User:Egm6321.f10.team2.oztekin/HW5

=Problem 5.1: Finding fifth derivative of y =

From the lecture slide Mtg 26-2

5.1.1 Given
Solving Euler L2_ODE_VC has to methods of solution. For the first method we have 2 stages. In mtheod one the first stage is transformation of variables in which we change dependent variable. The second stage we try to find solution for dependent variable. In second method we directly try to find solution for dependent variable. But know we discuss first stage for the first method.


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$$  \displaystyle \sum_{i=0}^{n}a_{i}x^{i}y^{(i)}=0 $$     (Eq 1.1)
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This equation involves coefficients need to determined.After changing independent variable


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$$  \displaystyle y=y(x(t)) $$     (Eq 1.2)
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Let's take
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$$  \displaystyle x=e^{t} $$     (Eq 1.3)
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$$  \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} x} $$ (Eq 1.4)
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$$  \displaystyle y_{x}=y_{t}e^{-t} $$     (Eq 1.5)
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$$  \displaystyle y_{xxxx}=e^{-4t}\left [ y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t} \right ] $$     (Eq 1.6)
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5.1.2 Solution

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$$  \displaystyle y_{xxxxx}=\frac{\mathrm{d} }{\mathrm{d} x}y_{xxxx}=\frac{\mathrm{d} t}{\mathrm{d} x}\frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-4t}y_{tttt}-6e^{-4t}y_{ttt}+11e^{-4t}y_{tt}-6e^{-4t}y_{t} \right ) $$     (Eq 1.7)
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$$  \displaystyle y_{xxxxx}=e^{-t}\left [ e^{-4t}y_{ttttt}-10e^{-4t}y_{tttt}+35e^{-4t}y_{ttt}-50e^{-4t}y_{tt}+24e^{-4t}y_{t} \right ] $$     (Eq 1.8)
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$$  \displaystyle y_{xxxxx}=e^{-5t}\left [ y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_{t} \right ] $$     (Eq 1.9)
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Also note that the coefficients before those derivatives are typical in polynominals like $$\displaystyle a(a-1)(a-2)......$$, then we can have a general form of derivatives of $$\displaystyle y$$ as,


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$$ \displaystyle {{y}^{(n)}}={{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y$$ (Eq 1.10)
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where $$\displaystyle D$$ denotes the operator "$$\displaystyle \frac{d}{dt}$$" and $$\displaystyle n\in \mathbb{N}$$.

We can prove Eq 1.10 by mathematical induction.

When $$\displaystyle n=1$$, Eq 1.5 has prove this case.

Now let's assume Eq 1.10 is valid at some $$\displaystyle n\in \mathbb{N}$$, meaning,


 * $$\displaystyle {{y}^{(n)}}={{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y$$

then,


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$$ \displaystyle

\begin{align} {{y}^{(n+1)}} & =\frac{d}{dx}{{y}^{(n)}} \\ & =\frac{1}\frac{d}{dt}{{y}^{(n)}} \\ & ={{e}^{-t}}\frac{d}{dt}\left[ {{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}\left[ (-n){{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right. \\ & \left. {{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }^ – +{{e}^{-nt}}{{D}^{2}}(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}{{e}^{-nt}}({{D}^{2}}-nD)(D-1)(D-2)......(D-n+1)y \\ & ={{e}^{-(n+1)t}}D(D-1)(D-2)......(D-(n+1)+1)y \\ \end{align}$$

(Eq 1.11)
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So we have proved Eq 1.10 is valid throughout any $$\displaystyle n\in \mathbb{N}$$.

=Problem 5.2:Using Trial Solution to Solve Euler's Equation=

From the lecture slide Mtg 26-3

5.2.1 Given
Given the ODE:
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$$  \displaystyle x^{2}{y}''+2x{y}'+2y=0 $$     (Eq 2.1) and boundary conditions:
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$$  \displaystyle y(1)=-2 $$     (Eq 2.3)
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$$  \displaystyle y(2)=5 $$     (Eq 2.4) solve using
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$$  \displaystyle y=x^{r} $$     (Eq 2.5) and the Trial Solution method.
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5.2.2 Solution
In the trial solution use the given equation and take the first two derivatives:
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$$  \displaystyle y=x^{r} $$     (Eq 2.6)
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$$  \displaystyle y^{'}=rx^{r-1} $$     (Eq 2.7)
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$$  \displaystyle y^{''}=r(r-1)x^{r-2} $$     (Eq 2.8) Substitute these values into Eq. 2.1 above:
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$$  \displaystyle x^{2}r(r-1)x^{r-2}-2xrx^{r-1}+2y=0 $$     (Eq 2.9) From this equation it is easy to eliminate constants and we end up with:
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$$  \displaystyle x^{r}(r^{2}-3r+2)=0 $$     (Eq 2.10) Meaning that our equation of interest is:
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$$  \displaystyle (r^{2}-3r+2)=0 $$     (Eq 2.11) Solving this the answers are:
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$$  \displaystyle r_{1}=1 $$     (Eq 2.12)
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$$  \displaystyle r_{2}=2 $$     (Eq 2.13) Using the equation:
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$$  \displaystyle y(x)=C_{1}x^{r_{1}}+C_{2}x^{r_{2}} $$     (Eq 2.14) we plug in r_{1} and r_{2} to get:
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$$  \displaystyle y(x)=C_{1}x+C_{2}x^{2} $$     (Eq 2.15) Using the above boundary conditions we get two equations:
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$$  \displaystyle -2=C_{1}+C_{2} $$     (Eq 2.16) and
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$$  \displaystyle 5=2C_{1}+4C_{2} $$     (Eq 2.17) These are then solved:
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$$  \displaystyle \frac{9}{2}=C_{2} $$     (Eq 2.18)
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$$  \displaystyle -\frac{13}{2}=C_{1} $$     (Eq 2.19) Plugging back in we get:
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$$  \displaystyle y(x)=-\frac{13}{2}x+\frac{9}{2}x^{2} $$     (Eq 2.20)
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=Problem 5.3: Euler L2_ODE_VC=

