User:Egm6321.f11.team1/report5

R*5.1 2nd Exactness Condition For n=2 Case of Nn-ODE (Method 2)

Given

Equivalent form of 2nd Exactness Condition for N2-ODEs:

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

$\displaystyle (Equation\;4.7.1)$

Coefficient equalities, located in previous reference above.

$\displaystyle g_0=f_yq+g_y$

$\displaystyle g_1=f_pq+g_p$

$\displaystyle g_2=f$

Note also that $\displaystyle p(x):=y'(x)$ and $\displaystyle q(x):=y''(x).$

Find

Show the following equality is true:

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

Solution

Solved on our own

Applying the chain rule to $\frac{dg_1}{dx}$ yields the following:

$\frac{dg_1}{dx}=\frac{\partial g_1}{\partial x}+\frac{\partial g_1}{\partial y}\frac{dy}{dx}+\frac{\partial g_1}{\partial y^{(1)}}\frac{dy^{(1)}}{dx}$

We know what $\displaystyle g_1$ is defined to be, we can substitute in $\displaystyle p(x):=y'(x)$ and $\displaystyle q(x):=y''(x)$ and carry out the derivatives:

$\frac{dg_1}{dx}=\frac{\partial g_1}{\partial x}+\frac{\partial g_1}{\partial y}p+\frac{\partial g_1}{\partial p}q$

$\frac{dg_1}{dx}=\left(f_{xp}q+g_{xp}\right)+\left(f_{yp}q+g_{yp}\right)p+\left(f_{pp}+g_{pp}\right)q$

We can apply the same approach to $\frac{d^2g_2}{dx^2}$ :

$\frac{d^2g_2}{dx^2}=\frac{\partial^2 g_2}{\partial x^2}+\frac{\partial^2 g_2}{\partial y\partial x}\frac{dy}{dx}+\frac{\partial^2 g_2}{\partial x\partial y}\frac{dy}{dx}+\frac{\partial^2 g_2}{\partial x\partial y^{(1)}}\frac{dy^{(1)}}{dx}+\frac{\partial^2 g_2}{\partial y^{(1)}\partial x}\frac{dy^{(1)}}{dx}+\frac{\partial^2 g_2}{\partial^2 y}\frac{dy^2}{dx^2}$

$\frac{d^2g_2}{dx^2}=f_{xx}+f_{yx}p+f_{xy}p+f_{xp}q+f_{px}q+f_{yy}p^2$

Adding $\displaystyle g_0$ , $\displaystyle \frac{dg_1}{dx}$ , and $\displaystyle \frac{d^2g_2}{dx^2}$ according to Equation 4.7.1 yields our final solution:

$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=\left(f_yq+g_y\right)-\left(f_{xp}q+g_{xp}+f_{yp}pq+g_{yp}p+g_{pp}q\right)+\left(f_{xx}+f_{yy}p^2+2f_{xy}p+2f_{xp}q\right)=0$

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

R*5.2 Exactness of Bessel, Legendre, and Hermite

Given

Legendre L2-ODE-VC:

$\displaystyle G=(1-x^2)y''-2xy'+n(n+1)y=0$

$\displaystyle (Equation\;5.2.1)$

Bessel L2-ODE-VC:

$\displaystyle (1-x^2)y''-2xy'+(x^2-\nu^2)y=0$

$\displaystyle (Equation\;5.2.2)$

Hermite L2-ODE-VC:

$\displaystyle y''-2xy'+2ny=0$

$\displaystyle (Equation\;5.2.3)$

Exactness Conditions

$\displaystyle ({Condition}\;1)\;\;G(x,y,y',y'')=g(x,y,p)+f(x,y,p)y''=0$

where $\displaystyle p=y'$

$\displaystyle (Condition\;2a)\;\;f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y\;\mathbf{AND}\;f_{xp}+pf_{yp}+2f_y=g_{pp}$

$\displaystyle (Condition\;2b)\;\;g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=0$

$\displaystyle g_0=f_yq+g_y$

$\displaystyle g_1=f_pq+g_p$

$\displaystyle g_2=f$

where $\displaystyle q=y''$

Find

(1) Verify the exactness conditions of Legendre, Bessel, and Hermite, using both methods for the 2nd exactness condition.

(2) If Equation 5.2.3 is not exact, check whether it is in power form, and see whether if it can be made exact using IFM with $\displaystyle h(x,y)=x^my^n$.

(3) Verify the following Hermite polynomials are homogenous solutions of Equation 5.2.3.

