User:Egm6321.f11.team3/Hwk6

Given
A nonhomogeneous L2-ODE-CC :

$$\displaystyle a_2y''+a_1y'+a_0y=f(t)$$ $$ (6.1.1) $$

Find
1) Find the PDE's that govern the integrating factor h(x,y)for equation 6.1.1. Is it possible to solve these PDE's for h(x,y)?

2) Trial solution for the integrating factor $$\displaystyle h(t)=e^{\alpha t}$$ where $$\displaystyle \alpha$$ is unknown to be determined.
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$$ Because of the integrating factor in exponential form, assume the LHS of equation (6.1.2)to take the form :
 * $$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= \int \, e^{\alpha t}f(t)dt$$
 * $$ (6.1.2)
 * $$ (6.1.2)
 * }
 * }
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$$
 * $$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= e^{\alpha t}[\bar a_1y'+\bar a_0y]$$
 * $$(6.1.3)
 * $$(6.1.3)
 * }
 * }

2.1) Find $$ (\bar a_1,\bar a_0)$$ in terms of $$ (a_0,\ a_1,\ a_2) $$.

2.2) Find the quadratic equation for $$\displaystyle \alpha $$.

2.3) Reduced-order equation: (6.1.3) and (6.1.4) lead to
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$$
 * $$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$$
 * $$ (6.1.4)
 * $$ (6.1.4)
 * }
 * }


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

which is easily solvable by IFM.

2.4) Use the IFM to solve equation 6.1.5.

2.5) Show that

$$\displaystyle  \alpha \beta = \frac {a_0}{a_2}$$.

$$\displaystyle  \alpha+\beta=\frac{a_1}{a_2}$$

2.6) Deduce the particular solution yP(t) for general excitation f(t)

2.7) Verify result with table of particular solutions for

$$\displaystyle f(t) = t exp(bt) $$

2.8) Solve equation 6.1.1 with $$\displaystyle f(t) = tanh t $$

For the coefficients (a0,a1,a2),consider 2 characteristic equations:

2.8.1) $$\displaystyle (r+1)(r-2) = 0$$

2.8.2) $$\displaystyle (r-4)^2 = 0 $$

2.9) For each case in questions 2.8.1 and 2.8.2, determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Solution
Solved the problem on our own. 1) Multiplying equation (6.1.1) by the integrating factor h(x,y),we get:

$$\displaystyle {h(x,y)a_2}y''+{h(x,y)a_1y'+h(x,y)a_0y-h(x,y)f(t)}=0 $$

The 2nd exactness condition for N2-ODEs:

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

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

To apply the 2nd exactness condition,we must calculate the following:

$$\displaystyle f_t=a_2h_t\,;\,f_{tt}=a_2h_{tt}\,;\,f_{ty}=a_2h_{ty}\,;\,f_{tp}=0$$

$$\displaystyle f_y=a_2h_y\,;\,f_{yy}=a_2h_{yy}\,;\,f_{yp}=0$$

$$\displaystyle g_t=a_1ph_t+a_0yh_t-f(t)\cdot h_t-h\cdot f'(t)$$

$$\displaystyle g_{tp}=a_1h_t$$

$$\displaystyle g_y=a_1ph_y+a_0yh_y+a_0h-f(t)h_y$$

$$\displaystyle g_{yp}=a_1h_y\,;\,g_p=a_1h\,;\,g_{pp}=0$$

Substituting these values in 2nd exactness condition:

$$\displaystyle a_2h_{tt}+2p\cdot a_2 h_{ty}+p^2\cdot a_2h_{yy}=a_1h_t-a_0yh_y-a_0h+f(t)h_y$$

$$\displaystyle 2\cdot a_2h_y=0$$

$$\displaystyle h_y=0$$

Hence,

$$\displaystyle a_2h_{tt}-a_1h_t+a_0h=0$$

Thus the PDE is reduced to a ODE which can be solved.

