User:Egm6321.f11.team3/Hwk5

Find
Find the following equation

Solution
We solved this problem on our own.

We know that Thus, put (5.1.2), (5.1.7) and (5.1.9) into (5.1.10) We can get

Find
(1)Using two methods to verify the exactness of the following L2-ODE-VC. Legendre: Bessel: Hermite: (2)If Hermite differential equation is not exact, check whether it is in power form, and see whether if it can be made exact using IFM with $$\displaystyle h(x,y)=x^{m}y^{n}$$. (3)The first few Hermite polynomials are Verify the (5.2.11), (5.2.12) and (5.2.13) are homogeneous solutions of the Hermite differential equation.

Solution
We solved this problem on our own.

Part 1.1.a
According to Legendre differential equation (5.2.8) Thus, put equation (5.2.15) and (5.2.16) into equation (5.2.6) and (5.2.7) we can get Thus, when only n = 0 or n = -1, Legendre differential equation will be exact.

Part 1.1.b
According R*5.1 work, we know that Put equation (5.2.15), (5.2.16), (5.2.21) and (5.2.22) into equation (5.2.2), (5.2.19) and (5.2.20)

Thus equation (5.2.5) becomes Thus, when only n = 0 or n = -1, Legendre differential equation will be exact.

Part 1.2.a
According to Hermite differential equation (5.2.10) Thus, put equation (5.2.25) and (5.2.26) into equation (5.2.6) and (5.2.7) we can get Thus, Hermite differential equation is exact only when n = -1.

Part 1.2.b
Put equation (5.2.25), (5.2.26), (5.2.29) and (5.2.30) into equation (5.2.2), (5.2.19) and (5.2.20) we can get Thus equation (5.2.5) becomes Thus, Hermite differential equation is exact only when n = -1.

Part 2
The power form for L2-ODE-VC is : We can choose Thus, Hermite differential equation is in power form In order to make Hermite differential equation be exact, multiply equation (5.2.10) by Put equation (5.2.37) and (5.2.38) into (5.2.7) Put equation (5.2.37) ,(5.2.38) and (5.2.39) into (5.2.6) Because b = 0, a cannot be zero so  Thus, Hermite differential equation becomes Thus after multiplying x, Hermite differential equation is exact.

Part3
The polynomial solution for Hermite differential equation is  Put equation (5.2.44), (5.2.45) and (5.2.46) into Hermite differential equation. Choose the initial condition as  Thus Choose the initial condition as  Thus, Choose the initial condition as  Thus,

Find
Find the expressions for X(x) in terms of cosKx,sinKx,coshKx,sinhKx.

Solution
We solved this problem on our own.

Recall Euler's formula:

Since X(x)is a real-valued function so the left hand side must also be real. Let:

The 4 real constants are c1,c2,b3,b4.

2.Another way:

We can use the same method above,so we can get:

R*5.4 - Finding the 5th derivatives of y with respect to t
From the lecture slide Mtg 31-1

Given
Transformation of variables is the first stage of the method 1 to solve Euler Ln-ODE-VC.

Euler Ln-ODE-VC:

$$\begin{align} x=\text{e}^{t}\end{align}$$

We need to transform x to new variable t.

Chain Rule:


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$$\begin{align} \underbrace{\frac{d y}{dt}}_{\displaystyle y_t}= {} & \underbrace{\frac{dy}{dx}}_{\displaystyle y_{x} }\underbrace{\frac{dx}{dt}}_{\displaystyle \text{e}^{t}} \\ \end{align}$$ (5.4.1)
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$$\begin{align} y_t = y_x e^{t} \end{align}$$


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and we obtain
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$$\begin{align} y_x = y_t e^{-t}\end{align}$$

(5.4.2)
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Find
Find the fifth derivative $$\begin{align}y(x(t))_{xxxxx}\end{align}$$ in terms of derivative of y with respect to t

Solution
Solved on our own


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$$\displaystyle y_{x}= \frac{dy}{dx} = \left(\frac{dt}{dx} \frac{d}{dt}\right)y=y_t e^{-t}$$ (5.4.3)
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$$\begin{align} y_{xx}=\frac{d}{dx}\frac{d}{dx}y = \frac{dt}{dx}\frac{d}{dt}\left(y_t e^{-t}\right) \end{align}$$ (5.4.4)
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$$\displaystyle y_{xx}= e^{-t}\left(-e^{-t}y_t+e^{-t}y_{tt}\right)=e^{-2t}\left(y_{tt}-y_t \right) $$ (5.4.5)
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$$\begin{align} y_{xxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-2t}\left(y_{tt}-y_t \right)\right) \end{align}$$ (5.4.6)
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$$\begin{align} y_{xxx}= e^{-t}\left(-2e^{-2t}\left(y_{tt}-y_t\right)+e^{-2t}\left(y_{ttt}-y_{tt}  \right) \right)=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t \right) \end{align}$$ (5.4.7)
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$$\begin{align} y_{xxxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-3t}(y_{ttt}-3y_{tt}+2y_t) \right)\end{align}$$ (5.4.8)
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$$\begin{align} y_{xxxx}= e^{-t}\left(-3e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt})\right)=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t} \right) \end{align}$$ (5.4.9)
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$$\begin{align} y_{xxxxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}) \right)\end{align}$$ (5.4.10)
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$$\begin{align} y_{xxxxx}= e^{-t}\left(-4e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}+e^{-4t}(y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt})\right)