$\displaystyle H_0(x)=1$

$\displaystyle H_1(x)=2x$

$\displaystyle H_2(x)=4x^2-2$

Solution

Solved on our own

Part (1)

Verify Exactness of Legendre L2-ODE-VC
Condition 1:

$\displaystyle G(x,y,y',y'')=\underbrace{n(n+1)y-2xy'}_{g(x,y,p)}+\underbrace{(1-x^2)}_{f(x,y,p)}y''=0$

Condition 2a:

$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=-2+0+0=-2$

$\displaystyle g_{xp}+pg_{yp}-g_y=-2+0-n(n+1)$

Letting

$\displaystyle n=0$

allows the first part of Condition 2a to be satisfied.

$\displaystyle f_{xp}+pf_{yp}+2f_y=0+0+0$

$\displaystyle g_{pp}=0$

$\displaystyle 0=0$

Thus Condition 2 (first method) is satisfied.
Condition 2b:

$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=n(n+1)-\frac{d}{dx}\left(0-2x\right)+(-2)=0+2-2=0$

Thus Condition 2a (second part) is satisfied.

Verify Exactness of Bessel L2-ODE-VC
Condition 1:

$\displaystyle G(x,y,y',y'')=\underbrace{(x^2-\nu^2)y-2xy'}_{g(x,y,p)}+\underbrace{(1-x^2)}_{f(x,y,p)}y''=0$

Condition 2a:

$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=-2+0+0=-2$

$\displaystyle g_{xp}+pg_{yp}-g_y=-2+0-(x^2-\nu^2)$

$\displaystyle -2=-2-(x^2-\nu^2)$

Thus the Bessel equation satisfies exactness only if we assume

$\displaystyle x^2=\nu^2$

$\displaystyle f_{xp}+pf_{yp}+2f_y=0+0+0$

$\displaystyle g_{pp}=0$

$\displaystyle 0=0$

Thus Condition 2a(second part) is satisfied.
Condition 2b:

$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=(x^2-\nu^2)-\frac{d}{dx}\left(0-2x\right)+(-2)=0+2-2=0$

Thus Condition 2a (second part) is satisfied.

Verify Exactness of Hermite L2-ODE-VC
Condition 1:

$\displaystyle G(x,y,y',y'')=\underbrace{2ny-2xy'}_{g(x,y,p)}+\underbrace{(1)}_{f(x,y,p)}y''=0$

Condition 2a:

$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=0+0+0=0$

$\displaystyle g_{xp}+pg_{yp}-g_y=-2+0+2n$

Letting

$\displaystyle n=1$

allows the first part of Condition 2a to be satisfied.

$\displaystyle f_{xp}+pf_{yp}+2f_y=0+0+0$

$\displaystyle g_{pp}=0$

$\displaystyle 0=0$

Thus Condition 2 (first method) is satisfied.
Condition 2b:

$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=(2n)-\frac{d}{dx}\left(0-2x\right)+0=2-2+0=0$

Thus Condition 2a (second part) is satisfied.

Part (2)
Hermite differential equation

$\displaystyle y''-2xy'+2ny=0$

is not exact if $\displaystyle n\neq -1$.
And it is in the power form of L2-ODE-VC:

$\displaystyle \alpha x^ry''+\beta x^sy'+\gamma x^ty$

with

$\displaystyle \alpha=1;\beta=-2;\gamma=2n.$

$\displaystyle r=0;s=1;t=0.$

Assuming that we can find the integrating factor which is in power form,

$\displaystyle h(x,y)=x^my^n$.

Use the exactness condition to find $m,n\in \mathbb{E}$, such that the following N2-ODE is exact:

$\displaystyle x^m y^n(y''-2xy'+2\alpha y)=0$

$\displaystyle (Equation\;5.2.4)$

Rewrite Equation(5,2,1) as

$\displaystyle G(x,y,y',y'')=\underbrace{2\alpha x^m y^{n+1}-2x^{m+1}y^ny'}_{g(x,y,p)}+\underbrace{x^my^n}_{f(x,y,p)}y''$

Calculate the following terms in the exactness condition:

$\displaystyle f_x=mx^{m-1}y^n;\,f_{xx}=m(m-1)x^{m-2}y^n$

$\displaystyle f_{xy}=mnx^{m-1}y^{n-1};\,f_{xp}=0$

$\displaystyle f_y=nx^my^{n-1}$

$\displaystyle f_{yy}=n(n-1)x^my^{n-2};\,f_{yp}=0$

$\displaystyle g_x=2m\alpha x^{m-1}y^{n+1}-2(m+1)x^my^np;\,g_{xp}=-2(m+1)x^my^n$

$\displaystyle g_y=2\alpha (n+1)x^{m}y^{n-1}-2nx^{m+1}y^{n-1}p;\,g_{yp}=-2nx^{m+1}y^{n+1}$

$\displaystyle g_p=-2x^{m+1}y^n;\,g_{pp}=0$

Recall the 2nd exactness condition for L2-ODE-VC:

$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$

$\displaystyle (Equation\;5.2.5)$

$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$

$\displaystyle (Equation\;5.2.6)$

Substituting those calculated terms into Equation(5,2,5) and Equation(5.2.6).
The 1st exactness condition Equation(5,2,5) is

$\displaystyle m(m-1)x^{m-2}y^n+2p\cdot mnx^{m-1}y^{n-1}+p^2n(n-1)x^my^{n-2}=-2(m+1)x^my^n+p(-2nx^{m+1}y^{n-1})-2\alpha (n+1)x^my^n+2nx^{m+1}y^{n-1}p$

The 2nd exactness conditionEquation(5.2.6) is

$\displaystyle 0+p\cdot 0+2nx^my^{n-1}=0$

It is obvious that only if $\displaystyle n=0$, the 2nd exactness condition Equation(5.2.6) is satisfied.
Substitute $\displaystyle n=0$ into Equation(5,2,5), we can obtain

$\displaystyle m(m-1)x^{m-2}=-2(m+1)x^m-2\alpha x^m$

It means

$\displaystyle m(m-1)=0$

and

$\displaystyle m+1+\alpha =0$

We choose $\displaystyle m=1,\alpha =-2 \,$ to make the factor h(x,y) not zero.

Therefore, Equation(5,2,4) can be exact using IFM with $\displaystyle h(x,y)=x$

Part (3)
Verify the first three terms of Hermite polynomials.
Letting

$\displaystyle n=0$

$\displaystyle y=H_0(x)=1$

$\displaystyle y'=H_0'(x)=0$

$\displaystyle y''=H_0''(x)=0$

Substitute them into the Hermite differential equation Equation(5.2.3):

$\displaystyle 0-2x\cdot 0+0\cdot 1=0$

Therefore, the first term of Hermite polynomial $\displaystyle H_0(x)=1$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting

$\displaystyle n=1$

$\displaystyle y=H_1(x)=2x$

$\displaystyle y'=H_1'(x)=2$

$\displaystyle y''=H_1''(x)=0$

Substitute them into the Hermite differential equation Equation(5.2.3):

$\displaystyle 0-2x\cdot 2+2\cdot 2x=0$

Therefore, the second term of Hermite polynomial $\displaystyle H_1(x)=2x$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting

$\displaystyle n=2$

$\displaystyle y=H_2(x)=4x^2-2$

$\displaystyle y'=H_2'(x)=8x$

$\displaystyle y''=H_2''(x)=8$

Substitute them into the Hermite differential equation Equation(5.2.3):

$\displaystyle 8-2x\cdot 8x+4\cdot (4x^2-2)=0$

Therefore, the third term of Hermite polynomial $\displaystyle H_2(x)=4x^2-2$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

R*5.3 Find the expressions for X(x)in terms of $\cos Kx,\ \sin Kx,\ \cosh Kx,\ \sinh Kx$

Given

$\displaystyle X^{(4)}-K^4X=0$

$\displaystyle K:= \frac{\omega^2m}{EI}$

Assume:

$\displaystyle X_{(x)}=e^{(rx)}$

characteristic equation:

$\displaystyle r^4=K^4$

then:

$\displaystyle r_1=K,\ r_2=-K,\ r_3=iK,\ r_4=-iK,\ i:=\sqrt{-1}$

Find

the expressions for $X_{(x)}$ in terms of:

$\displaystyle \cos Kx, \ \sin Kx,\ \cosh Kx,\ \sinh Kx$

Solution

Solved on my own

$\displaystyle X_{(x)}=c_1e^{Kx}+c_2e^{-Kx}+c_3e^{iKx}+c_4e^{-iKx}$

$\displaystyle (Equation\;5.3.1 )$

$\displaystyle \begin{align} A_1: &=c_1e^{Kx}+c_2e^{-Kx}\ \in \mathbb{R} \\ &= c_1(\frac{e^{Kx}+e^{-Kx}}{2}+\frac{e^{Kx}-e^{-Kx}}{2}) + c_2(\frac{e^{Kx}+e^{-Kx}}{2}-\frac{e^{Kx}-e^{-Kx}}{2})\\ &= c_1(\cosh Kx+ \sinh Kx) + c_2(\cosh Kx- \sinh Kx)\\ &= (c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx \end{align}$

We can rewrite it to:

$\displaystyle A_1=b_1\cosh Kx + b_2\sinh Kx$

$\displaystyle (Equation\;5.3.2 )$

$\displaystyle \begin{align} A_2: &=c_3e^{iKx}+c_4e^{-iKx} \\ &=c_3(\cos Kx +i\sin Kx)+c_4(\cos Kx -i\sin Kx) \\ &= (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx \ \in \mathbb{R} \end{align}$

We can rewrite it to:

$\displaystyle A_2=b_3\cos Kx- b_4\sin Kx$

$\displaystyle (Equation\;5.3.3 )$

So, add up the two terms from equation 5.3.2 and equation 5.3.3:

$\displaystyle X_{x}=A_1+A_2=b_1\cosh Kx + b_2\sinh Kx + b_3\cos Kx- b_4\sin Kx$

$\displaystyle (Equation\;5.3.4 )$

Where:

$\displaystyle b_1=c_1+c_2,\ b_2=c_1-c_2,\ b_3=c_3+c_4,\ b_4=-i(c_3-c_4)$

R*5.4 Find $y_{xxxxx}$ in terms of derivatives of y with respect to time

Given

Transformation of variables:

$\displaystyle x=e^t$.

