User:Egm4313.s12.team3.rountree

=Problem R2.6=

Problem Statement
R2.5 (sec5 page 7) Find an ODE in the form $$y''+ay'+by=0$$ for the given basis.

Part A:
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$$\displaystyle e^{2.6x}, \; e^{-4.6x} $$
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Part B:
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$$\displaystyle e^{-\sqrt{5}x},\; xe^{-\sqrt{5}x} $$
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Part A
Given the general solutions of $$y_{h,n}$$


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$$\displaystyle y_{h,1}=e^{\lambda_1 x}    \;\;\;\;\;\text{and}\;\;\;\;\;        y_{h,2}=e^{\lambda_2 x} $$
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We see that...


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$$\displaystyle \lambda_1 = 2.6   \;\;\;\;\;\text{and}\;\;\;\;\;        \lambda_2 = -4.3 $$
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Subbing the equations above into the characteristic equation...


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$$\displaystyle (\lambda + 2.6)(\lambda - 4.3)=0 $$
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Expanding that equation...


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$$\displaystyle \lambda^2 -1.7\lambda - 11.18 = 0 $$
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Therefore, the ODE is...


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$$\displaystyle y'' - 1.7y' - 11.18y =0 $$     (6.1)
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Part B
Given the general solutions of $$y_{h,n}$$


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$$\displaystyle y_{h,1}=e^{\lambda_1 x}    \;\;\;\;\;\text{and}\;\;\;\;\;        y_{h,2}=xe^{\lambda_2 x} $$
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We see that, in this case (See "Case 2" in notes - [[media:iea.s12.sec5.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 5]])...


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$$\displaystyle \lambda_1 = \lambda_2 = -\sqrt{5} $$
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Subbing the equations above into the characteristic equation...


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$$\displaystyle (\lambda -\sqrt{5})(\lambda - \sqrt{5})=0 $$
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Expanding that equation...


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$$\displaystyle \lambda^2 -2\sqrt{5}\lambda +5 = 0 $$
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Therefore, the ODE is...


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$$\displaystyle y'' - 2\sqrt{5}y' +5y =0 $$     (6.2)
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=Problem R1.4=

Problem Statement
Derive equations (4.1) and (4.2) from equation (4.3)


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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C} Q $$ (4.1)
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C} I $$ (4.2)
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$$\displaystyle V = LC \frac{d^2 v_c}{dt^2} + RC \frac{dv_c}{dt} +v_c $$     (4.3)
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Problem Solution
Given:


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$$\displaystyle V = v_R +v_L+v_C $$     (4.4)
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$$\displaystyle v_R = RI $$ (4.5)
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$$\displaystyle v_L = L \frac{dI}{dt} = LI' $$     (4.6)
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$$\displaystyle Q = C v_C \rightarrow v_C = \frac{Q}{C} $$     (4.7)
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$$\displaystyle I = \frac{dQ}{dt} = Q' \therefore Q = \int I dt $$ (4.8)
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Part A
Derivation of Equation 4.1

From Equation 4.7
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$$\displaystyle v_C = \frac{Q}{C} $$
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Therefore, taking the first derivative...
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$$\displaystyle \frac{dv_c}{dt} = \frac{d}{dt} (\frac{Q}{C}) = \frac{Q'}{C} $$     (4.10)
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and taking the second derivative...
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$$\displaystyle \frac{d^2 v_c}{dt^2} = \frac{d}{dt} (\frac{Q'}{C}) = \frac{Q''}{C} $$     (4.11)
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Substituting Equations 4.7, 4.10, and 4.11 into Equation 4.3...
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$$\displaystyle V = LC (\frac{Q''}{C}) + RC (\frac{Q'}{C}) + (\frac{Q}{C}) $$
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which simplifies to...
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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C}Q $$     (4.1)
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Part B
Derivation of Equation 4.2

Taking the derivative from Equation 4.1 (Previously derived from Equation 4.3)
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$$\displaystyle V' = LQ' + RQ + \frac{1}{C}Q' $$     (4.12)
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Taking the first derivative of Equation 4.8...
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$$\displaystyle Q' = I $$ (4.13)
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and taking the second derivative...
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$$\displaystyle Q'' = I' $$ (4.14)
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and taking the third derivative...
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$$\displaystyle Q' = I $$     (4.15)
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Then substituting Equations 4.13, 4.14, and 4.15 into Equation 4.12...
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C}I $$     (4.2)
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