From the lecture slide Mtg 27

5.3.1.Given
Characteristic equation is given in meeting 27-1 as:


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$$  \displaystyle \left ( r-\lambda \right )^2=0 $$     (Eq 3.1)
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Where $$\displaystyle \lambda=5$$

The form of equation with unknown coeffficients to be determined


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$$  \displaystyle a_{2}x^{2}{y}''+a_{1}x{y}'+a_{0}y=0 $$     (Eq 3.2)
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And the given first homogeneous solution :


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$$  \displaystyle y_{1}=x^{\lambda} $$     (Eq 3.3)
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5.3.2.1. Homogeneous equation with varying coefficients

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$$  \displaystyle \left ( r-5 \right )^{2}=0 $$     (Eq 3.4)
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$$  \displaystyle r^{2}-10r+25=0 $$     (Eq 3.5)
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$$  \displaystyle y=x^{r} $$     (Eq 3.6)
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$$  \displaystyle {y}'=rx^{(r-1)} $$     (Eq 3.7)
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$$  \displaystyle {y}''=r(r-1)x^{(r-2)} $$     (Eq 3.8) Subtituting the trial solution and its derivatives into Eq 3.2 yields :
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$$  \displaystyle a_{2}x^{2}r(r-1)x^{(r-2)}+a_{1}x^{1}rx^{(r-1)}+a_{0}x^{r}=0 $$     (Eq 3.9)
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$$  \displaystyle x^{r}\left \lfloor a_{2}r(r-1)+a_{1}r+a_{0} \right \rfloor=0 $$     (Eq 3.10)
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$$  \displaystyle a_{2}r^{2}+(a_{1}-a_{2})r+a_{0}=0 $$     (Eq 3.11)
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Comparing with Eq 3.5 we can see that :


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$$  \displaystyle a_{2}=1 ,a_{1}=-9 ,a_{0}=25 $$     (Eq 3.12)
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$$  \displaystyle x^{2}{y}''-9x{y}'+25y=0 $$     (Eq 3.13)
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Let the general solution of homogeneous solution to be,
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$$  \displaystyle y=U(x)x^{5} $$     (Eq 3.14)
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$$  \displaystyle {y}'={U(x)}'x^{5}+5x^{4}U(x) $$     (Eq 3.15)
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$$  \displaystyle {y}={U(x)}x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) $$     (Eq 3.16)
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$$  \displaystyle x^{2}\left [ {U(x)}''x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) \right ]-9x\left [ x^{5}{U(x)}'+5x^{4}U(x) \right ]+25x^{5}U(x)=0 $$     (Eq 3.17)
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$$  \displaystyle {U(x)}''x^{5}+{U(x)}'\left [ 10x^{4}-9x^{4} \right ]+U(x)\left [ 20x^{3}-45x^{3}+25x^{3} \right ]=0 $$     (Eq 3.18)
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$$  \displaystyle {U(x)}''+\frac{1}{x}{U(x)}'=0 $$     (Eq 3.19)
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Now we can use the variable separation to reduce the order.See:Separation Of Variables To decrease our equation order.


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$$  \displaystyle {Z}'+\frac{1}{x}Z=0 $$     (Eq 3.20)
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Where $$ \displaystyle Z={U}'$$