The derivatives of y with respect to x are:

$\displaystyle y_x=e^{-t} \, y_t$ $\displaystyle y_{xx}=e^{-2t} \, (y_{tt}-y_t)$ $\displaystyle y_{xxx}=e^{-3t} \, (y_{ttt}-3t_{tt}+2y_t)$ $\displaystyle y_{xxxx}=e^{-4t} \,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$

Find

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

Solution

Solved on our own

Recall that

$\displaystyle \frac{dt}{dx}=\left(\frac{dx}{dt}\right)^{-1}=e^{(-t)}$

The derivative is

$\displaystyle y_{xxxxx}=\frac{d}{dx}\,(y_{xxxx})=\frac{dy_{xxxx}}{dt}\cdot \frac{dt}{dx}$

$\displaystyle =\frac{d}{dt}\left[e^{(-4t)}\,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]\cdot e^{-t}$

$\displaystyle =e^{-t}\cdot\left[\frac{d}{dt}(e^{-4t}\cdot(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot \frac{d}{dt} (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]$

$\displaystyle =e^{-t}\cdot[-4e^{-4t}\cdot (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot (y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt})$

$\displaystyle =e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t)$

$\displaystyle$

$\displaystyle y_{xxxxx}=e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t)$

R*5.5 Solving Euler's 2^nd-order differential equation using the method of trial solution

Given

A second order linear ordinary differential equation with varying coefficients(L2-ODE-VC)

$\begin{align} {{x}^{2}}y''-2xy'+2y= 0 \end{align}$

(5.5.1)

With Boundary Coditions,

$\begin{align} & y(1)=-4 \\ & y(2)=7 \end{align}$

(5.5.2)

Find

The solution using the method of trial solution

Solution

Solved on our own

Using the given trial solution

$\displaystyle y(x)={ {x}^{r} }$,

(5.5.3)

where r is constant. Then

$\begin{align} & y'=r{{x}^{r-1}} \\ & y''=r(r-1){{x}^{r-2}} \\ \end{align}$

(5.5.4)

substituting (5.5.4) into (5.5.1)

$\begin{align} &{{x}^{2}} r(r-1){{x}^{r-2}} - 2x r{{x}^{r-1}} + 2 {x}^{r} = 0 \\ &{x}^{r} ({{r}^{2}}-r -2r +2) = 0 \end{align}$

(5.5.5)

this can be simplified to,

$\begin{align} & r(r-1)-2r+2=0 \\ & {{r}^{2}}-3r+2=0 \\ & (r-1)(r-2)=0 \\ & {{r}_{1}}=1 ;{~} {{r}_{2}}=2 \\ \end{align}$

(5.5.6)

IN general a homogeneous solution takes the following form

$y(x)= \sum_{i=1}^{n_{roots}} {c}_{i} {x}^{ri}$

(5.5.7)

Since we have two roots 5.5.7 is written as,

$y(x)={{c}_{1}}x^{r_1}+{{c}_{2}}{{x}^{r_2}}$

(5.5.8)

Using the given roots in 5.5.6 the general solution is

$\displaystyle y(x)={{c}_{1}}x+{{c}_{2}}{{x}^{2}}$

(5.5.9)

Using the boundary conditions given in 5.5.2, we get

$\begin{align} & {{c}_{1}}+{{c}_{2}}=-4 \\ & 2{{c}_{1}}+4{{c}_{2}}=7 \\ \end{align}$

(5.5.10)

These equations can be solved by simple substitution which yields c1=-11.5 and c2=7.5, therefore the solution of this BVP is

$\displaystyle y(x)=-11.5x+7.5{{x}^{2}}$

(5.5.11)

The plot of the homogeneous solution is shown below

R5.6 Show that the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1

Given

Show the equivalence of two methods using Euler L3-ODE-VC as an example:

$\displaystyle a_3x^3y^{(3)}+a_2x^2y^{(2)}+a_1xy'+a_0y=0,\ y'=y^{(1)},\ y=y^{(0)}$

$\displaystyle (Equation\;5.6.1)$

Find

That the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1.

