User:Egm6321.f11.team2/HW5

=R*5.1 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 2=

Given
The second exactness condition for N2-ODEs is given by

$$g_0 - \frac{dg_1}{dx} + \frac{d^2g_2}{dx^2} = 0$$

where

$$g_j := \frac{\partial G}{\partial y^{(j)}}, \quad y^{(i)} := \frac{\partial^i y}{\partial x^i}, \quad\text{and}\quad G := g + fy''$$

Find
Show that the form above is equivalent to the form of the second exactness condition for N2-ODEs given by

$$\displaystyle \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} &= g_{xp} + pg_{yp} - g_y\\ f_{xp} + pf_{yp} + 2f_y &= g_{pp} \end{align} $$

Solution
Let $$p := y'$$ and $$q:= y''$$.

Then,

$$\displaystyle g_1 = \frac{\partial}{\partial p} (g + fy'') = g_p + f_pq$$

$$\displaystyle g_2 = \frac{\partial}{\partial q} (g + fy'') = f$$

$$\displaystyle \begin{align} \frac{d^2g_2}{dx^2} =& \frac{d}{dx} (f_x + f_yp + f_pq) \\ =&\qquad f_{xx} + f_{xy}p + f_{xp}q\\ &+ f_{yx}p + f_{yy}p^2 + f_{yp}pq + f_yq \\ &+ f_{px}q + f_{py}pq + f_{pp}q^2 + f_pq' \end{align} $$

Plugging 5.1.1, 5.1.2, and 5.1.3 into

$$\displaystyle g_0 - \frac{dg_1}{dx} + \frac{d^2g_2}{dx^2} = 0 $$

yields

$$\displaystyle ( f_{xx} + 2pf_{xy} + p^2f_{yy} - g_{xp} - pg_{yp} + g_y ) + ( f_{xp} + pf_{yp} + 2f_y - g_{pp} ) q = 0$$

As noted in the lecture notes, since 1 and q are linearly independent, the only way for the above equation to be true is for their coefficients to be equal to zero.

Thus,

$$\displaystyle \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} &= g_{xp} + pg_{yp} - g_y \\ f_{xp} + pf_{yp} + 2f_y &= g_{pp} \end{align} $$

as required.

=R*5.2 - Verify the second exactness for Legendre and Hermite equations =

Given
Legendre equation: $$\displaystyle G = (1 - x^2)y'' - 2xy' + n(n+1)y = 0 $$ (5.2.1)  Hermite equation: $$\displaystyle y'' - 2xy' + 2ny = 0 $$ (5.2.2)

Find
1. Verify the exactness of the designated L2-ODE-VC. 2. If eq (5.2.2) is not exact, check whether it is in power form, and see whether if it can be made exact using IFM $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2 \end{align} $$ (5.2.2.2) 3. Verify the above equations are homogeneous solutions of the eq (5.2.2).

First Part
For Legendre equation by using first method: The first exactness condition is that

$$\displaystyle G= g(x,y,p) + f(x,y,p)y'' = 0 $$ $$\displaystyle G = \underbrace {(1 - x^2)}_{f(x,y,p)}y'' - \underbrace {2xp + n(n+1)y}_{g(x,y,p)} = 0 $$ this equation satisties the first exactness condition. In order to satisfy 2nd exactness condition, $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ (5.2.3) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ (5.2.4) $$\displaystyle f(x,y,p) = (1-x^2) $$  $$\displaystyle g(x,y,p) = -2xp + n(n+1)y $$ $$\displaystyle \begin{align} f_{xx} &=-2 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= n(n+1) \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, eq (5.2.3) and eq (5.2.4) are following that $$\displaystyle -2 = -2 + 0 - n(n+1), \,\,\,\,0=0 $$

