User:Egm4313.s12.team18.crete

=Andrew Crete=

Useful Links
Team 18 Course Wiki Course Website Report Table

Report 1
R1 #1.1 pg1.4

Given: Spring-dashpot system in parallel with a mass and applied force f(t). Derive the equation of motion.

Solution:

Using methods of kinematics to note that the displacement of the dashpot is equivalent to that of the spring:
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$$  \displaystyle y=y_k=y_c\! $$     (1)
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Using methods of kinetics, specifically Newton's 2nd Law:
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$$  \displaystyle my''+f_I=f(t)\! $$     (2)
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$$  \displaystyle f_I=f_k+f_c\! $$     (3)
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Constitutive Relations of forces for spring and dashpot:
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$$  \displaystyle f_k=ky_k=ky\! $$     (4)
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$$  \displaystyle f_c=cv=cy_c'\!=cy' $$     (5)
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From (3), (4), and (5):
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$$  \displaystyle f_I=ky+cy' $$     (6)
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Inserting (2) into (6):
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$$  \displaystyle my''+cy'+ky=f(t) $$     (7)
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Report 2
R2.5 K 2011 p.59 pbs. 16-17 Find an ODA of form $$y''+ay'+by=0$$ for the given basis:  #16 
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Given: $$ \displaystyle e^{2.6x}, e^{-4.3x} $$
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From the given, we can see that this is a scenario of two distinct roots of form:
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$$  \displaystyle \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6 $$     (1)
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$$  \displaystyle \lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3 $$     (2)
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Introducing a variable $$c$$ to simplify calculations:


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$$  \displaystyle c=\sqrt{a^2-4b} $$     (3)
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Substituting (3) into (1) and (2):
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$$  \displaystyle \lambda_1=\frac{1}{2}(-a+c)=2.6 $$     (4)
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$$  \displaystyle \lambda_2=\frac{1}{2}(-a-c)=-4.3 $$     (5)
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From (4) and (5) we have two equations and two unknowns. Begin by solving for $$a$$ via adding (4) and (5):


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$$  \displaystyle -a=-1.7 $$
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$$  \displaystyle \rightarrow a=1.7 $$     (6)
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Substituting the value from (6) into (4) and solving for $$c$$ :


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$$  \displaystyle \frac{1}{2}(-a+c)=2.6 $$
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$$  \displaystyle \rightarrow c=6.9 $$     (7)
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Substituting the value from (7) into (3) and solving for $$b$$ :


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$$  \displaystyle 6.9=\sqrt{1.7^2-4b} $$
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$$  \displaystyle b=\frac{1.7^2-6.9^2}{4} $$
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$$  \displaystyle \rightarrow b=-11.18 $$     (8)
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Substituting the values from (6) and (8) into the general form yields:


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$$  \displaystyle y''+1.6y'-11.18y=0 $$
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 #17 
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Given: $$ \displaystyle e^{-\sqrt{5}x}, xe^{-\sqrt{5}x} $$
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From the given we can see that this is a scenario of a real double root:


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$$  \displaystyle \lambda=-\frac{1}{2}a $$     (1)
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$$  \displaystyle a^2-4b=0 $$     (2)
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From the given basis:


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$$  \displaystyle e^{\frac{-ax}{2}} $$     (3)
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Setting (3) equal to (1) and solving for $$a$$ :


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$$  \displaystyle e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $$
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$$  \displaystyle \frac{a}{2}=\sqrt{5} $$
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$$  \displaystyle \rightarrow a=2\sqrt{5} $$     (4)
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Solving for $$b$$ by substituting (4) into (2):


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$$  \displaystyle {2\sqrt{5}}^2-4b=0 $$
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$$  \displaystyle b=\frac{{2\sqrt{5}}^2}{4} $$
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$$  \displaystyle \rightarrow b=\sqrt{5} $$     (5)
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Substituting the results from (4) and (5) into the general form yields:


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$$  \displaystyle y''+2\sqrt{5}y'+5y=0 $$
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Report 3
R3.6 Given:
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$$  \displaystyle y_{p,1}''-3y_{p,y}'+2y_{p,1}=r_1(x) := 4x^2 $$     (1)
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And


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$$  \displaystyle y_{p,1}''-3y_{p,y}'+2y_{p,1}=r_2(x) := -6x^5 $$     (2)
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Determining the characteristic equation:


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$$  \displaystyle r^2-3r+2=0 $$
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$$  \displaystyle \rightarrow r_1=1 $$     (3.a)
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$$  \displaystyle \rightarrow r_2=2 $$     (3.b)
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Creating a complete set of expressions for $$y_p$$:


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$$  \displaystyle y_p(x)=\sum_{j=0}^5c_jx^j $$     (4.a)
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$$  \displaystyle y_p'(x)=\sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} $$     (4.b)
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$$  \displaystyle y_p'(x)=\sum_{j=2}^5 c_j \cdot j \cdot (j-1) \cdot x^{j-2} $$     (4.c)
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6 unknown coefficients, must modify (4.b) and (4.c) to get $$x^j$$ as a common variable. Using the dummy variable k:

From (4.c):


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$$  \displaystyle k=j-2 \rightarrow j=k+2 $$
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$$  \displaystyle \sum_{k=0}^3 c_{k+2} \cdot (k+2) \cdot (k+1) \cdot x^k = \sum_{j=0}^3 c_{j+2} \cdot (j+2) \cdot (j+1) \cdot x^j $$     (5.a)
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From (4.b):


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$$  \displaystyle k=j-1 \rightarrow j=k+1 $$
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$$  \displaystyle \sum_{k=0}^4 c_{k+1} \cdot (k+1) \cdot x^k = \sum_{j=0}^4 c_{j+1} \cdot (j+1) \cdot x^j $$     (5.b)
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The next step involves adding (4.a), (5.a) and (5.b). Since all three equations are summed through 3, we can combine terms to a constant summation of 3 and add the 4-5 terms individually:


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$$  \displaystyle \sum_{j=0}^3 [c_{j+2}(j+2)(j+1) - 3c_{j+1}(j+1)+2c_j]x^j - 3c_5(5)x^4 + 2[c_4x^4 + 5c_5x^5] = -6x^5 $$     (6)
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We can now equate the coefficients of the basis function for each respective order of x:


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$$  \displaystyle x^0(j=0): (2c_2-3c_1+2c_0)x^0 $$     (7.a)
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$$  \displaystyle x^1(j=1): (6c_3 - 6c_2 + 2c_1)x^1=0 $$     (7.b)
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$$  \displaystyle x^2(j=2): (12c_4 - 9c_3 + 2c_2)x^2=0 $$     (7.c)
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$$  \displaystyle x^3(j=3): (20c_5 - 12c_4 + 2c_3)x^3=0 $$     (7.d)
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$$  \displaystyle x^4: -15c_5x^4+2c_4x^4=0 $$     (7.e)
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$$  \displaystyle x^4: 2c_5x^5=-6x^5 $$     (7.f)
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From (7.f):


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$$  \displaystyle 2c_5=-6 \rightarrow c_5=-3 $$     (8)
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Substituting (8) into (7.e):
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$$  \displaystyle -15(-3) + 2c_4=0 \rightarrow c_4=-22.5 $$     (9)
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Substituting (8) and (9) into (7.d):
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$$  \displaystyle 20(-3)-12(-22.5)+2c_3=0 \rightarrow c_3=-105 $$     (10)
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Substituting (8), (9), and (10) into (7.c):
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$$  \displaystyle 12(-22.5)-9(-105)+2c_2=0 \rightarrow c_2=-337.5 $$     (11)
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Substituting (8), (9), (10), and (11) into (7.b):
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$$  \displaystyle 6(-105)-6(-337.5)+2c_2=0 \rightarrow c_1=-697.5 $$     (12)
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Substituting (8), (9), (10), (11), and (12) into (7.a):
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$$  \displaystyle 2(-337.5)-3(-697.5)+2c_2=0 \rightarrow c_0=-708.75 $$     (13)
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Thus, the particular solution of (2) is:


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$$  \displaystyle y_p(x)=-3x^5-22.5x^4-105x^3-337.5x^2-697.5x-708.75 $$
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 Part 2  For the initial conditions given in (1), (7.a)-(7.e) remain unchanged. (7.f) becomes:
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$$  \displaystyle 2c_5=4x^5 \rightarrow c_5=2 $$     (14)
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Solving for the remaining constants using the same method as before yields:
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$$  \displaystyle c_0=472.5 ; c_1=465 ; c_2=225 ; c_3=70 ; c_4=15 ; c_5=2 $$     (15)
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And thus using the coefficients calculated in (15), the particular solution for (1) is: Solving for the remaining constants using the same method as before yields:
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$$  \displaystyle y_p(x)=2x^5+15x^4+70x^3+225x^2+465x+472.5 $$
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R3.9 K 2011 p.84 pbs. 13-14  Given: 
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$$  \displaystyle 8y''-6y'+y=6\cosh{x} $$     (1)
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$$  \displaystyle y(0)=0.2 $$
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$$  \displaystyle y'(0)=0.05 $$
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We begin by determining the general solution of the homogeneous ODE of (1):


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$$  \displaystyle 8\lambda^2-6\lambda+1=0 $$     (2)
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By factoring we obtain:


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$$  \displaystyle (4\lambda-1)(2\lambda-1) \rightarrow \lambda=\frac{1}{4},\frac{1}{2} $$     (3)
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From (3), we can construct a general solution to (1):


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$$  \displaystyle y_h=c_1 e^{\frac{x}{4}} + e^{\frac{x}{2}} $$     (4)
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According to K. 2011 pg. 633, we can express the right hand side of (1) as:


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$$  \displaystyle r(x)=6\cosh{x}=(6)[\frac{1}{2}(e^x+e^{-x})]=r_1(x)+r_2(x) $$
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$$  \displaystyle r_1(x)=3e^x $$     (5.a)
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$$  \displaystyle r_2(x)=3e^{-x} $$     (5.b)
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From this, it is evident that we may use the sum rule. From Table 2.1, in the first line, our particular solution then becomes:


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$$  \displaystyle y_p(x)=C_1e^x+C_2e^{-x} $$     (6)
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Taking the first and second derivatives of $$y_p(x)$$:


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$$  \displaystyle y_p'=C_1e^x-C_2e^{-x} $$     (7.a)
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$$  \displaystyle y_p''=C_1e^x+C_2e^{-x} $$     (7.b)
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In order to use the Method of Undetermined Coefficients, we must put (1) into the form of $$y''+ay'+b=r(x)$$. Dividing (1) by 8 and substituting (5.a) and (5.b):


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$$  \displaystyle y''-\frac{6}{8}y'+\frac{1}{8}y=\frac{3e^x+3e^{-x}}{8} $$     (8)
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We can now substitute (6), (7.a), and (7.b) into (8):


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$$  \displaystyle (C_1e^x+C_2e^{-x})-\frac{6}{8}(C_1e^x-C_2e^{-x})+\frac{1}{8}(C_1e^x+C_2e^{-x})=\frac{3}{8}(e^x+e^{-x}) $$     (9)
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This form allows us to solve for $$C_1$$ and $$C_2$$:


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$$  \displaystyle e^x: C_1-\frac{6}{8}C_1+\frac{1}{8}C_1=\frac{3}{8} $$
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$$  \displaystyle \rightarrow C_1=1 $$     (10.a)
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$$  \displaystyle e^{-x}: C_2+\frac{6}{8}C_2+\frac{1}{8}C_2=\frac{3}{8} $$
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$$  \displaystyle \rightarrow C_2=\frac{1}{5} $$     (10.b)
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Substituting (10.a) and (10.b) into (6) yields the particular solution to (1):