Solution

Solved on my own

1.Method I Stage 1,

$\displaystyle x=e^t$

and we have derived in class that:

$\displaystyle y'=e^{-t}y_t$

$\displaystyle (Equation\; 5.6.2)$

$\displaystyle y^{(2)}=e^{-2t}(y_{tt}-y_t)$

$\displaystyle (Equation\; 5.6.3)$

$\displaystyle y^{(3)}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$

$\displaystyle (Equation\; 5.6.4)$

plug equations 2,3,4 into 1, we obtain:

$\displaystyle a_3e^{3t}\cdot e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+a_2e^{2t}\cdot e^{-2t}(y_{tt}-y_t)+a_1e^t\cdot e^{-t}y_t+a_0y=0$

$\displaystyle (Equation\; 5.6.5)$

Rearranging:

$\displaystyle a_3y_{ttt}+(a_2-3a_3)y_{tt}+(2a_3-a_2+a_1)y_t+a_0y=0$

$\displaystyle (Equation\; 5.6.6)$

Stage 2,

$\displaystyle y=e^{rt},\ r=constant$

substitute into equation 5.6.6, we get:

$\displaystyle a_3r^3e^{rt}+(a_2-3a_3)r^2e^{rt}+(a_1-a_2+2a_3)re^{rt}+a_0e^{rt}=0$

$\displaystyle (Equation\; 5.6.7)$

Finally:

$\displaystyle a_3r^3+(a_2-3a_3)r^2+(a_1-a_2+2a_3)r+a_0=0$

2.Method II

$\displaystyle y=x^r,\ r=constant$

substitute into equation 1:

$\displaystyle a_3x^3r(r-1)(r-2)x^{r-3}+a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0$

$\displaystyle (Equation\; 5.6.8)$

$\displaystyle a_3r(r-1)(r-2)x^r+a_2r(r-1)x^r+a_1rx^r+a_0x^r=0$

$\displaystyle (Equation\; 5.6.9)$

rearranging:

$\displaystyle a_3(r^3-3r^2+2r)+a_2(r^2-r)+a_1r+a_0=0$

$\displaystyle (Equation\; 5.6.10)$

And, finally:

$\displaystyle a_3r^3+(a_2-3a_3)r^2+(a_1-a_2+2a_3)r+a_0=0$

So, using whatever method I or method II, we can derive the exact same characteristic equation

thus, they are just the same.

R*5.7 Euler L2-ODE With Multiple Roots

Given

Characteristic Equation

$\displaystyle (r-\lambda)^2=0$

$\displaystyle (Equation\;5.7.1)$

Euler L2-ODE-VC

$\displaystyle a_2x^2y''+a_1xy'+a_0y=0$

$\displaystyle (Equation\;5.7.2)$

Euler L2-ODE-CC

$\displaystyle b_2y''+b_1y'+b_0y=0$

$\displaystyle (Equation\;5.7.3)$

Find

(1.1) Find $\displaystyle a_2,a_1,a_0$ so that Equation 5.7.1 is a characteristic equation of Equation 5.7.2.

(1.2) Show the 1st homogenous solution is $\displaystyle y_1(x)=x^\lambda$.

(1.3) Find $\displaystyle c(x)$ such that $\displaystyle y(x)=c(x)y_1(x)$.

(1.4) Find the 2nd homogenous solution $\displaystyle y_2(x)$.

(2) Repeat steps 1.1 thru 1.4 for Equation 5.7.3.

Solution

Solved on our own

Part 1

Section 1

Solve for the coefficients using the trial solution method. We shall assume the solution is in the form $\displaystyle y=e^r$ where $\displaystyle r$ is a constant. The following derivatives are calculated:

$\displaystyle y'=rx^{r-1}$

$\displaystyle y''=r(r-1)x^{r-2}$

Substituting the above derivatives into Equation 5.7.2 yields the following:

$\displaystyle a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0$

$\displaystyle a_2r^2x^r-a_2rx^r+a_1rx^r+a_0x^r=0$

$\displaystyle a_2r^2+(a_1-a_2)r+a_0=0$

$\displaystyle$

Now we can compare the above result to the expanded characteristic equation:

$\displaystyle r^2-2\lambda r+\lambda^2=0$

A simple comparison of coefficients implies the following:

$\displaystyle a_2=1$ $\displaystyle a_1=1-2\lambda$ $\displaystyle a_0=\lambda^2$

Section 2

We assumed a form of $\displaystyle y=x^r$ and the characteristic equation is $\displaystyle (r-\lambda)^2=0$, thus $\displaystyle r$ must equal $\displaystyle \lambda$ to satisfy the characteristic equation, and thus the first homogenous solution is

$\displaystyle y_1(x)=x^\lambda$

Section 3

The complete solution will be the form such that

$\displaystyle y(x)=c(x)y_1(x)=c(x)x^\lambda$

This $\displaystyle y(x)$ will be inserted into Equation 5.7.2, so lets calculate derivatives.