It shows that eq (5.2.1) satisfies the 2nd exactness condition when n=0 or n= -1 For legendre equation by using second method:  $$\displaystyle g_0 - \frac {dg_1}{dx} + \frac {d^2g_2}{dx^2}=0$$  (5.2.5) $$\displaystyle \begin{align} g_0 &= \frac {\partial G}{\partial y^{(0)}} = n(n+1) \\ g_1 &= \frac {\partial G}{\partial y^{(1)}} = -2x \\ g_2 &= \frac {\partial G}{\partial y^{(2)}} = (1-x^2) \\ \frac {dg_1}{dx} &= -2 \\ \frac {d^2g_2}{dx^2} &= -2 \\ \end{align} $$ From above values, eq 5.2.5 is following that n(n+1) + 2 - 2 = 0 It also shows that this equation satisfies the second exactness when n=0. For Hermite equation by using first method: Hermite equation satisfies the first exactness condition due to following form. $$\displaystyle \underbrace {1}_{f(x,y,p)}\cdot y'' \underbrace {-2xp + 2ny}_{g(x,y,p)} = 0 $$ In order to satisfy 2nd exactness condition, the equation satisfies eq (5.2.3) and eq (5.2.4). $$\displaystyle f(x,y,p) = 1 \,\,\,\, g(x,y,p) = -2xp + 2np $$ $$\displaystyle \begin{align} f_{xx} &= 0 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= 2n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ Eq (5.2.3) and Eq(5.2.4) is following that $$\displaystyle 0 + 0 + 0 = -2 + 0 + -2n \,\,\,\,2n = -2 \, \,\,\,\, n= -1 $$ $$\displaystyle 0=0 $$ It shows that eq (5.2.2) satisfies the 2nd exactness condition when n=-1. For Hermite equation by using second method: $$\displaystyle f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0$$ (5.2.6) $$\displaystyle \begin{align} f_0 &= \frac {\partial F}{\partial y^{(0)}} = 2n \\ f_1 &= \frac {\partial F}{\partial y^{(1)}} = -2x \\ f_2 &= \frac {\partial F}{\partial y^{(2)}} = 1 \\ \frac {df_1}{dx} &= -2 \\ \frac {d^2f_2}{dx^2} &= 0 \\ \end{align} $$ From the above values, eq 5.2.6 is following that 2n + 2 = 0 n= -1 It shows that this equation satisfies 2nd exactness condition when n= -1

Second Part
$$\displaystyle h(x,y) = x^m y^n$$ n in the eq 5.2.2 is replaced with b to avoid some mistakes. $$\displaystyle x^m y^n[y'' - 2xy' + 2ny]= 0 $$ $$\displaystyle \underbrace {x^m y^n}_{f(x,y,p)}y'' \underbrace {-2x^{m+1}y^ny'+2bx^my^{n+1}}_{g(x,y,p)}=0 $$ This equation satisfies the first exactness condition due to form of above equation. $$\displaystyle f(x,y,p) = x^my^n$$ $$\displaystyle g(x,y,p) = -2x^{m+1}y^np + 2bx^my^{n_1}$$ $$\displaystyle \begin{align} f_x &= mx^{m-1}y^n \\ f_{xx} &=m(m-1)x^{m-2}y^n \\ f_{xy} &= nmx^{m-1}y^{n-1} \\ f_{yy} &= n(n-1)x^my^{n-2} \\ g_x &= -2(m+1)x^my^np + 2bmx^{m-1}y^{n+1} \\ g_{xp} &= -2(m+1)x^my^n \\ g_{yp} &= -2nx^{m+1}y^{n-1} \\ g_y   &= -2nx^{m+1}y^{n-1}p+2b(n+1)x^my^n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= nx^my^{n-1} \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, Eq 5.2.4 is following that $$\displaystyle 2nx^my^{n-1}=0 $$ To satisfy the above equation, n=0. From the above values with n=0, Eq 2.5.3 is following that $$\displaystyle m(m-1)x^{m-2} = -2(m+1)x^m -2bx^m $$ $$\displaystyle \frac {m(m-1)}{x^2}x^m = -2(m+1)x^m - 2bx^m $$ $$\displaystyle m(m-1) = -2(m+1)x^2 - 2bx^2 $$ $$\displaystyle m^2-m = -2mx^2 - 2x^2 - 2bx^2 $$ $$\displaystyle m^2-m+2mx^2+2x^2+2bx^2 $$ $$\displaystyle m^2 + (2x^2-1)m + 2x^2(1+b) = 0 $$ $$\displaystyle m = \frac {-(2x^2-1)\pm \sqrt{(2x^2-1)^2-4\cdot 2x^2(1+b)}}{2} $$ When m satisfies the above relationship, Eq 5.2.2 would be exact. Generic form of power series : $$\displaystyle y(x) = \sum^{oo}_{n=0} a_n\cdot x^n $$ (5.2.7) $$ y'(x) = \sum^{oo}_{n=1} a_n\cdot nx^{n-1}$$  (5.2.8) $$ y''(x) = \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2}$$ (5.2.9) From eq (5.2.7)~(5.2.9), eq (5.2.2) is following that $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -2x\sum^{oo}_{n=1} a_n\cdot nx^{n-1}+2b\sum^{oo}_{n=0} a_n\cdot x^n =0 $$ $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -\sum^{oo}_{n=1} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ $$ \displaystyle \sum^{oo}_{n=0} (n+2)(n+1)a_{n+2}\cdot x^{n} -\sum^{oo}_{n=0} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ (5.2.10) From Eq (5.2.10), $$\displaystyle a_{n+2}= \frac {2(n-b)}{(n+2)(n+1)}a_n = 0$$ (5.2.11) For n=0,1,2,3,...... This is valid for only positive integer of b (1,2,3,4,.......)