\end{align}$$ (5.4.11)
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Find
Solve 5.5.1 using Method 2 with a trail solution of 5.5.2 and boundary conditions: $$\displaystyle y(1)=-4, y(2)=7 $$ Plot the solution.

Solution
We solved this on our own. First we find the first and second derivative of the trial solution 5.5.2, And now we substitute 5.5.2, 5.5.3 and 5.5.4 into 5.5.1, $$\displaystyle x^2(r(r-1)x^{r-2})-2x(rx^{r-1})+2(x^r)=0 $$ And simplify, $$\displaystyle (r^2-r)x^r-2rx^r+2x^r=0 $$ $$\displaystyle r^2-3r+2=0 $$ $$\displaystyle r_1=1, r_2=2 $$ Therefore, the following form applies, $$\displaystyle y(x)=c_1x+c_2x^2 $$ Now we apply the given boundary conditions to solve the constants, $$\displaystyle y(1)=c_1(1)+c_2(1^2)=-4 $$ $$\displaystyle y(1)=c_1(2)+c_2(2^2)=7 $$ The equations are now solved simulataneously to detemine the constant values, $$\displaystyle c_1=-11.5 $$ $$\displaystyle c_2=7.5 $$ And the equation to plot becomes, $$\displaystyle y(x)=7.5x^2-11.5x $$ And the equation is plotted in Matlab over an arbitrary interval,

Find
Show the trial solution in Method 1 and Method 2 are equal

Solution
We solved this problem on our own.

Method 2: Characteristic equation:

So

Method 1

Let: substituting in 5.6.1

Let:

So So the two methods are the same.

Find
1.1 Find $$a_2, a_1, a_0$$ such that 5.7.1 is the characteristic equation of 5.7.2 1.2 1st homogeneous solution: $$y_1(x)=x^\lambda$$ 1.3 Complete solution: Find $$c(x)$$ such that $$y(x)=c(x)y_1(x)$$ 1.4 Find the 2nd homogeneous solution $$y_2(x)$$ 2 Repeat steps 1.1 through 1.4 for 5.7.3

Solution
We solved this on our own.

Part1.1
First we find the first and second derivative of the trial solution, Substituting these into 5.7.2 yields, $$\displaystyle a_2x^2(r(r-1)x^{r-2})+a_1x(rx^{r-1})+a_0x^r=0$$ $$\displaystyle a_2(r^2-r)x^r+a_1rx^r+a_0x^r=0 $$ $$\displaystyle a_2r^2x^r+(a_1-a_2)rx^r+a_0x^r=0 $$ $$\displaystyle a_2r^2+(a_1-a_2)r+a_0=0 $$ And using 5.7.1, $$\displaystyle a_2r^2+(a_1-a_2)r+a_0=(r-\lambda)^2=r^2-2\lambda r+\lambda^2 $$ The coefficients are matched, resulting in, $$\displaystyle a_2=1$$ $$\displaystyle (a_1-a_2)=2\lambda$$ $$\displaystyle a_0=\lambda^2 $$ Using $$\displaystyle\lambda=7$$, $$\displaystyle a_2=1$$ $$\displaystyle a_1=1-2\lambda=15$$ $$\displaystyle a_0=49 $$

Part1.2
The 1st homogeneous solution is found by evaluating 5.7.1 and seeing that $$\displaystyle r=\lambda$$. By using the trial solution of 5.7.4, we see that $$\displaystyle y_1(x)=x^\lambda$$.