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$$  \displaystyle \int\frac{dZ}{Z}=-\int \frac{dx}{x} $$     (Eq 3.21)
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$$  \displaystyle logz=logk_{1}-logx $$     (Eq 3.22)
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$$  \displaystyle z=\frac{k_{1}}{x} $$     (Eq 3.23)
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$$  \displaystyle U=\int Z dx+k_{2} $$     (Eq 3.24)
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$$  \displaystyle U=k_{1}logx+k_{2} $$     (Eq 3.25)
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$$  \displaystyle y=U(x)y_{1}(x)=\left ( k_{1}logx+k_{2} \right )x^{5} $$     (Eq 3.26)
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$$  \displaystyle y_{2}(x)=k_{1}x^{5}logx $$     (Eq 3.27)
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5.3.2.2Homogeneous function with constant coefficients
Euler L2_ODE_CC is given as follows ;
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$$  \displaystyle b_{2}{y}''+b_{1}{y}'+b_{0}y=0 $$     (Eq 3.28) Let the trial solution to be
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$$  \displaystyle y=e^{rx} $$     (Eq 3.29)
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$$  \displaystyle {y}'=re^{rx} $$     (Eq 3.30)
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$$  \displaystyle {y}''=r^{2}e^{rx} $$     (Eq 3.31)
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$$  \displaystyle e^{rx}\left [ b_{2}r^{2}+b_{1}r+b_{0} \right ]=0 $$     (Eq 3.32)
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$$  \displaystyle b_{2}r^{2}+b_{1}r+b_{0}=r^{2}-10r+25 $$     (Eq 3.33) It can obciously seen that the unknowns are
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$$  \displaystyle b_{2}=1 ,b_{1}=-10 ,b_{0}=25 $$     (Eq 3.34)
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$$  \displaystyle y=U(x)e^{5x} $$     (Eq 3.35)
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$$  \displaystyle {y}'={U(x)}'e^{5x}+5U(x)e^{5x} $$     (Eq 3.36)
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$$  \displaystyle {y}=e^{5x}{U(x)}+10e^{5x}{U(x)}'+25e^{5x}U(x) $$     (Eq 3.37)
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$$  \displaystyle e^{5x}{U(x)}''+10e^{5x}{U(x)}'+25e^{5x}U(x)-10\left [ e^{5x}{U(x)}'+5e^{5x}U(x) \right ]+25e^{5x}U(x)=0 $$     (Eq 3.38)
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$$  \displaystyle {U(x)}''+{U(x)}'\left [ 10-10 \right ]+U(x)\left [ 25+25-50 \right ]=0 $$     (Eq 3.39)
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$$  \displaystyle {U(x)}''=0 $$     (Eq 3.40)
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$$  \displaystyle U=k_{1}x+k_{2} $$     (Eq 3.41)
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$$  \displaystyle y(x)=U(x)y_{1}(x)=k_{1}xe^{5x}+k_{2}e^{5x} $$     (Eq 3.42)
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$$  \displaystyle y_{2}=k_{1}xe^{5x} $$     (Eq 3.43)
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= Problem 5.4 Particular Solution by using Variation of Parameters Method =

From the lecture slide Mtg 27-2

Given
Linear, nonhomogeneous, first order differential equation


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$$ \displaystyle \frac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$$ (Eq 4.1)
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Find
Particular solution using $$y\displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)$$

To find homogeneous solution:
Homogeneous differential equation


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$$ \displaystyle \frac{d{{y}_{h}}}{dx}+P\left( x \right){{y}_{h}}=0$$ (Eq 4.2)
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and the solution of homogeneous differential equation is


 * $${{y}_{h}}=A{{e}^{-\int\limits_ – ^{x}{P\left( t \right)dt}}}=A{{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$

where A is a constant of integration (King, page 512).

To find particular solution:
Assuming


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$$ \displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.3)
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is solution of nonhomogeneous ODE (Eq4.1).

We determine A(x) by substituting Eq4.3 and its first derivative into Eq4.1. Differentiating Eq4.3, we obtain,


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$$ \displaystyle \frac{d}{dx}=={A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.4)
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Now $$\displaystyle $$ must satisfy (Eq4.1). We need to substitute $$\displaystyle $$ and $$\displaystyle $$ according to Eq4.3 and Eq4.4 into Eq4.1.


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$$ \displaystyle \frac{d}{dx}+P\left( x \right){{y}_{p}}=Q\left( x \right)$$ (Eq 4.5)
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$$ \displaystyle \left[ {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right]+P(x)\left( A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right)=Q(x)$$ (Eq 4.6)
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Second term and third term in the left side of Eq4.7 eliminate each other and we obtain


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$$ \displaystyle {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}=Q(x)$$ (Eq 4.7)
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We rearrange Eq4.7


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$$ \displaystyle {A}'(x)=Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.8)
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If we integrate both sides Eq4.8, A(x) will be


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$$ \displaystyle A(x)=\int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx}$$ (Eq 4.9)
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If we put Eq4.9 into Eq4.3 in order to get the particular solution, we obtain


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$$ \displaystyle \left( x \right)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\left( \int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx+C} \right)$$ (Eq 4.10)
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Eliminate the constant $$\displaystyle C$$ we can derive particular solution,


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$$ \displaystyle {{y}_{P}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\int Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx$$ (Eq 4.11)
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Now we can solve homework equation in Mtg10-4 as an example,


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$$ \displaystyle \frac{dy}{dx}+xy=2x+3$$ (Eq 4.12)
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Thus we have,


 * $$\displaystyle \left\{ \begin{align}

& P\left( x \right)=x \\ & Q\left( x \right)=2x+3 \\ \end{align} \right.$$

If we solve the homogenous equation for $$\displaystyle P\left( x \right)=x$$,


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$$ \displaystyle {{y}_{h}}=A{{e}^{-\frac{2}}}$$
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then we have(see Kreyszig's book ),


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$$ \displaystyle ={{e}^{-\int\limits_ – ^{x}{tdt}}}\int\limits_ – ^{x}{(2s+3){{e}^{\int\limits_ – ^{s}{tdt}}}ds}={{e}^{-\frac{2}}}\left[ \int\limits_ – ^{x}{2s{{e}^{^{\frac{2}}}}ds}+\int\limits_ – ^{x}{3{{e}^{^{\frac{2}}}}ds} \right]$$ (Eq 4.13)
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Integrating inside of bracket by using Wolfram Alpha, we obtain $$\displaystyle $$ shown below


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$$ \displaystyle =2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)+{{C}_{1}}{{e}^{-\frac{2}}}$$ (Eq 4.14)
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where erfi(x) is the imaginary error function.