Third Part
From eq 5.2.2.2 Boundary condition of eq 5.2.2.2 is that where b=0, y(0)= 1, y'(0)=0 where b=1, y(0)= 0, y'(0)=2 where b=2, y(0)= -2, y'(0)= 0 where b=1, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 0, y'(0) = a_1 = 2$$ It shows that all even coefficients will be equal to zero because each coefficient is a multiple of its second predecessor. $$\displaystyle a_0 = a_2 = a_4 = a_6 = a_8 = a_10 ..... = 0$$ From eq (5.2.11), $$\displaystyle a_3 = \frac {2(1-1)}{(1+2)(1+1)}a_1 = 0 $$ It shows that $$\displaystyle a_5 = a_7 = a_9 = .... = 0$$ Only a1 = 2. Thus, from the eq 5.2.7, $$\displaystyle y(x) = H_1(x) = a_0\cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ...... a_n\cdot x^n = x $$ Where b=2, and the initial conditions are given as $$\displaystyle y(0)= a_0 = -2, y'(0) = a_1 = 0$$ eq 5.2.11 is following that $$\displaystyle a_2 = \frac {2(0-2)}{(0+2)(0+1)}a_0 = 4$$ $$\displaystyle a_4 = \frac {2(2-2)}{(2+2)(2+1)}a_2 = 0 $$ It shows that $$\displaystyle a_4 = a_6 = a_8 = .... = 0$$ Only a0 = -2 and a2 = 4

Thus, from the eq 5.2.7 $$\displaystyle y(x) = H_2(x) = a_0 \cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ........ a_n\cdot x^n = 4x^2 -2 $$

Where b=0, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 1, y'(0) = a_1 = 0$$ We can not use eq 5.2.11 because the eq 5.2.11 is only valid for positive integer number. So We can calculate other ways. $$\displaystyle y'' - 2xy' = 0$$ $$\displaystyle y(x)= c1 + \frac {1}{2}\sqrt{\pi}\cdot c2\cdot erfi(x) $$ (5.2.12) From the initial values, c1= 1, c2=0. eq 5.2.12 is that <Br/> $$\displaystyle y(x)=H_0(x) = 1$$ $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2\\ \end{align} $$

These results show that the equations in (5.2.2.2) are homogeneos solutions of this Hermite differentail equation.

