User:Egm6321.f10.team4.Yoon/HW5

= Problem 1 -  Solving multiple y derivative wrt x =

From

Given
The multiple y derivatives wrt x that is given from F&9
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$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }


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$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-4)
 * }
 * }


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$$\displaystyle y_{xx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-5)
 * }
 * }


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$$\displaystyle y_{xxxx}=e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-6)
 * }
 * }

Two approach to obtain y derivatives
method 1
 * $$\displaystyle y(x(t)) = \underbrace{\frac{dy}{dt}}_{y_t=} = \underbrace{\frac{dx}{dt}}_{e^t} \cdot \underbrace{\frac{dy}{dx}}_{y_x} $$

rewriting the equation wrt y derivate with respect to x,
 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

method 2
 * $$\displaystyle y_{x}= \underbrace{\frac{dy}{dx}}_{y_x=} = \underbrace{\frac{dy}{dt}}_{y_t} \cdot \underbrace{\frac{dt}{dx}}_{\frac{dx}{dt}=e^{-t}}$$


 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

Find
Part a find $$\displaystyle y_{xxxxx}$$ in terms of derivative of y with respect to t

Part b Plug (eq 1.3) and (eq 1.4) into (eq 1.1) to obtain $$\displaystyle y_{tt} - 3y_{t} + 2y = 0$$

Solution for the part a
Solving for the 5th Derivative


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$$\displaystyle y_{xxxxx} = \frac{d}{dx} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-7)
 * }
 * }


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$$\displaystyle = \frac{dt}{dx} \cdot \frac{d}{dt} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-8)
 * }
 * }

Substituting (Eq 1-6) into the previous (Eq 1-8) yields,
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$$\displaystyle y_{xxxxx} = \underbrace {\frac{dt}{dx}}_{e^{-t}} \cdot \frac{d}{dt} \left[e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) \right] $$ $$
 * $$\displaystyle (Eq. 1-9)
 * }
 * }


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$$\displaystyle = e^{-t} \left[-4e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) + e^{-4t}(y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}) \right] $$ $$
 * $$\displaystyle (Eq. 1-10)
 * }
 * }


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$$\displaystyle = e^{-5t} \left[-4y_{tttt} + 24y_{ttt} -44y_{tt} + 24y_{t} + y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}\right] $$ $$
 * $$\displaystyle (Eq. 1-11)
 * }
 * }

Collecting terms with reduction order of y derivate wrt t yields,
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$$\displaystyle y_{xxxxx} = e^{-5t}(y_{ttttt} - 10y_{tttt} + 35y_{ttt} - 50y_{tt} + 24y_{t}) $$ $$
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 * $$\displaystyle (Eq. 1-12)
 * }
 * }

solution for the part b
Plugging the obtained y deribatives into the given equation to obtain the following:

First, taking a look at what we have to use,
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$$\displaystyle x=e^t $$ $$
 * $$\displaystyle
 * }
 * }


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$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-14)
 * }
 * }


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$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle
 * }
 * }

Substituting what we have into the coresponding variables yields,

reorganizing the equation by reduction order, we can solve the desired equation

= Problem 10: Method of Trial Solutions =

Given
Solve the given L2_ODE_VC((4)P25.3 using Method 2 i.e.) Trial Solution $$\displaystyle y=x^r$$ with the boundary condition as followed


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x^2y'' - 2xy' + 2y = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

The given boundary conditions
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y(_{x=}1) = -2, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-2)
 * }
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y(_{x=}2) = 5. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

required

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y(x) = \sum^{n}_{i=1} C_ix^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

Find
Solve (Eq. 2-1) using method2(trial solution) to make it have the shown format below.


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y(x) = C_1x^{r_1} + C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-4)
 * }
 * }

Solution
First, taking the given trial solution y's 1st and 2nd deriviates,
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y=x^r $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-5)
 * }
 * }


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y'=rx^{r-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-6)
 * }
 * }


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y''=r(r-1)x^{r-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-7)
 * }
 * }

Substiuting (Eq 2-5), (Eq 2-6), and (Eq 2-7) into the given L2_ODE(Eq 2-1) yields
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\cancelto{_1}{x^2} r \cdot x^{r- \cancelto{1}{2}}(r-1) - 2 \cancelto{_1}{x} \cdot x^{r-\cancelto{1}{1}}r + 2x^r = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-8)
 * }
 * }

Simplifying and eliminating non-zero constant ahead of polynomial,
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x^r \left( r^2 -r - 2r + 2 \right) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-9)
 * }
 * }


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r^2- 3r + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-10)
 * }
 * }

Solving for the unknown 'r's yields
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r_1 = 1, r_2 = 2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-11)
 * }
 * }

The solution takes the form of a homogeneous solution, so substituting the rs into the equation,
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y(x) = C_1x^{r_1}+C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-12)
 * }
 * }

Substituting the roots of the polynomial into the homogeneous format,
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y(x) = C_1x^{1}+C_2x^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-13)
 * }
 * }

plugging the given boundary conditions yields
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y(_{x1=}1)= C_1 \cdot (_{x1=} 1)^{1}+C_2 \cdot (_{x1=}1)^{2} = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-14)
 * }
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y(_{x2=}2)= C_1 \cdot (_{x2=}2)^{1}+C_2 \cdot (_{x2=}2)^{2} = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-15)
 * }
 * }

Reorganizing the equation wrt $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
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C_1 + C_2 = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-16)
 * }
 * }


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2C_1 + 4C_2 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-17)
 * }
 * }

Solving the above equations, we obtain $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
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C_1 = - \frac{13}{2}; C_2= \frac{9}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-18)
 * }
 * }

Therefore substiting the results from (Eq. 2-18) into the homogeneous solution format(Eq. 2-13)
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 * style="width:20%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-19)
 * }
 * }

Given
The particular solution for a L2-ODE-VC developed in class during Mtg 30 is given as:
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$${{y}_{p}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}$$ Where $$h(x)$$ is equal to:
 * <p style="text-align:right;">$$\,(Eq. 6.1) $$
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 * }
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$$h(x)=\frac{{{u}_{1}}^{2}(x)}{{{u}_{2}}^{\prime }(x)\cdot {{u}_{1}}(x)-{{u}_{2}}(x)\cdot {{u}_{1}}^{\prime }(x)}$$
 * <p style="text-align:right;">$$\,(Eq. 6.2) $$
 * style= |
 * }