Then we extract particular solution from above,


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$$ \displaystyle {{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)$$ (Eq 4.15)
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= Problem 5.5 Solving Non-homogeneous Linear Second Order ODE with Constant Coefficients=

From the lecture slide Mtg 28 and Mtg 29-1

Given
The spring-dashpot system has the Equation of Motion like follows:


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$$ \displaystyle {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y=f(t)$$ (Eq 5.1)
 * <p style="text-align:right">
 * }

where $$\displaystyle y$$ is a function of $$\displaystyle t$$, and $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ are all constants making Eq 5.1 a linear second-order ODE with constant coeffiients.

Find
(1)Find PDEs for integrating factor $$\displaystyle H(t,y)$$ using condition 2nd of exactness stated in Mtg15

(2)Solve Eq 5.1 using trial solution for integrating factor such that $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha $$ is some constant. .

(1) Finding PDEs for integrating factor $$\displaystyle H(t,y)$$
Multiply $$\displaystyle H(t,y)$$ to both sides of the Eq 5.1,


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$$ \displaystyle H(t,y){{a}_{2}}{y}''+H(t,y){{a}_{1}}{y}'+H(t,y){{a}_{0}}y=H(t,y)f(t)$$ (Eq 5.2)
 * <p style="text-align:right">
 * }

If we assume firstly $$\displaystyle H(t,y)$$ to be a function of only $$\displaystyle t$$(see Mtg 10), then we can just find some $$\displaystyle H(t)$$ to make the L.H.S of Eq 5.2 exact in order to change its form as,


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$$ \displaystyle \frac{d\phi }{dt}=H(t)f(t)$$ (Eq 5.3)
 * <p style="text-align:right">
 * }

then we can integrate both sides to reduce the order of Eq 5.1. However, if the assumption "$$\displaystyle H(t,y)$$ is a function of only $$\displaystyle t$$" is not satisfid, we can't integrate the R.H.S of Eq 5.3. In this case, we should move $$\displaystyle f(t)$$ from R.H.S to L.H.S and try to find any $$\displaystyle H(t,y)$$ to make the whole thing exact.

Thus let


 * $$\displaystyle F=H(t){{a}_{2}}{y}+H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y=f(t,y,p){y}+g(t,y,p)$$

in which,


 * $$\displaystyle \begin{align}

& f=H(t){{a}_{2}} \\ & g=H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y \\ & p={y}' \\ \end{align}$$

Then we have,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & {{f}_{tt}}={{a}_{2}}{{H}_{tt}} \\ & {{f}_{ty}}={{a}_{2}}{{H}_{ty}} \\ & {{f}_{tp}}={{f}_{yp}}=0 \\ & {{f}_{y}}={{a}_{2}}{{H}_{y}} \\ & {{f}_{yy}}={{a}_{2}}{{H}_{yy}} \\ & {{g}_{tp}}={{a}_{1}}{{H}_{t}} \\ & {{g}_{yp}}={{a}_{1}}{{H}_{y}} \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{a}_{1}}p{{H}_{y}}+{{a}_{0}}H+{{a}_{0}}y{{H}_{y}} \\ \end{align}$$ (Eqs 5.4)
 * <p style="text-align:right">
 * }

Substituting Eqs 5.4 into (1)&(2) in Mtg15, we have,


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$$ \displaystyle {{a}_{2}}{{H}_{tt}}+2p{{a}_{2}}{{H}_{ty}}+{{p}^{2}}{{a}_{2}}{{H}_{yy}}={{a}_{1}}{{H}_{t}}-{{a}_{0}}H-{{a}_{0}}y{{H}_{y}}$$ (Eq 5.5)
 * <p style="text-align:right">
 * }


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$$ \displaystyle 2{{a}_{2}}{{H}_{y}}=0$$ (Eq 5.6)
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 * }

From Eq 5.6 we can see that the basic assumption is satisfied and if we substitute Eq 5.6 in to Eq 5.5, we have,


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$$ \displaystyle {{a}_{2}}{{H}_{tt}}-{{a}_{1}}{{H}_{t}}+{{a}_{0}}H=0$$ (Eq 5.7)
 * <p style="text-align:right">
 * }

or


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$$ \displaystyle {{a}_{2}}{H}''-{{a}_{1}}{H}'+{{a}_{0}}H=0$$ (Eq 5.8)
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 * <p style="text-align:right">
 * }

Since the integrating factor $$\displaystyle H(t,y)$$ turns out to be $$\displaystyle H(t)$$, the PDE governing $$\displaystyle H(t,y)$$ become an ODE given by Eq 5.8.

(2) Solve Eq 5.1 using trial solution for integrating factor $$\displaystyle h(t)={{e}^{\alpha t}}$$
Let's try an integrating factor like $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha$$ is a constant.

Multiply Eq 5.1 by $$\displaystyle h(t)={{e}^{\alpha t}}$$, we have,


 * {| style="width:100%" border="0"

$$ \displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y \right)={{e}^{\alpha t}}f(t)$$ (Eq 5.9)
 * <p style="text-align:right">
 * }

In order to reduce the order, we integrating Eq 5.9 according to $$\displaystyle t$$ and assume that the result can be written in such form:


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$$ \displaystyle {{e}^{\alpha t}}\left( {{{\bar{a}}}_{1}}{y}'+{{{\bar{a}}}_{2}}y \right)=\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.10)
 * <p style="text-align:right">
 * }

where $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{2}}$$ are assumed to be constants.