Given
Given L4-ODE-CC,

$$\displaystyle X^{(4)} - K^4 X = 0 $$ <p style="text-align:center"> (5.3.1)

Assuming

$$\displaystyle X(x) = e^{(rx)} $$ <p style="text-align:center"> (5.3.2)

Where

$$\displaystyle r_{1,2} = \pm K $$

$$\displaystyle r_{3,4} = \pm i \, K $$ <p style="text-align:center"> (5.3.3)

Find
In terms of $$ \displaystyle \cos Kx, \sin Kx, \cosh Kx, \sinh kh$$

Find expression for $$ \displaystyle X(x)$$

Solution
According to the (5.3.3),

$$\displaystyle X(x) = C_1 e^{Kx} + C_2 e^{-Kx} + C_3 e^{iKx} + C_4 e^{-iKx} $$ <p style="text-align:center"> (5.3.4)

Since

$$ \displaystyle e^{iKx} = \cos Kx +i\sin Kx$$

$$ \displaystyle e^{Kx} = \cosh Kx + \sinh Kx$$

Substituting the above relations to (5.3.4)

$$\displaystyle X(x) = C_1 (\cosh Kx \ + \sinh Kx) + C_2 (\cosh Kx \ - \sinh Kx) + C_3 (\cos Kx + i \sin Kx) + C_4 (\cos Kx - i\sin Kx) $$ <p style="text-align:center"> (5.3.5) $$\displaystyle X(x) = c_1 \cosh Kx \ + c_2 \sinh Kx + c_3 \cos Kx  + c_4 \sin Kx $$ <p style="text-align:center"> (5.3.6) Where

$$\displaystyle c_1 = C_1 + C_2 \in \mathbb R $$

$$\displaystyle c_2 = C_1 - C_2 \in \mathbb R $$

$$\displaystyle c_3 = C_3 + C_4 \in \mathbb R $$

$$\displaystyle c_4 = i(C_3 - C_4) \in \mathbb R$$

=R*5.4- Find $$\displaystyle y_{xxxxx} $$=

Given
A new variable (i.e. t) is introduced from current variable (i.e. x) in Euler Ln-ODE-CC. The transformation of variables is

Find
Find $$\displaystyle y_{xxxxx} $$ in terms of the derivatives of y with respect to t.

Solution
With the transformation of variables in Euler Ln-ODE-CC

We can obtain the derivatives of y with respect to x in terms of the derivatives of y with respect to t.

Finally,

=R*5.5 - Solve L2-ODE-VC with Method 2 Trial Solution =

Given
With Boundary Conditions as:

Find
Find the solution assuming trial solution as: where r = constant and plot the solution.

Solution
Substituting the value of $$ \displaystyle y' $$ and $$ \displaystyle y'' $$ in equation 5.5.1. Now $$ \displaystyle x^r \neq 0 $$ therefore : Equation 5.5.6 is the characteristic equation whose roots are: Hence the solution can be expressed as: Where $$ \displaystyle C_1 $$ and $$ \displaystyle C_2 $$ are constants need to be determined by applying boundary conditions. On applying boundary conditions we get: Solving these simultaneous equations we get: Hence the solution becomes as: Plotting the solution with 'x' varying from [-10:10]

=R5.6 - Equivalence of the Two Trial Solutions for an Euler Ln-ODE-VC=

Given
An Euler Ln-ODE-VC takes the form
 * $$\displaystyle a_n x^n y^{(n)} + a_{(n-1)} x^{(n-1)} y^{(n-1)} + \cdots + a_1 x y' + a_0 y = 0$$

<p style="text-align:right">(5.6.1) Or equivalently,
 * $$\displaystyle \sum_{i=0}^n a_i x^i y^{(i)} = 0$$

<p style="text-align:right">(5.6.2) There are two methods for solving an Euler Ln-ODE-VC. Method 1: Use a transformation of variables to transform the Euler Ln-ODE-VC into an Euler Ln-ODE-CC
 * $$\displaystyle x = e^t$$

<p style="text-align:right">(5.6.3) Then use the trial solution
 * $$\displaystyle y = e^{rt}$$

<p style="text-align:right">(5.6.4) Where r is a constant. Method 2: Use the trial solution
 * $$\displaystyle y = x^r$$

<p style="text-align:right">(5.6.5)

Find
Show that the trial solution for Method 2 is equivalent to the combined trial solution for Method 1.

