User:Egm6321.f10.team2.oztekin/hw2

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)
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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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$$ \displaystyle y'(x) = r e^{rx} $$     (6.5.3)
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$$ \displaystyle y''(x) = r^2 e^{rx} $$    (6.5.4)
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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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$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$     (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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$$\begin{align} u_2(x)=e^{2x}\int\left(\frac{1}{e^{2x}}(x+1)\right)\,dx \end{align}$$ (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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