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$$  \displaystyle y_p(x)=e^{-x}+\frac{1}{5}e^{-x} $$     (11)
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The complete solution to (1) is the sum of the particular solution to (1) given in (11) plus the general solution to the homogeneous ODE of (1) given in (4):


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$$  \displaystyle y(x)=c_1e^{\frac{x}{4}}+c_2e^{\frac{x}{2}}+e^x+\frac{1}{5}e^{-x} $$     (12)
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Taking the first derivative of (12):


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$$  \displaystyle y'=\frac{c_1}{4}e^{\frac{x}{4}}+\frac{c_2}{2}e^{\frac{x}{2}}+e^x-\frac{1}{5}e^{-x} $$     (13)
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Using the initial conditions given in (1) with (12) and (13):


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$$  \displaystyle y(0)=c_1+c_2+1+\frac{1}{5}=0.2 $$     (14.a)
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$$  \displaystyle y'(0)=\frac{c_1}{4}+\frac{c_2}{2}+1-\frac{1}{5}=0.05 $$     (14.b)
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From (14.a):


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$$  \displaystyle c_2=-(c_1+1) $$     (15)
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Substituting (15) into (14.b):


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$$  \displaystyle \frac{c_1}{4}-\frac{c_1}{2}-\frac{1}{2}+1-\frac{1}{5}=0.05 $$
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$$  \displaystyle \rightarrow c_1=1 $$     (16)
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Substituting (16) into (14.a):


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$$  \displaystyle 1+c_2+1+\frac{1}{5}=0.02 $$
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$$  \displaystyle \rightarrow c_2=-2 $$     (17)
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Substituting the values from (16) and (17) into (12) gives us the complete solution to the nonhomogeneous ODE given in (1):


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$$  \displaystyle y(x)=e^{\frac{x}{4}}-2e^{\frac{x}{2}}+e^x+\frac{1}{5}e^{-x} $$
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Given
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$$  \displaystyle y''+4y'+4y=e^{-2x}\sin{2x} $$     (1)
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$$  \displaystyle y(0)=1 $$
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$$  \displaystyle y'(0)=-1.5 $$
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Begin by determining the general solution of the homogeneous ODE of (1):


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$$  \displaystyle \lambda^2 + 4 \lambda + y = 0 $$     (2)
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By factoring we obtain:


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$$  \displaystyle (\lambda+2)(\lambda+2)=0 \rightarrow \lambda=-2,-2 $$     (3)
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From (3) we can construct a general solution to the homogeneous ODE of (1):


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$$  \displaystyle y_h=c_1e^{-2x}+c_2xe^{-2x} $$     (4)
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$$r(x)$$ is given in (1) as a workable solution as:


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$$  \displaystyle r(x)=e^{-2x}\sin{2x} $$     (5)
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From this, it is evident that we may use the basic rule. From Table 2.1, in the sixth line our particular solution then becomes:


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$$  \displaystyle y_p=e^{-2x}(K\cos{2x}+M\sin{2x}) $$     (6)
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Taking the first derivative of (6):


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$$  \displaystyle y'_p=e^{-2x}(-2K\sin{2x}+2M \cos{2x})-2e^{-2x}(K\cos{2x}+M\sin{2x}) $$     (7.a)
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(7.a) can be arranged as:


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$$  \displaystyle y'_p=-2e^{-2x}(K\sin{2x}-M\cos{2x}+K\cos{2x}+M\sin{2x}) $$     (7.b)
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$$  \displaystyle \rightarrow y'_p=e^{-2x}[2(M-K)\cos{2x}-2(M+K)\sin{2x}] $$     (7.c)
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Taking the second derivative of (6):


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$$  \displaystyle y_{p}''=-2e^{-2x}[2(K+M) \cos{2x} -2(K-M) \sin{2x} ]+4e^{-2x}[(K+M) \sin{2x} +(K-M) \cos{2x}] $$     (8.a)
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(8.a) can be arranged as:


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$$  \displaystyle y''_p=-4e^{-2x}[(K+M)\cos{2x}+(M-K)\sin{2x}-(K+M)\sin{2x}+(M-K)\cos{2x}] $$     (8.b)
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$$  \displaystyle \rightarrow y''_p=4e^{-2x}(2K\sin{2x}-2M\cos{2x}) $$     (8.c)
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Since (1) is in a plausible form for the Method of Undetermined Coefficients, we can substitute (6), (7.c), and (8.c) into (1):


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$$  \displaystyle 4e^{-2x}(2K\sin{2x}-2M\cos{2x})+(4)e^{-2x}[2(M-K)\cos{2x}-2(K+M)\sin{2x}]+(4)e^{-2x}(K\cos{2x}+M\sin{2x})=e^{-2x}\sin{2x} $$     (9)
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To solve for K and M easier, we can distribute the terms of 4 on the left hand side of (9):


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$$  \displaystyle e^{-2x}(8K\sin{2x}-8M\cos{2x})+e^{-2x}(8(M-K)\cos{2x}-8(K+M)\sin{2x})+e^{-2x}(4K\cos{2x}+4M\sin{2x})=e^{-2x}\sin{2x} $$     (10)
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Consolidating terms further:


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$$  \displaystyle e^{-2x}[(8K-8K-8M+4M)\sin{2x}+(-8M+8M-8K+4K)\cos{2x}]=e^{-2x}sin{2x} $$     (11)
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Which finally becomes:


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$$  \displaystyle e^{-2x}(-4M\sin{2x}-4K\cos{2x})=e^{-2x}\sin{2x} $$     (12)
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From (12), we can math $$e^{-2x}sin{2}$$ and $$e^{-2x}cos{2}$$ terms to solve for K and M:


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$$  \displaystyle e^{-2x}\sin{2x}: -4M=1 $$     (13.a)
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$$  \displaystyle e^{-2x}\cos{2x}: -4K=0 $$     (13.b)
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From (13.a):


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$$  \displaystyle M=-\frac{1}{4} $$     (14.a)
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From (13.b):


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$$  \displaystyle K=0 $$     (14.b)
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Substituting (14.a) and (14.b) into (6) yields the particular solution to (1):


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$$  \displaystyle y_p(x)=e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (15)
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The complete solution to (1) is the sum of the particular solution to (1) given in (15) plus the general solution to the homogeneous ODE of (1) given in (4):


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$$  \displaystyle y(x)=c_1e^{-2x}+c_2xe^{-2x}+e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (16)
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Taking the first derivative of (16):


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$$  \displaystyle y'=-2c_1e^{-2x}-2c_2xe^{-2x}+c_2e^{-2x}+2e^{-2x}(-\frac{1}{4}\cos{2x})-2e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (17)
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Consolidating terms of (17):


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$$  \displaystyle y'=-2c_1e^{-2x}+(1-2x)c_2e^{-2x}+\frac{1}{2}e^{-2x}(\sin{2x}-\cos{2x}) $$     (18)
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Using the initial conditions given in (1) with (16) and (18) to solve for $$c_1$$ and $$c_2$$:


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$$  \displaystyle y(0)=c_1=1 $$     (19.a)
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$$  \displaystyle -2c_1+c_2-\frac{1}{2}=-1.5 $$     (19.b)
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$$  \displaystyle \rightarrow c_2=1 $$     (19.c)
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Thus we have the complete solution to the nonhomogeneous ODE given in (1). Substituting (19.a) and (19.c) into (16) yields:


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$$  \displaystyle y(x)=e^{-2x}+xe^{-2x}+e^{-2x}+e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (20)
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Which can be arranged for the final solution as:


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$$  \displaystyle y(x)=e^{-2x}(2+x-\frac{1}{4}\sin{2x}) $$
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