Solution
It can be seen that the two methods are equivalent by combining equations (5.6.3) and (5.6.4) to form equation (5.6.5).
 * $$\displaystyle y = \left ( e^t \right )^r$$
 * $$\displaystyle y = \left ( x \right )^r$$

Which is equivalent to equation (5.6.5).

=R5.7 - Derivation of Homogeneous Solutions to the Euler-Cauchy Differential Equation with Repeated Real Roots of the Characteristic Equation &mdash; Variable &amp; Constant Coefficient Cases=

Find
<ol> <li>Find a2, a1, and a0 such that 5.7.1 is the characteristic equation of 5.7.2.</li> <li>Show that $$\displaystyle y_1 = x^\lambda$$ is a solution to 5.7.2 with the coefficients determined in part 1.</li> <li>Consider y = c(x) y1(x). Determine c(x) so that y is a solution of 5.7.2</li> <li>Use the results above to determine the second homogeneous solution of 5.7.2 </ol>

Repeat the above for 5.7.3 with $$\displaystyle y_1 = e^{\lambda x}$$

Solution
We begin by determining the characteristic equation of 5.7.2. Suppose, for some r yet to be determined,

$$\displaystyle \begin{align} \mathrm{Let}\quad y &= x^r \\ \mathrm{then}\quad y' &= r\, x^{r-1} \\ y'' &= r(r-1)\,x^{r-2} \end{align} $$

Substituting these into 5.7.2 yields

$$\displaystyle a_2\, x^2 \left[ r(r-1)\,x^{r-2} \right] + a_1\, x \left[r x^{r-1} \right] + a_0\, x^r = \left[ a_2\, r^2 + (a_1 - a_2)\, r + a_0 \right] x^r$$

The expression in brackets on the right hand side is the characteristic equation of 5.7.2. Expanding 5.7.1 as

$$\displaystyle r^2 - 2\lambda\, r + \lambda^2$$

and equating coefficients of r yields

$$\displaystyle \begin{align} a_2 &= 1 \\ a_1 &= 1 - 2\lambda \\ a_0 &= \lambda^2 \end{align} $$

Substituting these values back into 5.7.2 gives us our equation

The standard procedure for solving the Euler-Cauchy differential equations with distinct roots $$\lambda_1$$ and $$\lambda_2$$ is to use two solutions of the form $$y_1 = x^{\lambda_1}$$ and $$y_2 = x^{\lambda_2}$$. The general solution to the homogeneous equation would then be a linear combination of these two solutions. When there are real, repeated roots to the characteristic equation, as in this example, we must determine another homogeneous solution since $$\lambda_1 = \lambda_2$$ and the two solutions above are not linearly independent (indeed, they are identical!).

We begin by showing that $$\displaystyle y_1 = x^\lambda$$ is a solution of 5.7.2b.

$$\displaystyle \begin{align} y_1 &= x^\lambda \\ y_1' &= \lambda\, x^{\lambda - 1} \\ y_1'' &= \lambda\, (\lambda - 1)\, x^{\lambda - 2} \end{align} $$

Upon substituting these values into 5.7.2b, we get

$$\displaystyle \left[ (\lambda^2 - \lambda) + (1 - 2\lambda)\, \lambda + \lambda^2 \right] x^\lambda = \cancelto{0}{(\lambda^2 - 2\lambda^2 + \lambda^2 - \lambda + \lambda)}\, x^\lambda = 0$$

so y1 is indeed a solution of 5.7.2b.

We next seek a solution of the form y = c(x) y1(x).

$$\displaystyle \begin{align} y = c(x)\, y_1(x) &= c\, x^\lambda \\ y' = c'\, y_1 + c\, y_1' &= c'\, x^\lambda +c\, \lambda\,x^{\lambda - 1} \\ y = c\, y_1 + 2c'\,y_1' + c\,y_1 &= c\,x^\lambda + 2c'\, \lambda\, x^{\lambda-1} + c\,(\lambda^2 - \lambda)\, x^{\lambda - 2} \end{align} $$