(2.1) Determine $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{1}}$$ according to $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$
Differentiate Eq 5.10 we have,


 * {| style="width:100%" border="0"

$$ \displaystyle {{e}^{\alpha t}}\left[ {{{\bar{a}}}_{1}}{y}''+(\alpha {{{\bar{a}}}_{1}}+{{{\bar{a}}}_{0}}){y}'+\alpha {{{\bar{a}}}_{0}}y \right]={{e}^{\alpha t}}f(t)$$ (Eq 5.11)
 * <p style="text-align:right">
 * }

Compare 5.11 and 5.9 we have,


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$$ \displaystyle \begin{align} & {{{\bar{a}}}_{1}}={{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ \end{align}$$ (Eqs 5.12)
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

from which we can see that the validity of Eq 5.10 relies on the value of $$\displaystyle \alpha $$.

(2.2) Find quadratic equation for $$\displaystyle \alpha $$
From Eqs 5.12 we have,


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$$ \displaystyle \begin{align} & {{a}_{1}}-\alpha {{a}_{2}}={{a}_{0}}/\alpha \\ \Rightarrow & {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0 \\ \end{align}$$ (Eq 5.13)
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 * <p style="text-align:right">
 * }

(2.3) Derive a reduced - order equation from Eq 5.10
From Eq 5.10 we have,


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$$ \displaystyle {y}'+\beta y=\frac{1}\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.14)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

where $$\displaystyle \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}}$$

(2.4) Solve Eq 5.14 using Integrating Factor Method
From Mtg10 we know the integrating factor(for Eq 5.14) is,


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$$ \displaystyle i(t)=\exp \int{\beta dt}={{e}^{\beta t}}$$ (Eq 5.15)
 * <p style="text-align:right">
 * }

and the solution is:


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} y & =\frac{1}{i(t)}\int_ – ^{t}{\frac{i(\tau )}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \end{align}$$ (Eq 5.16)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

(2.5) Show the relation between $$\displaystyle \alpha $$ and $$\displaystyle \beta $$
Since,


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$$ \displaystyle \begin{align} & \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ & {{{\bar{a}}}_{1}}={{a}_{2}} \\ \end{align}$$ (Eqs 5.17)
 * <p style="text-align:right">
 * }

So we have,


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$$ \displaystyle \begin{align} & \alpha \beta =\frac \\ & \alpha +\beta =\frac \\ \end{align}$$ (Eq 5.18)
 * <p style="text-align:right">
 * }

which means $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the two roots of,


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$$ \displaystyle {{\lambda }^{2}}-(\alpha +\beta )\lambda +\alpha \beta =0$$
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 * }

or


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$$ \displaystyle {{\lambda }^{2}}-\frac\lambda +\frac=0$$ (Eq 5.19)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

which is identical to Eq 5.13. And from Eq 5.17~Eq 5.19 we can conclude that $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are inseparable pairs so that we can't distiguish them from each other through the their governing equation and the sequense of two integrating factor $$\displaystyle h(t)$$ and $$\displaystyle i(t)$$ is arbitrary. We can also conclude that the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, meaning exchanging the position of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ won't change the result.

When $$\displaystyle \alpha \ne \beta $$
Eq 5.16 can be expanded as,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} y & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{(\beta -\alpha ){_{1}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{\bar{a}}}_{1}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align}$$ (Eq 5.20)
 * <p style="text-align:right">
 * }

Note that Eq 5.20 is under the assumption that $$\alpha \ne \beta $$.

After conbination of constants, we have,


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$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ (Eq 5.21)
 * <p style="text-align:right">
 * }

In which we can derive,


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$$ \displaystyle \begin{align} & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ \end{align}$$
 * <p style="text-align:right">
 * }

However, $$\displaystyle {{y}_{P}}$$ can be furthuer deduced as,


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$$ \displaystyle \begin{align} {{y}_{P}} & =\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)d\left( {{e}^{(\beta -\alpha )t}} \right)} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left[ {{e}^{(\beta -\alpha )t}}\int{{{e}^{\alpha t}}f(t)dt}-\int{{{e}^{(\beta -\alpha )t}}{{e}^{\alpha t}}f(t)dt} \right] \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \end{align}$$
 * <p style="text-align:right">
 * }

This proved my statement "the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$".

In sum we have (when $$\displaystyle \alpha \ne \beta $$),


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \\ \end{align}$$ (Eq 5.22)
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 * <p style="text-align:right">
 * }

When $$\displaystyle \alpha =\beta $$
Substitute the relation $$\displaystyle \alpha =\beta $$ into Eq 5.16,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} y= & \frac\int_ – ^{t}{{{e}^{(\alpha -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+{{k}_{1}}t \right] \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \end{align}$$ (Eq 5.20a)
 * <p style="text-align:right">
 * }

After conbination of constans we have


 * {| style="width:100%" border="0"

$$ \displaystyle y={{C}_{1}}t{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right)$$ (Eq 5.21a)
 * <p style="text-align:right">
 * }

In which we can derive,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$
 * <p style="text-align:right">
 * }

In sum we have (when $$\displaystyle \alpha =\beta $$),


 * {| style="width:100%" border="0"


 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$ (Eq 5.22a)
 * <p style="text-align:right">
 * }