$\displaystyle y'(x)=c'y_1+cy_1'$

$\displaystyle y''(x)=c''y_1+c'y_1'+c'y_1'+cy_1''=c''y_1+2c'y_1'+cy_1''$

Entering these derivatives into Equation 5.7.2 yields:

$a_2x^2\left[c''y_1+2c'y_1'+cy_1''\right]+a_1x\left[c'y_1+cy_1'\right]+a_0\left[cy_1\right]=0$

Now group by $\displaystyle c(x)$:

$\left[a_0x^2y_1\right]c''+\left[2a_2x^2y_1'+a_1xy_1\right]c'+\left[a_2x^2y_1''+a_1xy_1'+a_0y_1\right]c=0$

Note the last set of terms will be zero by Equation 5.7.2. Then enter in the derivatives and constants and simplify.

$c''(x)\left[x^{\lambda+2}\right]+c'(x)\left[2\lambda x^{\lambda+1}+\left(1-2\lambda\right)x^{\lambda+1}\right]=0$

$\displaystyle c''(x)x^{\lambda+2}+c'(x)x^{\lambda+1}=0$

$\displaystyle \left[c''(x)x+c'(x)\right]x^{\lambda+1}=0$

$\displaystyle c''(x)x+c'(x)=0$

This is a fairly straighforward ODE:

$\displaystyle c''x+c'=(c'x)'=0$

$\displaystyle c'x=k_1$

$\displaystyle c'=\frac{k_1}{x}$

$\displaystyle c(x)=k_1\ln(x)+k_2$

Therefore it is clear that

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

Section 4

We have found the complete solution $\displaystyle y(x)$ in the previous section. We know this complete solution is equivalent to the linear combination of two homogeneous solutions:

$\displaystyle y(x)=C_1y_1(x)+C_2y_2(x)$

Thus it is a matter of comparing the two equations to extract the 2nd homogeneous solution $\displaystyle y_2(x)$. From Section 3 it is true that

$\displaystyle y(x)=k_1\ln(x)x^\lambda+k_2x^\lambda$

Knowing that $\displaystyle y_1(x)=x^\lambda$ it can be deduced that

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

Part 2

Section 1

The coefficients of 5.7.3 can be found by using the trial solution method. The given trial solution for the Euler equation with constant coefficient is:

$\begin{align} y={{e}^{xr}} \end{align}$

The derivatives are calculated:

$\begin{align} y'=r{{e}^{rx}} \end{align}$

$\begin{align} y''={{r}^{2}}{{e}^{rx}} \end{align}$

Substituting the above derivatives into Equation 5.7.3 we get:

$\begin{align} {{b}_{2}}{{r}^{2}}{{e}^{rx}}+{{b}_{1}}r{{e}^{rx}}+{{b}_{0}}{{e}^{xr}}=0 \end{align}$

$\begin{align} \left[ {{b}_{2}}{{r}^{2}}+{{b}_{1}}r+{{b}_{0}} \right]{{e}^{xr}}=0 \end{align}$

Now we can cancel out ${e}^{xr}$

$\begin{align} {{b}_{2}}{{r}^{2}}+{{b}_{1}}r+{{b}_{0}}=0 \end{align}$

Comparing the above equation to the characteristic equation given in 5.7.3 we have:

$\begin{align} {{b}_{2}}{{r}^{2}}+{{b}_{1}}r+{{b}_{0}}={{r}^{2}}-2r\lambda +{{\lambda }^{2}} \end{align}$

By comparing the coefficients of both sides of the above equation we get:

$\begin{align} {{b}_{2}}=1 \end{align}$ $\begin{align} {{b}_{1}}=-2\lambda \end{align}$ $\begin{align} {{b}_{0}}={{\lambda }^{2}} \end{align}$

Section 2

From the Part 2 Section 1, the characteristic equation of 5.7.3 is 5.7.1. There are two identical roots of equation 5.7.1 equal to $\begin{align}r=\lambda \end{align}$. The 1st homogeneous solution can be calculated by:

$\begin{align} {{y}_{1}}={{e}^{x{{r}_{1}}}} \end{align}$

$\begin{align} {{y}_{1}}={{e}^{x\lambda }} \end{align}$

Section 3

The complete solution will be of the form such that:

$\begin{align} {{y}_{2}}\left( x \right)=c\left( x \right)\cdot {{y}_{1}}\left( x \right) \end{align}$

This $\displaystyle y(x)$ will be inserted into Equation 5.7.3, so lets calculate derivatives.