Substituting these into 5.7.2b and simplifying gives

$$\displaystyle \begin{align} &x^2 \left[ c''\,x^\lambda + 2c'\, \lambda\, x^{\lambda-1} + c\,(\lambda^2 - \lambda)\, x^{\lambda - 2} \right] + (1 - 2\lambda)\,x\left[ c'\, x^\lambda +c\, \lambda\,x^{\lambda - 1} \right] + \lambda^2\,\left[ c\,x^\lambda\right] \\ &= c''\,x^{\lambda+2} + \left[ 2\lambda\, c' + (1 - 2\lambda)\,c'\right]x^{\lambda+1} +\cancelto{0}{\left[(\lambda^2-\lambda)\, c + (1-2\lambda)\,\lambda\, c + \lambda^2\,c\right]} x^\lambda \\ &= x^{\lambda+2}\, c'' + x^{\lambda+1}\, c' = 0 \end{align} $$

which is a L2-ODE-VC for c.

Since c does not appear in this equation, we can set p = c' to reduce the order of the equation by one. We are left with

5.7.4 is not exact, since

$$\displaystyle M_p = x^{\lambda+1} \neq (\lambda+2)\,x^{\lambda+1} = N_x$$

It can, however, be made exact with the IFM:

$$\displaystyle \begin{align} \frac{h_x}{h} &= \frac{1}{N}(M_p - N_x) \\ &= x^{-(\lambda+2)}\left[ x^{\lambda+1} - (\lambda+2)\,x^{\lambda+1}\right] \\ &= x^{-(\lambda+2)} (-\lambda - 1)\, x^{\lambda+1} \\ \frac{h_x}{h} &= -(\lambda + 1)\, x^{-1} \\ \int \frac{dh}{h} &= -\int (\lambda + 1)\, x^{-1}\,dx \\ \ln h &= -(\lambda + 1)\,\ln x \\ h &= x^{-(\lambda+1)} \end{align} $$

Multiplying 5.7.4 by h yields

$$\displaystyle \begin{align} p + x\, p' &+ 0 \\ \frac{d}{dx}(xp) &= 0 \\ xp &= k_1 \\ p &= k_1\,x^{-1} = c' \\ \therefore\, c &= k_1 \ln x + k_2 \end{align} $$

Plugging this value of c into the definition of y gives

$$\displaystyle y(x) = (k_1\ln x + k_2)\,x^\lambda = k_1\, x^\lambda \ln x + k_2\, x^\lambda$$

Inspection of this expression reveals that it is composed of a linear combination of the functions x&#x03bb; ln x and x&#x03bb;. But the second function is the first homogeneous solution y1, which means that the first function must be the second homogeneous solution we were seeking.

$$\displaystyle y_2 = x^\lambda\ln x$$

Since the method of variation of parameters has proven to be such a success with the second-order Euler-Cauchy differential equation with varying coefficients, we shall now turn our attention to the second order Euler differential equation with constant coefficients, where we hope to have similar success.

As before, we will consider the case where the characteristic equation has repeated real roots and attempt to determine a second homogeneous solution via the method of variation of parameters. As we are following the same steps as above, the exposition will be brief.

We begin by determining the values of b2, b1, and b0 such that the characteristic equation of 5.7.3 is equal to 5.7.1, the definition of repeated real roots.

$$\displaystyle \begin{align} \mathrm{Let}\quad y &= e^{rx} \\ \mathrm{then}\quad y' &= r\,e^{rx} \\ y'' &= r^2\,e^{rx} \end{align} $$

Plugging these into 5.7.3, expanding 5.7.1 as before, and matching coefficients

$$\displaystyle b_2\,r^2\,e^{rx} + b_1\, r\, e^{rx} + b_0\,e^{rx} = \left( b_2\,r^2 + b_1\, r + b_0 \right) e^{rx}$$

Thus

$$\displaystyle \begin{align} b_2 &= 1 \\ b_1 &= -2\lambda \\ b_0 &= \lambda^2 \end{align} $$

which, upon substitution into 5.7.3 gives

Let $$\displaystyle y_1 = e^{\lambda x}$$

y1 satisfies 5.7.3b since

$$\lambda^2\,e^{\lambda x} - 2\lambda^2\, e^{\lambda x} + \lambda^2\,e^{\lambda x} = 0$$