2.1) The trial solution is $$\displaystyle h(t)=\exp(\alpha t) $$. So, $$\displaystyle \int \exp(\alpha t)[a_2y+a_1y'+a_0y]dt=\int \exp(\alpha t) F(t)dt $$. Assume $$\displaystyle \int \exp(\alpha t)[a_2y+a_1y'+a_0y]dt=\exp(\alpha t)[\cancel{\bar a_2}y+\bar a_1y'+\bar a_0y] $$. Differentiating the RHS of the above equation, $$\displaystyle \int \exp(\alpha t)[a_2y+a_1y'+a_0y]dt=\int \exp(\alpha t)[\bar a_1y''+(\bar a_1\alpha+\bar a_0)y'+(\bar a_0\alpha)y]dt $$. Comparing the integrands, $$\displaystyle \bar a_1= a_2 $$. $$\displaystyle \bar a_0=a_1-\alpha a_2 $$. $$\displaystyle \bar a_0=\frac{a_0}{\alpha} $$.

2.2) Using the equations found in the 2.1,we can form a quadratic equation for $$\displaystyle \alpha $$

$$\displaystyle a_1-\alpha a_2=\frac{a_0}{\alpha} $$. $$\displaystyle \alpha a_1-\alpha^2 a_2=a_0 $$ $$\displaystyle a_2 \alpha^2 - a_1 \alpha + a_0 = 0 $$

2.3) The reduced order equation:

$$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$$ can be rewritten as

$$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\,\int e^{\alpha t}f(t)dt$$

2.4) Using IFM to solve the reduced-order equation:

The integrating factor is:

$$\displaystyle \bar h(t)=exp[\int\,\frac{\bar a_0}{\bar a_1} dt] =exp[\int\,\beta dt]$$

$$\displaystyle y(t)=\frac{1}{\bar h(t)}\int^t \bar h(s)B(s)ds $$ To obtain B(s), $$\displaystyle y'+\beta y = \frac{\exp(-\alpha t)}{\bar a_1} \int \exp(\alpha t)F(t)dt = B(s) $$ Therefore,

$$\displaystyle y(t)=\frac{1}{[\exp(\beta t)]}\int^t \frac{[\exp((\beta-\alpha) s)]}{\bar a_1} \left[ \int \exp(\alpha t)F(t)dt \right] ds$$

2.5) Substitutng $$\displaystyle \beta = \frac {\bar a_0}{\bar a_1}; a_0 = \alpha \bar a_0; \bar a_1 = a_0 $$ in $$\displaystyle \alpha,\beta $$

We get, $$\displaystyle \alpha\beta = \frac{a_0}{a_2} $$

$$\displaystyle \alpha+\beta = \frac{a_1}{a_2} $$

Given
and

Find
Show that the two expressions are equal.

Solution
(6.2.3) can be writen in follows,

According to (6.2.1) and the hint given by teacher

Differentiation rules shows that

So, we can get Therefore (6.2.5) becomes, Integration by parts for (6.2.2)

Find (6.2.9) and (6.2.10) are equivalent.

Given
1st homogeneous solution, pretend not knowing: Trial solution: $$\displaystyle y=e^{rx}$$, r=constant The characteristic equation: $$\displaystyle r_1=1$$, r = constant $$\displaystyle r_2(x)=\frac {1}{x-1}$$, r = non-constant

Find
Explain why $$\displaystyle r_2(x)$$ is not a valid root, i.e. $$\displaystyle u_2(x)=e^{xr_2(x)} $$ is not a valid solution.

Solution
We solved this problem on our own. Using the trial solution, we see that, $$\displaystyle y=e^{rx}$$ $$\displaystyle y'=re^{rx}$$ $$\displaystyle y''=r^2e^{rx}$$ This leads to a characteristic equation of $$\displaystyle ((x-1)r^2-xr+1)e^{rx}=0$$ Since the second term represents the trial solution, only equation 6.3.3 remains. $$\displaystyle(x-1)r^2-xr+1=0$$ Since $$\displaystyle {r}_{1} $$ was applied to the 1st homogeneous solution, resulting in $$\displaystyle u_1(x)=e^x$$ Then $$\displaystyle {r}_{2} $$ must be applied to the 6.3.3 as follows, $$\displaystyle(x-1)(\frac{1}{x-1})^2-x\frac{1}{x-1}+1=0$$ $$\displaystyle \frac{1}{x-1}-\frac{x}{x-1}+1=0$$ $$\displaystyle 1-x+x-1=0$$ $$\displaystyle 0=0$$ Since the left side of the equation cancels, we are not able to solve for x and this makes it a non valid root and solution.