$\begin{align} {{y}_{2}}'\left( x \right)=c'\left( x \right)\cdot {{y}_{1}}\left( x \right)+c\left( x \right)\cdot {{y}_{1}}'\left( x \right) \end{align}$

$\begin{align} {{y}_{2}}''\left( x \right)=c''\left( x \right)\cdot {{y}_{1}}\left( x \right)+c'\left( x \right)\cdot {{y}_{1}}'\left( x \right)+c\left( x \right)\cdot {{y}_{1}}''\left( x \right) \end{align}$

Entering these derivatives into Equation 5.7.3 yields:

$\begin{align} {{b}_{2}}\left[ c''\left( x \right)\cdot {{y}_{1}}\left( x \right)+2c'\left( x \right)\cdot {{y}_{1}}'\left( x \right)+c\left( x \right)\cdot {{y}_{1}}''\left( x \right) \right]+{{b}_{1}}\left[ c'\left( x \right)\cdot {{y}_{1}}\left( x \right)+c\left( x \right)\cdot {{y}_{1}}'\left( x \right) \right]+{{b}_{0}}c\left( x \right)\cdot {{y}_{1}}\left( x \right)=0 \end{align}$

Now group by $c(x)$ terms:

$\begin{align} c\left( x \right)\left[ {{{b}_{2}}{{y}_{1}}''\left( x \right)+{{b}_{1}}{{y}_{1}}'\left( x \right)+{{b}_{0}}{{y}_{1}}\left( x \right)}\right]+c'\left( x \right)\left[ 2{{b}_{2}}{{y}_{1}}'\left( x \right)+{{b}_{1}}{{y}_{1}}\left( x \right) \right]+c''\left( x \right)\cdot {{b}_{2}}{{y}_{1}}\left( x \right)=0 \end{align}$

Note that the first term with $c(x)$ will be zero by equation 5.7.3, then enter the derivatives and constants and simplify:

$\begin{align} c'\left( x \right)\left[ 2{{b}_{2}}{{y}_{1}}'\left( x \right)+{{b}_{1}}{{y}_{1}}\left( x \right) \right]+c''\left( x \right)\cdot {{b}_{2}}{{y}_{1}}\left( x \right)=0 \end{align}$

$\begin{align} U''\left( x \right)\cdot {{e}^{\lambda x}}=0 \end{align}$

In order for this to be zero the following must be true:

$\begin{align} c''\left( x \right)=0 \end{align}$

Integrating both sides of the above we have:

$\begin{align} c'\left( x \right)={{k}_{1}} \end{align}$

Integrating the above we have c(x) as:

$\begin{align} c\left( x \right)={{k}_{1}}x+{{k}_{2}} \end{align}$

Finally, using the above equation and the 1st homogeneous solution from part 2 section 2 substituted into the form of the 2nd homogeneous solution stated above yields the 2nd homogeneous solution:

$\begin{align} {{y}_{2}}\left( x \right)=\left( {{k}_{1}}x+{{k}_{2}} \right){{e}^{\lambda x}} \end{align}$

Section 4

The generalized homogeneous solution takes the following form:

$\begin{align} y\left( x \right)={{C}_{1}}{{y}_{1}}\left( x \right)+{{C}_{2}}{{y}_{2}}\left( x \right) \end{align}$

Substituting the 1st and 2nd homogeneous solution into the general form above yields:

$\begin{align} y\left( x \right)={{C}_{1}}{{e}^{\lambda x}}+{{C}_{2}}\left( {{k}_{1}}x+{{k}_{2}} \right){{e}^{\lambda x}} \end{align}$

$\begin{align} y\left( x \right)={{C}_{1}}{{e}^{\lambda x}}+{{C}_{2}}{{k}_{1}}x{{e}^{\lambda x}}+{{C}_{2}}{{k}_{2}}{{e}^{\lambda x}} \end{align}$

$\begin{align} y\left( x \right)={\left( {{C}_{1}}+{{C}_{2}}{{k}_{2}} \right)}{{e}^{\lambda x}}+{{{C}_{2}}{{k}_{1}}}x{{e}^{\lambda x}} \end{align}$

with

$\begin{align} C_3 = {{C}_{1}}+{{C}_{2}}{{k}_{2}} \end{align}$

$\begin{align} C_4 = {{C}_{2}}{{k}_{1}} \end{align}$

Substituting these constants yields the generalized homogeneous solution:

$\begin{align} y\left( x \right)={{C}_{3}}{{e}^{\lambda x}}+{{C}_{4}}x{{e}^{\lambda x}} \end{align}$

R*5.8

Question from meeting 32, page 32-2:

Given: recall p. 12-2, R*2.17

$y' + a_0(x)y = 0$

$\displaystyle (Equation\;5.8.1)$

Find: use the same idea of variation of constants (parameters) to find the particular solution $Y_P(x)$ after knowing the homogeneous solution $Y_H(x)$, i.e., let y(x) = A(x)$y_H(x)$, with A(x) being the unknown to be found.

From lecture notes Mtg 32

Solution

Solved on our own

The key is to start with equation (3) on p. 11-3 below.