We now begin the method of variation of parameters to determine a second homogeneous solution of 5.7.3b.

$$\displaystyle \begin{align} y &= c\,e^{\lambda x} \\ y' &= c'\,e^{\lambda x} + \lambda\, c\, e^{\lambda x} \\ y &= c\, e^{\lambda x} + 2\lambda\,c'\, e^{\lambda x} + \lambda^2\,c\, e^{\lambda x} \end{align} $$

Plugging these into 5.7.3b gives

$$\displaystyle \begin{align} &\left[ c''\, e^{\lambda x} + 2\lambda\,c'\, e^{\lambda x} + \lambda^2\,c\, e^{\lambda x} \right] - 2\lambda\left[c'\,e^{\lambda x} + \lambda\, c\, e^{\lambda x}\right] + \lambda^2\,c\,e^{\lambda x} \\ &= c''\,e^{\lambda x} + \cancel{2\lambda\,c'\,e^{\lambda x}} + \cancel{\lambda^2\,c\, e^{\lambda x}} - \cancel{2\lambda\,c'\,e^{\lambda x}} - \cancel{2\lambda^2\,c\, e^{\lambda x}} + \cancel{\lambda^2\,c\,e^{\lambda x}} \\ &= c''\,e^{\lambda x} = 0 \end{align} $$

The only way this expression can be true for all x is if $$\displaystyle c'' = 0$$. Thus,

$$\displaystyle c = k_1\,x + k_2$$

Thus,

$$\displaystyle y = (k_1\,x + k_2)\,e^{\lambda x} = k_1\,x\,e^{\lambda x} + k_2\,e^{\lambda x}$$

Using the same logic as in part 1, we conclude that

$$\displaystyle y_2 = x\,e^{\lambda x}$$

is the second homogeneous solution we are seeking.

=R*5.8 - Particular solution by using variation parameters method =

Given
$$\displaystyle y(x) = A(x)y_H(x)$$<p style="text-align:center"> (5.8.1) $$\displaystyle y' + a_0(x)y = 0 $$ $$\displaystyle y_H(x) = exp[-\int^x a_0(s)ds] $$

Differential eq is that $$\displaystyle y' + P(x)y = Q(x) $$<p style="text-align:center"> (5.8.2) $$\displaystyle y_H(x) = exp[-\int P(x)dx] $$<p style="text-align:center"> (5.8.3)

Find
Find particular solution Yp <Br/>

Solution
$$\displaystyle y(x) = A(x)y_H(x) $$ $$\displaystyle y(x)' = A(x)'y_H(x) + A(x)y_H(x)' $$<p style="text-align:center"> (5.8.4) From the eq 5.8.3, $$\displaystyle y_H(x)'= -P(x)\cdot exp[-\int P(x)dx]=-P(x)y_H(x)$$ Thus, eq 5.8.4 is following that <Br/> $$\displaystyle y(x)' = A(x)'\cdot exp[-\int P(x)dx] - P(x)\cdot exp[-\int P(x)dx]A(x) = exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)]$$ <Br/> Plugging the above equation into Eq 5.8.2, $$\displaystyle exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)] + P(x)A(x)\cdot exp[-\int P(x)dx]= Q(x) $$ $$\displaystyle exp[-\int P(x)dx](A(x)'-P(x)A(x)+A(x)P(x))=Q(x)$$ $$\displaystyle exp[-\int P(x)dx]\cdot A(x)' = Q(x)$$ $$\displaystyle A(x)' = \frac {Q(x)}{exp[-\int P(x)dx]} = Q(x)\cdot exp[\int P(x)dx] $$ $$\displaystyle A(x) = \int Q(x) \cdot exp[\int P(x)dx]dx $$ Thus, paricular solution is that $$\displaystyle y(x) = \int Q(x) \cdot exp[\int P(x)dx]dx \cdot exp[-\int P(x)dx] $$

=R*5.9 -Special IFM=

Given
Nonhomogeneous L2-ODE-CC

$$\displaystyle a_2 y'' + a_1 y' +a_0 y = f(t)$$ <p style="text-align:center">(5.9.1)