Part1.3
$$\displaystyle y(x)=c(x)y_1(x)$$ Now we find the first and second derivative of the complete solution using the chain rule, $$\displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x)=c'x^\lambda + c\lambda x^{\lambda-1} $$ $$\displaystyle y(x)=c(x)y_1(x) +c'(x)y_1'(x)+c'(x)y_1'(x)+c''(x)y_1(x) $$ $$\displaystyle y(x)=c(x)y_1(x) +2c'(x)y_1'(x)+c(x)y_1(x)=cx^\lambda + 2c'\lambda x^{\lambda-1} + c(\lambda^2-\lambda)x^{\lambda-2} $$ And now subsituting these into 5.7.2 yields, $$\displaystyle a_2x^2(cy_1+2c'y_1'+cy_1)+a_1x(cy_1'+c'y_1)+a_0y_1=0$$ Distributed to, $$\displaystyle a_2x^2y_1c+a_2x^2cy_1+2a_2x^2c'y_1'+a_1xcy_1'+a_1xc'y_1+a_0cy_1=0$$ And grouped, $$\displaystyle (a_2x^2y_1)c+(2a_2x^2y_1'+a_1xy_1)c'+(a_2x^2y_1+a_1xy_1'+a_0y_1)c=0$$ The last term is equivalent to 5.7.2 and equals zero. Now we apply the first homoegeneous solution and the coefficients, $$\displaystyle (x^2x^\lambda)c''+(2x^2 \lambda x^{\lambda-1}+x(1-2\lambda)x^\lambda)c'=0$$ $$\displaystyle (x^{\lambda+2})c''+x^{\lambda+1}(2\lambda+(1-2\lambda))c'=0$$ $$\displaystyle xc''+c'=0$$ Integrating the equation yields, $$\displaystyle c(x)=k_1 ln(x)+k_2$$ And subsituted back into y(x) gives us,

Part1.4
The 2nd homogenous equation is found by inspected the complete solution. Since we know the the first homogeneous equation is, $$\displaystyle y_1=x^\lambda $$ We can see the 2nd equation must be the other term in the complete solution, $$\displaystyle y_2=x^\lambda ln(x) $$

Part2.1
First we find the first and second derivative of the trial solution with a constant coefficient, $$\displaystyle y=e^{rx}$$ $$\displaystyle y'=re^{rx}$$ $$\displaystyle y''=r^2e^{rx}$$ Substituting these into 5.7.3 yields, $$\displaystyle b_2(r^2e^{rx})+b_1(re^{rx})+b_0e^{rx}=0$$ $$\displaystyle b_2r^2+b_1r+b_0=0$$

And using 5.7.1, $$\displaystyle b_2r^2+b_1r+b_0=(r-\lambda)^2=r^2-2\lambda r+\lambda^2 $$ The coefficients are matched, resulting in, $$\displaystyle b_2=1$$ $$\displaystyle b_1=-2\lambda$$ $$\displaystyle b_0=\lambda^2 $$ Using $$\displaystyle\lambda=7$$, $$\displaystyle b_2=1$$ $$\displaystyle b_1=-14$$ $$\displaystyle b_0=49 $$

Part2.2
The 1st homogeneous solution is found by evaluating 5.7.1 and seeing that $$\displaystyle r=\lambda$$. By using the trial solution, we see that $$\displaystyle y_1(x)=e^{\lambda x}$$.

Part2.3
$$\displaystyle y(x)=c(x)y_1(x)$$ Now we find the first and second derivative of the complete solution using the chain rule, $$\displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x) $$ $$\displaystyle y(x)=c(x)y_1(x) +2c'(x)y_1'(x)+c''(x)y_1(x)$$ And now subsituting these into 5.7.3 yields, $$\displaystyle b_2(cy_1+2c'y_1'+cy_1)+b_1(cy_1'+c'y_1)+b_0y_1=0$$ Distributed to, $$\displaystyle b_2y_1c+b_2cy_1+2b_2c'y_1'+b_1cy_1'+b_1c'y_1+b_0cy_1=0$$ And grouped, $$\displaystyle (b_2y_1)c+(2b_2y_1'+b_1y_1)c'+(b_2y_1+b_1y_1'+b_0y_1)c=0$$ The last term is equivalent to 5.7.3 and equals zero. Now we apply the first homoegeneous solution and the coefficients, $$\displaystyle e^{\lambda x}c''+(2\lambda e^{\lambda x}-2\lambda e^{\lambda x})c'=0$$ $$\displaystyle e^{\lambda x}c''=0$$ Integrating the equation yields, $$\displaystyle c(x)=k_1x+k_2$$ And subsituted back into y(x) gives us,

Part2.4
The 2nd homogenous equation is found by inspected the complete solution. Since we know the the first homogeneous equation is, $$\displaystyle y_1=e^{\lambda x} $$ We can see the 2nd equation must be the other term in the complete solution, $$\displaystyle y_2=xe^{\lambda x} $$

Given
$$\begin{align} & y'+P(x)y=Q(x) \\ & {{y}_{h}}=exp [-\int{P(x)dx}] \\ \end{align}$$  (5.8.1)

Find
The particular solution using Variation of Parameters method.

Solution
I referred to http://en.wikiversity.org/wiki/User:EGM6321.f10.team6.cook/hw5 for the solution of this problem.

Then

Substitution gives

Therefore, the particular solution is

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