R*6.4 Solution of Legendre equation with n=1
We solved this problem on our own.

Find
For the L2-ODE-VC in 6.3.1, select a valid homogeneous solution, and call it $$\displaystyle u_1$$. Find the 2nd homogeneous solution $$\displaystyle u_2(x)$$ by variation of parameters, and compare to $$\displaystyle e^{xr_2(x)}$$

Solution
First we select a trial solution of $$\displaystyle y=e^{rx}$$ to find $$ \displaystyle u_1(x)$$, $$\displaystyle y=e^{rx}$$ $$\displaystyle y'=re^{rx}$$ $$\displaystyle y''=r^2e^{rx}$$ From this we select $$\displaystyle u_1$$, $$\displaystyle u_1(x)=e^x$$ From King section 1.1, we have $$\displaystyle u_2(x)=u_1(x)\int{\frac{1}{u_1^2(t)}} exp \left \{ \int{a_1(s)ds} \right \} dt$$ By determining the $$\displaystyle a_1$$ term from equation 6.3.1 we have, $$\displaystyle u_2(x)=e^x\int{\frac{1}{e^{2x}}} exp \left \{ \int{\frac{x}{x-1} dx} \right \} dx$$ By using integration by parts for the $$\displaystyle a_1$$ term, we have, $$\displaystyle u_2(x)=e^x\int{e^{-2x} exp \left \{x+log(x-1) \ \right \}} dx$$ $$\displaystyle u_2(x)=e^x\int{e^{-2x} (e^x)(e^{log(x-1)})} dx$$ $$\displaystyle u_2(x)=e^x\int{e^{-x}(x-1)} dx$$ $$\displaystyle u_2(x)=e^x\int({e^{-x}x-e^{-x}}) dx$$ $$\displaystyle u_2(x)=e^x(-xe^{-x}-e^{-x}+e^{-x}) dx$$ By cancellation of terms we see, $$\displaystyle u_2(x)=-x$$

R*6.5 - Finding two homogeneous solutions by using trial solution and variation of parameters
From the lecture slide Mtg 36-3

Given
L2-ODE-VC:


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$$\begin{align} \left( x+1 \right)y''-\left( 2x+3 \right)y'+2y=0 \end{align}$$ (6.5.1)
 * 
 * }

Find
Find two homogeneous solutions $$\begin{align} u_1(x), u_2(x) \end{align}$$ by using trial solution $$\begin{align} y=e^{rx} \end{align}$$

Solution
Solved on our own

Differentiating the trial solution as follows:


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$$ \displaystyle y(x) = e^{rx} $$     (6.5.2)
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 * }


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$$ \displaystyle y'(x) = r e^{rx} $$     (6.5.3)
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$$ \displaystyle y''(x) = r^2 e^{rx} $$    (6.5.4)
 * 
 * }

Substituting Equation (6.5.2), (6.5.3), and (6.5.4) into equation (6.5.1) gives


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$$\begin{align} (x+1)\left(r^{2}e^{rx}\right)-(2x+3)\left(re^{rx}\right)+2\left(e^{rx}\right)=0 \end{align}$$ (6.5.5) Taking common parentheses,
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 * }


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$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$     (6.5.6)
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 * }

We obtain,
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$$ \displaystyle (x+1)r^2-(2x+3)r+2=0 $$     (6.5.7) Finding the roots by using the quadratic equation formula
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$$\displaystyle r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$    (6.5.8) This gives  two roots:
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$$\displaystyle {r}_{1} = 2 $$    (6.5.9)
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and


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$$\begin{align} {r}_{2}= \frac{1}{ x+1}

\end{align} $$ (6.5.10)
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There is only one valid root because the root $$\displaystyle {r}_{2} $$ is not a constant.