$\underbrace{\color{blue}{1}}_{\color{blue}{a_1(x)}} \cdot \underbrace{y'}_{\color{blue}{y^{(1)}}} + \underbrace{\frac{Q(x)}{P(x)}}_{\color{blue}{a_0(x)}}y^{\color{blue}{(0)}}=\underbrace{\frac{R(x)}{P(x)}}_{\color{blue}{b(x)}}$

$\displaystyle (Equation\;5.8.2)$

The equation is an L1-ODE-VC. Note that the non-homogeneous equation above utilizes the following integration factor for the solution of y. But we must find the homogenous counterpart to this solution

$y(x) = \frac{1}{h(x)} \left[\int^x h(s)b(s)ds + k_2 \right]$

$\displaystyle (Equation\;5.8.3)$

The homogeneous solution of the L1-ODE-VC was derived in R2.17 by team 1. The solution summary is repeated for clarity. Note that team 1 has already submitted this solution for R2.17 back in Report 2. The steps are from a simple cross-reference used to define the homogeneous solution of 5.8.2

$\displaystyle \frac{y'}{y} = -{a}_{0}$

$\displaystyle (Eq 2.17.1)$

Integration of both sides of this equation gives

$\ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C$

$\displaystyle (Eq 2.17.2)$

Solving for $y$ gives

$y_H= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]$

Now, the homogenous solution is known from 2.17. The next step is to find the complete solution for the homogenous equation- $y(x) = A(x)y_H(x)$. The technique demonstrated in the lecture is to substitute the chosen form of y(x) into the original Euler equation to find the complete solution below. This is a classic application of the variation of parameters: (1) find homogenous solution, (2) find complete solution of Euler form - the L1-ODE-VC, and (3) find the particular solution by using (1) and (2).

$y' + a_0(x)y = b(x) = 0$

$\displaystyle (Equation\;5.8.4)$

The derivative simplifies the Euler equation. First, it must be derived.

$y = A(x)y_H$

$y' = A_xy_H + Ay_{Hx}$

The values can be substituted into the L1-ODE-VC to give the form below.

$A_xy_H + Ay_{Hx} + a_0(x)Ay_H = 0$

The equation can be rerranged to give a simple form for integration by natural logarithms and subsequent exponentiation for the complete solution of the L1-ODE-VC.

$A_xy_H = -A(y_{Hx} + a_0(x)y_H$

$\frac{A_x}{A} = -\frac{y_{Hx} + a_0(x)y_H}{y_H} = -\frac{y_{Hx}}{y_H} + a_0(x)$

The rearranged equation can be integrated by standard application of natural logarithms.

$\int^x \frac{A_x}{A} = -\int^x \frac{y_{Hx}}{y_H} + \int^x a_0(s)ds$

Use the identity for logarithms: $\ln \frac{a}{b} = \ln{a} - \ln{b}$, which gives $\ln \frac{A(x)}{y_H} = \int^x a_0(s)ds$.

The equation then gives the next output form.

$A(x) = y_H{\exp \int^x a_0(s)ds } = {y_H}^2$

$\displaystyle (Equation\;5.8.5)$

The final complete solution is y(x) = A(x)$y_H$, which gives

$y(x) = {y_H}^3$

$\displaystyle (Equation\;5.8.6)$

The particular solution is simply found by substration the homogenous solution from the complete solution.

$y_p = y(x) - y_H = {y_H}^3 - y_H$

$\displaystyle (Equation\;5.8.7)$

Team Work Distribution

Problem Assignments Problem # Assigned To R*5.1 Fenner Colson R*5.2 Jing Pan Manuel Steele Fenner Colson R*5.3 Zexi Zheng Yi Zhao R*5.4 Ben Neri Jing Pan R*5.5 Ben Neri Yi Zhao R5.6 Zexi Zheng Manuel Steele R*5.7 Fenner Colson Ben Neri R*5.8 Yi Zhao Manuel Steele

Table of Contributions Name Problems Solved Problems Checked Signature Fenner Colson R*5.1,R*5.2,R*5.7 R*5.4 Egm6321.f11.team1.colsonfe 19:14, 2 November 2011 (UTC) Yi Zhao R*5.5 R*5.3 Yi Zhao Manuel Steele R*5.2, R5.6, R*5.8 R*5.1, R*5.3, R*5.7 Egm6321.f11.team1.steele.m 09:19, 2 November 2011 (UTC) Jing Pan R*5.2, R*5.4 R5.6 Egm6321.f11.team1.pan 19:41, 2 November 2011 (UTC) Benjamin Neri R*5.5, R*5.7 R*5.5, R*5.4, R*5.7 Egm6321.f11.team1.Benneri Zexi Zheng R*5.3, R5.6 R*5.1, R*5.2, R*5.5 Egm6321.f11.team1.zheng.zx 19:13, 2 November 2011 (UTC)