{-x} \ln (x-1)-\bar C (x-1) e^{-x} + \int^x \left[ e^{-\xi} \ln (\xi-1) + \cancel{e^{-\xi}} - \cancel{e^{-\xi}} + \bar C e^{-\xi} \right] {\rm d}\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-\cancel{1}) e^{-x} \ln (x-1)-\bar C (x-\cancel{1}) e^{-x} - \cancel{e^{-x} \ln (x-1)} - \cancel{\bar C e^{-x}} + \int^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = c_1 - x e^{-x} \ln (x-1)-\bar C x e^{-x}+ \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$ Then
 * <p style="text-align:right;">$$\displaystyle (Eq. 38)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = c_1 u_1(x) + c_2 x - x \ln (x-1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 39)
 * }
 * }

= Problem 4 = From lecture notes slide 22-1 / King et. al, p. 28, problem 1.3a.

Given
The non-homogeneous L2-ODE-CC equation
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'' - 2 y' + y = x^{3/2} e^x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution

Solution
Finding the homogeneous solution

Start with the trial solution
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_H(x) = e^{rx} $$ $$ Differentiate
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'_H(x) = r e^{rx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y''_H(x) = r^2 e^{rx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Substitute into the homogeneous version of Eq. 1
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left( r^2 - 2 r + 1 \right) e^{rx} = 0 $$ $$ The solutions to the characteristic equation
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle r = 1 \pm \sqrt{1 - 1} = 1 $$ $$ Since we have a double root at $$r=1$$, we get
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_H(x) = (c_1 + c_2 x) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Finding the particular solution

We begin with the trial solution


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

$$ \displaystyle y_P(x) = g(x) e^x $$ $$ Then we differentiate
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'_P(x) = g'(x) e^x + g(x) e^x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = g(x) e^x + 2 g'(x) e^x + g(x) e^x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Substituting Eqs. 8, 9, and 10 into Eq. 1 yields:
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[ g''(x) + \cancel{2 g'(x)} + \cancel{g(x)} - \cancel{2 g'(x)} - \cancel{2 g(x)} + \cancel{g(x)} \right]e^x = x^{3/2} e^x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) e^x = x^{3/2} e^x $$ $$ Divide by $$e^x$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) = x^{3/2} $$ $$ Integrate $$g''(x)$$ twice
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g'(x) = \frac{2}{5} x^{5/2} + c_3 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g(x) = \frac{4}{35} x^{7/2} + c_3 x + c_4 $$ $$ Hence
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ and the complete solution is
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = y_H(x) + y_P(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = (c_1 + c_2 x) e^x + \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ which simplifies to just
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = \left(c_1 + c_2 x + \frac{4}{35} x^{7/2}\right) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

= Solutions for problems 1-4 are verified with Maple =

= Problem 5 = From lecture slide 22-3

Find
Describe in words step-by-step method of attaching similar problems, given only

1. a homogeneous L2-ODE-VC

2. a non-homogeneous L2-ODE-VC

Solution
1. The homogeneous solution to a L2-ODE-VC can be found using the following scheme

Relevant slides

Reduction of order method 0 (missing dependent variable): 4-1, 9-4

Reduction of order method 1 (exactness conditions): 9-4, 12-3, Eqs. 4 and 5 10-3, Eq. 5 14-1

Trial solution (undetermined coefficients): 16-3,

Reduction of order method 2 (undetermined coefficient): 17-1,

2. If the ODE is non-homogeneous, first find the homogeneous solution according to 1., then proceed to find the particular solution with the following scheme

Note that the alternative method combines the reduction of order method 2 and finding the particular solution in a single step.

Relevant slides

Trial solutions: 20-2, 20-3

Alternative solution: 21-2

= References =

= Contributing Team Members = Egm6321.f09.Team1.sallstrom 18:17, 13 October 2009 (UTC)