Finding the first trial solution $$\displaystyle u_1(x)$$ by using $$\displaystyle y= {e}^{rx}$$ and $$\displaystyle {r}_{1} = 2$$


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$$\displaystyle {u}_{1}(x) = {e}^{2x} $$    (6.5.11)
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Rearranging the Eq. 6.5.1,
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$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$     (6.5.12)
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which includes


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$$\begin{align} a_1(x)=\frac{-(2x+3)}{x+1} \end{align}$$ (6.5.13)
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We can find 2nd homogenous solution by using the equation which is from Lecture slide (4) p. 34-5


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$$\begin{align} u_2(x)=u_1(x)\int\left(\frac{1}{u_{1}^{2}(x)}e^{-\int a_1(x)dx}\right)dx \end{align}$$ (6.5.14) Substituting Eq. (6.5.11) and (6.5.13) into above Eq. (6.5.14), we obtain
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$$\begin{align} u_2(x)=e^{2x}\int\left(\frac{1}{e^{4x}}e^{\int {\frac {2x+3}{x+1}\,dx}}\right)\,dx \end{align}$$ (6.5.15) Taking the inner integral gives,
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$$\begin{align} u_2(x)=e^{2x}\int\left(\frac{1}{e^{4x}}e^{ln(x+1)+2x}\right)\,dx \end{align}$$ (6.5.16) When it is simplified,
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 * }


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$$\begin{align} u_2(x)=e^{2x}\int\left(\frac{1}{e^{2x}}(x+1)\right)\,dx \end{align}$$ (6.5.17)
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Taking the integral above by using Wolfram Alpha


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$$\begin{align} \int\frac{1}{e^{2x}}(x+1)\,dx =\frac{-1}{4}e^{-2x}(2x+3)\end{align}$$ (6.5.18)
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If we put the Eq. (6.5.18) into Eq. (6.5.17), we can find $$\displaystyle u_2(x)$$,


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$$\begin{align} u_2(x)=\frac{-1}{4}(2x+3)\end{align}$$ (6.5.19)
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R*6.6 - Finding a homogenous L2-ODE-VC by using reverse engineering
From the lecture slide Mtg 36-10

Given
Trial solution:


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$$ y=\frac{e^{rx}}{sin x}, r=\text{constant}$$ (6.6.1)
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Characteristic equation:


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$$\displaystyle (r-r_1)[r-r_2(x)]=(r-2)\left[r-\frac {1}{x+1}\right]=r^2+r\left(-2-\frac{1}{x+1}\right)+1 \left(\frac{2}{x+1}\right)=0$$ (6.6.2) which are
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 * }

$$\begin{align} r_1=2 \end{align}$$

and

$$\begin{align} r_2(x)=\frac {1}{x+1} \end{align}$$

Find
Find a homogeneous L2-ODE-VC from the trial solution and characteristic equation given above by using reverse engineering

Solution
Solved on our own

Differentiating the trial solution as follows:


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$$ \displaystyle y=\frac{e^{rx}}{sin x} $$ (6.6.3)
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For the first differentiate with Wolfram Alpha:


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$$ \displaystyle y' =e^{rx}\csc x(r-\cot x) = \frac{e^{rx}}{sin x}(r-\cot x)$$ (6.6.4)
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For the second differentiate with Wolfram Alpha:


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$$ \displaystyle y'' =e^{rx}\csc x(r^2-2r \cot x+ \cot^{2}x+ \csc^{2}x)= \frac{e^{rx}}{sin x}(r^2-2r \cot x+ \cot^{2}x+ \csc^{2}x)$$ (6.6.5)
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 * }

A homogenous L2-ODE-VC is equal to the product of the trail solution by characteristic equation:


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$$ \displaystyle a_2 y''+{a_1}y'+{a_0}y=\frac{e^{rx}}{\sin x}(r-r_1)[r-r_2(x)] $$     (6.6.6)
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Substituting Equation (6.6.3), (6.6.4), and (6.6.5) into equation (6.6.6) gives


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$$ \displaystyle \frac{e^{rx}}{\sin x}\left[a_2(r^2-2r\cot x+\cot^2x+\csc^2x)+a_1(r-\cot x)+a_0\right]=\frac{e^{rx}}{\sin x}(r-2)\left[r-\frac {1}{x+1}\right]$$ (6.6.7)
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With expanding the inside of brackets and then taking common parentheses of $$\displaystyle {{r}^{2}}$$, $$\displaystyle r$$ and 1, we obtain


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

(6.6.8)
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So we get 3 equations for $$ \displaystyle a_2, a_1, a_0 $$ from the equality of both sides,

Solving for $$ \displaystyle a_2 $$


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$$ \displaystyle a_2=1 $$

(6.6.9)
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Solving for $$ \displaystyle a_1 $$


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$$ \displaystyle a_1-2\cot x = -2-\frac{1}{x+1} $$

(6.6.10)
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so we obtain


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$$ \displaystyle a_1=2\cot x -2-\frac{1}{x+1} $$

(6.6.11)
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Solving for $$ \displaystyle a_0 $$


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$$ \displaystyle a_0-2\cot^2 x+2\cot x+\frac{\cot x}{x+1}+\cot^2x+\csc^2x=\frac{2}{x+1} $$

(6.6.12)
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so we obtain


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$$ \displaystyle a_0=\cot^2 x-2\cot x-\frac{\cot x}{x+1}-\csc^2x+\frac{2}{x+1} $$    (6.6.13)
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If we substitute Eq. (6.6.9), Eq. (6.6.11) and Eq. (6.6.9) into a general form of homogenous L2-ODE-VC Eq. (6.6.6), we can have


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$$y''+\left(2\cot x-2-\frac{1}{x+1}\right)y'+ \left(\cot^2 x-2\cot x-\frac{\cot x}{x+1}-\csc^2x+\frac{2}{x+1}\right)y =0$$ (6.6.14)
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Multipling all equation with $$\displaystyle {(x+1)}$$ gives,


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$$y''+\left[2(x+1)(\cot x-1)-1\right]y'+ \left[(x+1)(\cot^2 x-2\cot x-\csc^2x)-\cot x+2\right]y =0$$ (6.6.15)
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Given
Legendre equation with n = 2

Find
Show:

Solution
We solved this problem on our own.

I checked from WolframAlpha that the first homogeneous solution should be Thus, I used (6.7.4) to do the following solution. Put (6.7.1) into standard form According to lecture note (4) p.34-5 According to Wolfram Alpha

Find
For equation(6.8.1) and (6.8.2), do  (a)  Find the 1st homogeneous solution by trial solution (see King 2003 p.28 Pb.1.1 ab) (b) Find the complete solution by variation of parameters with the following excitation:

Solution
We solved this problem on our own.

Part 1.a
Use trial solution Because $$\displaystyle r=\frac{1}{(x-1)}  $$ is not valid root. Thus, the 1st homogeneous solution is

Part 1.b
Use trial solution Put trial solution into equation (6.8.2). According Euler's formula Because we can change the constant for the homogeneous solution, equation(6.8.13) can be the homogeneous solution of equation(6.8.2)

Part 2.a
According to lecture note (4) p.34-5 and (1) p.34-6 The 2nd homogeneous solution: Particular solution: Put equation (6.8.1) into standard form

According to Wolfram Alpha Thus the complete solution for equation (6.8.1) is

Part 2.b
Put equation (6.8.2) into standard form. According to Wolfram Alpha Thus the complete solution of equation (6.8.2) is

Given
The 1st homogeneous solution

Find
Find the find solution y(x) by variation of parameters.

Solution
(6.9.1) can be written as:

= References =

= Contributing Members =

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