(2.7) Verify particular solution according to table in Mtg20, Fall2009
If $$\displaystyle \alpha \ne \beta$$, then from Eq 5.22 we know,


 * $$\displaystyle {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt}$$

If we substitute $$\displaystyle f(t)={{t}^{2}}$$ into the above equation, we have,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} {{y}_{P}} & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}f(t)dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}f(t)dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}{{t}^{2}}dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}{{t}^{2}}dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2} \right)-\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\beta }^{2}}{{t}^{2}}-2\beta t+2} \right) \\ & =\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-(2{{\alpha }^{2}}\beta +2\alpha {{\beta }^{2}})t+2{{\alpha }^{2}}+2{{\beta }^{2}}+2\alpha \beta } \\ & =\frac{{{(\alpha \beta )}^{2}}{{t}^{2}}-2(\alpha +\beta )t+2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{\frac{{{a}_{0}}^{2}}{{{a}_{2}}^{2}}{{t}^{2}}-2\frac{{{a}_{2}}^{2}}t+2\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}-2\frac{{{a}_{2}}^{2}}}{\frac{{{a}_{0}}^{3}}{{{a}_{2}}^{3}}{{a}_{2}}} \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23)
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 * <p style="text-align:right">
 * }

in which we utilize,


 * $$\displaystyle \begin{align}

\int{{{e}^{\alpha t}}{{t}^{2}}dt} & =\frac{1}{\alpha }\int{{{t}^{2}}d{{e}^{\alpha t}}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{1}{\alpha }\int{2t{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}\int{td}{{e}^{\alpha t}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}\int{{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}{{e}^{\alpha t}} \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2}{{e}^{\alpha t}} \\ \end{align}$$

Compare Eq 5.23 with Mtg20 slide of Fall 2009, we can see that Eq 5.23 has the same form with that in the slide which is a polynominal with at least order 2 according to $$\displaystyle t$$.

If $$\displaystyle \alpha=\beta$$, then from Eq 5.22a we know,


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & {{y}_{P}}=\frac\left( t\int {{e}^{\alpha t}}{{t}^{2}}dt-\int {{e}^{\alpha t}}{{t}^{3}}dt \right) \\ & =\frac\left( \frac{{{e}^{\alpha t}}({{\alpha }^{2}}{{t}^{3}}-2\alpha {{t}^{2}}+2t)}-\frac{{{e}^{\alpha t}}({{\alpha }^{3}}{{t}^{3}}-3{{\alpha }^{2}}{{t}^{2}}+6\alpha t-6)} \right) \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-4\alpha t+6} \\ & =\frac{\alpha \beta }{{t}^{2}}-\frac{2(\alpha +\beta )}t+\frac{2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23a)
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

We can see that Eq 5.23a still has the same form with Eq 5.23 and that in the slide Mtg20. Thus the particular solution is verified.

(2.8) Solve L2-ODE-CC with $$\displaystyle f(t)=\exp (-{{t}^{2}})$$
If $$\displaystyle f(t)=\exp (-{{t}^{2}})$$, then Eqs 5.22 and Eqs 5.22a we derive


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & y_{H}^{1}=\left\{ \begin{align} & {{e}^{-\alpha t}}....................................................................................(\alpha \ne \beta ) \\ & t{{e}^{-\alpha t}}...................................................................................(\alpha =\beta ) \\ \end{align} \right. \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\left\{ \begin{align} & \frac{\sqrt{\pi }\left[ {{e}^{\frac{4}-\alpha t}}erf(\frac{\alpha }{2}-t)-{{e}^{\frac{4}-\beta t}}erf(\frac{\beta }{2}-t) \right]}{2{{a}_{2}}(\alpha -\beta )}........................(\alpha \ne \beta ) \\ & -\frac{1}{2{{a}_{2}}}\sqrt{\pi }{{e}^{\frac{4}-\alpha t}}\left[ (t-\frac{\alpha }{2})erf(\frac{\alpha }{2}-t)-\frac{1}{\sqrt{\pi }}{{e}^{-\frac{4}}} \right].........(\alpha =\beta ) \\ \end{align} \right. \\ \end{align}$$ (Eq 5.24)
 * <p style="text-align:right">
 * }

It has been mentioned that governing equation of $$\displaystyle \alpha $$ is indentical to Eq 5.19, and it has also been mentioned that the sequence of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ is arbitrary.

Case 1: $$\displaystyle (\gamma +1)(\gamma -2)=0$$
From the characteristic equation we can get information of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, which are the ratio of $$\displaystyle {{a}_{0}}$$ and $$\displaystyle {{a}_{1}}$$ with respect to $$\displaystyle {{a}_{2}}$$. However we can't derive information of $$\displaystyle {{a}_{2}}$$. In fact $$\displaystyle {{a}_{2}}$$ has nothing to do with characteristic equation since in the homogeneous case $$\displaystyle {{a}_{2}}$$ can always be the divisor of both sides(the R.H.S is zero so $$\displaystyle {{a}_{2}}$$ disappeared only left some ratios) and has no effect on the results if $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are fixed. But in the non-homogeneous case $$\displaystyle {{a}_{2}}$$ can be seen as part of the R.H.S(non-zero) and do have influence on the final result. As we can't get enough information to determine the value of $$\displaystyle {{a}_{2}}$$, we take it as 1:


 * $$\displaystyle {{a}_{2}}=1$$

From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:


 * $$\displaystyle {y}''-{y}'-2y={{e}^{-{{t}^{2}}}}$$

then


 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-1 \\ & \alpha \beta =\frac=-2 \\ \end{align} \right. \\ & \Rightarrow \left\{ \begin{align} & \alpha =1 \\ & \beta =-2 \\ \end{align} \right. \\ \end{align}$$

Substitute these into Eq 5.24 we have the solution as,


 * {| style="width:100%" border="0"

$$ \displaystyle y={{C}_{1}}{{e}^{-t}}+{{C}_{2}}{{e}^{2t}}+\frac{\sqrt{\pi }}{6}\left[ {{e}^{0.25-t}}erf(0.5-t)-{{e}^{1+2t}}erf(-1-t) \right]$$ (Eq 5.25)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

Case 2: $$\displaystyle {{(\gamma -4)}^{2}}=0$$
From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:


 * $$\displaystyle {y}''-8{y}'+16y={{e}^{-{{t}^{2}}}}$$

then,


 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-8 \\ & \alpha \beta =\frac=16 \\ \end{align} \right. \\ & \Rightarrow \alpha =\beta =-4 \\ \end{align}$$

Substitute this into Eq 5.24 we have,


 * {| style="width:100%" border="0"

$$ \displaystyle y={{C}_{1}}t{{e}^{4t}}+{{C}_{2}}{{e}^{4t}}-\frac{\sqrt{\pi }}{2}{{e}^{4+4t}}\left[ (t+2)erf(2+t)-\frac{1}{\sqrt{\pi }}{{e}^{-{{(2+t)}^{2}}}} \right]$$ (Eq 5.26)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

= Problem 5.6: Show Agreement of Particular Integrals =

From the lecture slide Mtg 30-1

Given
Show that the following equation:


 * {| style="width:100%" border="0"

$$ \displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}

$$ (Eq 6.1)
 * <p style="text-align:right">
 * }

is equivalent to the following particular integral found in the Differential Equations text book by A.C.King:
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$$ \displaystyle

{{y}_{p}}\left( x \right)=\oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$ (Eq 6.2)
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given that the following equations holds true:
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$$ \displaystyle

W={{u}_{1}} {{u}_{2}}'-{{u}_{1}}' {{u}_{2}}=\left| \begin{matrix} {{u}_{1}} & {{u}_{2}} \\ {{u}_{1}}' & {{u}_{2}}' \\ \end{matrix} \right|

$$ (Eq 6.3)
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$$ \displaystyle

(\frac{u_{2}}{u_{1}})'=\frac{1}{h}

$$ (Eq 6.4)
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And considering the following conversion between the notation used in the lectures and A.C. King's notation:
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$$ \displaystyle

\oint^x{f\left( s \right)}\,ds=\int^x{f\left(s\right)}\,ds =\int{f\left(x\right)}\,dx

$$  w/o constant (Eq 6.5)
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Solution
The first step to tackling this problem requires the application of the Integration by Parts formula to equation 6.1


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$$\displaystyle \int w\,dv=wv-\int v\,dw $$

(Eq 6.6)
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$$\displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{w}\underbrace{\frac{1}{h\left( x \right)}\,dx}_{dv}}$$

(Eq 6.7) The remaining unknowns in the right hand side of equation 6.6 are shown below:
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$$\displaystyle dw=h\left( x \right) f\left( x \right)dx

$$

(Eq 6.8)
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$$\displaystyle v= \int {\frac{1}{h\left( x \right)}}\,dx

$$

(Eq 6.9) Equations 6.7, 6.8 and 6.9 can be combined and rewritten into the form shown in equation 6.5 to yield the following:
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$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}h\left( s \right)f\left( s \right)\,ds \right)\int \frac{1}{h\left( x \right)}\,dx \right. \\ & \left. -\mathop{\oint }^{x}\left( \mathop{\oint }^{x}\frac{1}{h\left( s \right)}\,ds \right)h\left( s \right)f\left( s \right)\,ds \right] \\ \end{align}$$

(Eq 6.10)
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The relation shown in equation 6.4 can be used to simplify equation 6.10 as follows:
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$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}\frac{u_{1}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right)\frac{{{u}_{2}}\left( x \right)}{{{u}_{1}}\left( x \right)} \right. \\ & \left. -\left[ \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right)}{{{u}_{1}}\left( s \right)}\frac{{{u}_{1}}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right] \right] \\ \end{align}$$ (Eq 6.11) Now, multiplying through by u1(x) and using the relation shown in equation 6.3 yields the following:
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$$ \displaystyle \begin{align} {{y}_{p}}\left( x \right) & =\left( \mathop{\oint }^{x}\frac{{{u}_{1}}\left( s \right){{u}_{2}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right)-\left( \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right){{u}_{1}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right) \\ & =\mathop{\oint }^{x}f\left( s \right)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}\,ds \\ \end{align}$$ (Eq 6.12)
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= Problem 5.7: Find Two Homogeneous Solutions using a Trial Solution =

From the lecture slide Mtg 31-1

Given
Find u1 and u2 of Equation 7.1 using the trial solution shown in equation 7.2:


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$$ \displaystyle

(x +1) y'' - (2x+3)y' + 2y = 0

$$ (Eq 7.1)
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$$ \displaystyle

y(x)=e^{rx}

$$ (Eq 7.2)
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Solution
The first step is to differentiate the trial solution as follows:


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$$ \displaystyle y(x) = e^{rx} $$ (Eq 7.3)
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$$ \displaystyle y'(x) = r e^{rx} $$ (Eq 7.4)
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$$ \displaystyle y''(x) = r^2 e^{rx} $$ (Eq 7.5)
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Next, plug the values into equation 7.1 to yield:


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$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ (Eq 7.6) Since erx will not equal zero for any value of x, divide equation 7.6 by this value and factor out the x to yield:
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$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0 $$ (Eq 7.7) In order to satisfy equation 7.7 for any x, the following equation is solved:
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$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ (Eq 7.8)
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$$ \displaystyle r=2 $$ (Eq 7.9) By plugging equation 7.9 into the trial solution in equation 7.3 we have obtained the first homogeneous solution:
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$$ \displaystyle u_1(x)=e^{2x} $$ (Eq 7.10)
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The second homogeneous solution is obtained by the following:


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$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$
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(Eq 7.11)
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Equation 7.1 is rearranged into the following form:


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$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$ (Eq 7.12)
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The value for a1 shown in equation 7.12 is then plugged into equation 7.11 along with the first homogeneous solution to solve for the second solution as shown:


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$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ (Eq 7.13)
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$$ \displaystyle \Rightarrow u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ (Eq 7.14)
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$$ \displaystyle \Rightarrow u_2(x)=\frac{-1}{4}(2x+3)$$ (Eq 7.15)
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= Problem 5.8 L2_ODE_VC (Homogeneous) by using Reverse Engineering =

From the lecture slide Mtg 31-4

Given
Trial Solution: $$\displaystyle y=\frac$$ and Characteristic Equation: $$\displaystyle {{r}^{2}}+3=0$$

Find
Find L2_ODE_VC according to a trial solution and the characteristic equation given above

Solution
Wolfram Alpha used to find the derivatives of the trial solution $$\displaystyle y$$


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$$ \displaystyle \left[ y=\frac \right].{{a}_{0}}(x)$$ (Eq 8.1)
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$$ \displaystyle \left[ \frac{d}{dx}\left( \frac \right)=\left( r-\frac{2}{x} \right)\left( \frac \right) \right].{{a}_{1}}(x)$$ (Eq 8.2)
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$$ \displaystyle \left[ \frac{d{{x}^{2}}}\left( \frac \right)=\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)\left( \frac \right) \right].{{a}_{2}}(x)$$ (Eq 8.3)
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Adding Eq8.1, Eq8.2 and Eq8.3 each other with taking common parenthesis of $$\displaystyle y$$ will be equal to $$\displaystyle y$$ multiply by characteristic equation.


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$$ \displaystyle \frac\left[ {{a}_{2}}\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)+{{a}_{1}}\left( r-\frac{2}{x} \right)+{{a}_{0}} \right]=\frac\left( {{r}^{2}}+3 \right)=0$$ (Eq 8.4)
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First expanding the inside of brackets and then rearranging and taking common parentheses of $$\displaystyle {{r}^{2}}$$, $$\displaystyle r$$ and 1, it will be


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$$ \displaystyle {{r}^{2}}({{a}_{2}})+r\left( -\frac{4}{x}{{a}_{2}}+{{a}_{1}} \right)+1\left( \frac{6}{{a}_{2}}-\frac{2}{x}{{a}_{1}}+{{a}_{0}} \right)={{r}^{2}}+0.r+3$$ (Eq 8.5)
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From the equality of both sides, we find the values of coefficients $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$.


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$$ \displaystyle \begin{align} & {{a}_{2}}=1 \\ & {{a}_{1}}=\frac{4}{x} \\ & {{a}_{0}}=3+\frac{2} \\ \end{align}$$
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$$ \displaystyle {{r}^{2}}+3=0\Rightarrow {{r}_{1,2}}=\mp i\sqrt{3}$$ (Eq 8.6)
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$$ \displaystyle {{a}_{2}}(x){y}''+{{a}_{1}}(x){y}'+{{a}_{0}}(x)y=0$$ (Eq 8.7)
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is the general form of a second-order ordinary differential equation with coefficients.

If we substitute $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ into Eq8.7, we obtain


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$$ \displaystyle {y}''+\frac{4}{x}{y}'+(3+\frac{2})y=0$$ (Eq 8.8)
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Multiply all equation with $$\displaystyle {{x}^{2}}$$ to eliminate denominators


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$$ \displaystyle {{x}^{2}}{y}''+4x{y}'+\left( 3{{x}^{2}}+2 \right)y=0$$ (Eq 8.9)
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=Contributions Of Team Members=


 * Problem 1:    Solved by: Oztekin,Zou

Author: Oztekin

Proofread by: Oztekin, Dube, Zou


 * Problem 2:    Solved by: Dube

Author: Dube

Proofread by: Oztekin,Ismail


 * Problem 3:    Solved by: Oztekin

Author: Oztekin

Proofread by: Dube


 * Problem 4:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin


 * Problem 5:    Solved by: Zou, Oztekin

Author: Zou

Proofread by: Oztekin,Yaakow


 * Problem 6:    Solved by: Patterson

Author: Patterson

Proofread by: Yaakov, Oztekin


 * Problem 7:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Yaakov,Zou


 * Problem 8:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin,Dube

= References =