Find
Using special IFM to solve (5.9.1) for general $$\displaystyle f(t)$$

Solution
Suppose the integrating factor has the form $$\begin{align} h(t,y) \end{align}$$

Multiply the integrating factor to the 5.9.1.

$$\displaystyle h(t,y) \cdot \left[ {{a}_{2}}y''+{{a}_{1}}y'+{{a}_{0}}y \right] = h(t,y) \cdot \left[{{f}_{\left( t \right)}}\right] $$ <p style="text-align:center"> (5.9.2)

rewrite the 5.9.2 to check the exactness condition.

$$\displaystyle F(x,y,y',y) = \underbrace{h(t,y){a}_{2}}_{f(x,y,p)}y+ \underbrace{h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)}_{g(x,y,p)} $$ <p style="text-align:center"> (5.9.3)

where $$\begin{align} p:= y' \end{align} $$

therefore,

$$\displaystyle f(x,y,p) = h(t,y){a}_{2} $$

$$\displaystyle g(x,y,p) = h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)$$

There are two relations for the 2nd condition of exactness.

$$\displaystyle {f}_{tt}+2p{f}_{ty}+{p}^{2}{f}_{yy}={g}_{tp}+p{g}_{yp}-{g}_{y} $$ <p style="text-align:center"> (5.9.4)

$$\displaystyle {f}_{tp}+p{f}_{yp}+2{f}_{y}={g}_{pp} $$ <p style="text-align:center"> (5.9.5)

From (5.9.5), we have

$$\displaystyle {h}_{y}=0$$

<p style="text-align:center"> (5.9.6) and from 5.9.4

$$\displaystyle {h}_{tt}{a}_{2}-{h}_{t}{a}_{1}+h{a}_{0}=0 $$ <p style="text-align:center"> (5.9.7) So

$$\displaystyle {a}_{2}h''-{a}_{1}h'+{a}_{0}h=0 $$ <p style="text-align:center"> (5.9.8)

thus, assuming $$ \displaystyle h(t)={e}^{\alpha t} $$

From 5.9.8, we have

$$\displaystyle {a}_{2}{\alpha}^{2}-{a}_{1}\alpha +{a}_{0}=0 $$ <p style="text-align:center"> (5.9.10)

Then, substituting (5.9.8) to (5.9.2) and integrating it yields

$$\displaystyle y(t)=\frac{{e}^{-\alpha t}}{2\alpha a_{2}-a_{1}}\left({e}^{(\frac{a_{1}}{a_{2}}-2\alpha)t}\,\int{e}^{-\alpha t}\, f\left(t\right)dt-\int{e}^{\alpha t}\, f\left(t\right)dt\right)$$ <p style="text-align:center"> ( 5.9.11) If $$\displaystyle a_{1}^{2}-4a_{0}a_{2}\neq0$$, substitute the solution of (5.9.10) to (5.9.11), we have

$$\displaystyle y(t)={C}_{1}\,{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}+{C}_{2}\,{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}+}$$

$$\displaystyle \frac{{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}}{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}}\left({e}^{\frac{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}\, t}{{a}_{2}}}\,\int{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt-\int{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt\right)$$

=Contributing Members= Egm6321.f11.team2.epps I solved problems R5.1 and R*5.7 02:11, 2 November 2011 (UTC) Egm6321.f11.team2.kim 15:11, 2 November 2011 (UTC) I solved problems R*5.2 and R*5.8 and proofread problems R*5.3, R*5.4, R*5.7, and R*5.9. Egm6321.f11.team2.rho 16:50, 2 November 2011 (UTC) I solved Problem R*5.4 and proofread problems R*5.2, R*5.3, R*5.8, and R*5.9. Egm6321.f11.team2.johri.y 17:36, 2 November 2011 (UTC) I solved R*5.5 and proofread R*5.4 Egm6321.f11.team2.Xia I solved problems R*5.3 and R*5.9, proofread R*5.1, R*5.5 04:08, 2 November 2011 (UTC)