User:Egm4313.s12.team18/R2

=Report 2=

Problem
Given the two roots and the initial conditions:
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$$  \displaystyle \lambda_1=-2\qquad\qquad\lambda_2 = +5 $$
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$$  \displaystyle y(0)=1\qquad\qquad y'(0)=0 $$ Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x)=0$$.
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Consider no excitation:
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$$  \displaystyle r(x)=0 $$
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Plot the solution.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values above as the 2 roots of the corresponding characteristic equation.

Solution

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$$  \displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) $$     (2.1.1)
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Now Insert the given roots into (1)
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$$  \displaystyle (\lambda-(-2))(\lambda-5) $$     (2.1.2) By the FOIL method,  (2) gives the characteristic equation for the unknown L2-ODE-CC.
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$$  \displaystyle \lambda^2-3\lambda-10=0 $$     (2.1.3)
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From this characteristic equation, (3), we can find the non-homogeneous L2-ODE-CC.
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$$  \displaystyle y''-3y'-10=r(x) $$     (2.1.4)
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Now It is known from the given roots that the homogeneous L2-ODE-CC has the solution
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$$  \displaystyle y_H(x)=C_1e^{-2x} + C_2e^{5x} $$     (2.1.5)
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Therefore, the overall solution is
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$$  \displaystyle y(x)=y_H(x)+y_P(x) $$     (2.1.6)
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$$  \displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}+y_P(x) $$     (2.1.7) where $$y_P(x)$$ is the particular solution.
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Now Using the given initial conditions
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$$  \displaystyle y(0)=1\qquad y'(0)=0 $$
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$$  \displaystyle y(0)=C_1+C_2+y_P(0)=1 $$     (2.1.8) Differentiating $$y(x)$$ and then inserting the initial condition.
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$$  \displaystyle y'(x)=-2C_1e^{-2x}+5C_2e^{5x}+y_P'(x) $$     (2.1.9)
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$$  \displaystyle y'(0)=-2C_1+5C_2+y_P'(0)=0 $$     (2.1.10)  Now Multiplying (2.1.8) by 2 and adding it to (2.1.10)
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$$  \displaystyle (2C_1+2_C2+2y_P(0)=2) + (-2C_1+5C_2+y_P'(0)=0) $$
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$$  \displaystyle C_2=\frac{2}{7}-\frac{2}{7}y_P(0)-\frac{1}{7}y_P'(0) $$     (2.1.11) By inserting (2.1.11) into (2.1.8) and combining like terms $$C_1$$ is also found.
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$$  \displaystyle C_1=\frac{5}{7}-\frac{5}{7}y_P(0)+\frac{1}{7}y_P'(0) $$     (2.1.12)  Now The overall solution $$y(x)$$ is given in terms of the initial conditions.
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$$  \displaystyle y(x)=\frac{1}{7}[(5-5y_P(0)+y_P'(0))e^{-2x}+(2-2y_P(0)-y_P'(0))e^{5x}] $$     (2.1.13)
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Now Considering no excitation


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$$  \displaystyle r(x)=0 $$ Which causes
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$$  \displaystyle y_P(x)=0\,\,,\,\,y_P'(x)=0 $$ So
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$$  \displaystyle C_1=\frac{5}{7} $$
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$$  \displaystyle C_2=\frac{2}{7} $$ The homogeneous solution to the L2-ODE-CC is therefore
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$$  \displaystyle y_H(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$
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This solution will now be plotted using MATLAB.



Now Generate 3 non-standard and non-homogeneous L2-ODE-CC that satisfy the roots given in the problem statement.
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$$  \displaystyle 2y''-6y'-20=r(x) $$
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$$  \displaystyle \frac{3}{2}y''-\frac{9}{2}y'-15y=r(x) $$
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$$  \displaystyle 3y''-9y'-30y=r(x) $$
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Author
Solved and typed by - Egm4313.s12.team18.roberts.jr 18:28, 7 February 2012 (UTC)

Problem
Initial Conditions: $$y(0)=1 \, \ y'(0)=0$$ No excitation: $$r(x)=0$$ Find and plot the solution for L2-ODE-CC (4) p.5-5.:
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$$ \displaystyle y''-10y'+25y=r(x) $$
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Solution
Replacing y with $$\lambda$$:
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$$ \displaystyle \lambda^{2}-10\lambda+25=0 $$     (1)
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This can be factored to:


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$$ \displaystyle (\lambda-5)^{2}=0 $$     (2)
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Meaning there is a real double-root of
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$$ \displaystyle \lambda=5 $$     (3)
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Therefore the equations can be set up as:


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$$ \displaystyle y_{g}=C_{1}e^{\lambda x}+C_{2}xe^{\lambda x} $$ (4)
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Using the double-root value for lambda:


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$$ \displaystyle y_{g}=C_{1}e^{5x}+C_{2}xe^{5x} $$     (5) Using the initial conditions given, solve for C1 and C2: y(0)=1 Using values: y=1, x=0
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$$ \displaystyle 1=C_{1}e^{5(0)}+C_{2}(0)e^{5(0)} $$     (6)
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$$ \displaystyle 1=C_{1} $$     (7)
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Using the derivative of y'(0)=0: y'=0, x=0, C1=1


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$$ \displaystyle y_{g}=C_{1}e^{5x}+C_{2}xe^{5x} $$     (8)
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$$ \displaystyle {y_{g}}'=5C_{1}e^{5x}+C_{2}e^{5x}(5X+1) $$     (9) Inputting values:
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$$ \displaystyle 0=5(1)e^{5(0)}+C_{2}e^{5(0)}(1) $$     (10)
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Therefore:
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$$ \displaystyle 0=5+C_{2} $$     (11)
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$$ \displaystyle C_{2}=-5 $$     (12) Final Solution:
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$$ \displaystyle y_{g}=e^{5x}-5xe^{5x} $$     (13)
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Author
Solved and typed by - Egm4313.s12.team18.alvinb 13:53, 6 February 2012 (UTC)

Problem 2.3
Solve problems 3 and 4 on page 59 in the textbook, Kreyszig 2011. "Find a general solution for the ODE and check by substitution."

3.$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$

4.$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$

Solution
Background information:

For homogeneous linear ODE's with constant coefficients, a general equation can be described as, $$ \displaystyle {y{}''+ay{}'+by=0.} $$

The function $$ \displaystyle {y=e^{\lambda x}} $$ may be tried as a solution for the general equation above.

After substituting this equation and its derivatives: $$ \displaystyle {y{}'=\lambda e^{\lambda x}} $$ and $$ \displaystyle {y{}''=\lambda ^{2}e^{\lambda x}} $$ into the general equation,the result is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0} $$

and the characteristic equation is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)=0} $$ which means that if $$ \displaystyle {\lambda} $$ is a solution of this characteristic equation, then the exponential function is a solution of the general equation. If we solve this characteristic equation for $$ \displaystyle {\lambda} $$, the result is$$ \displaystyle {\lambda _{1,2}=\frac{1}{2}(-a\pm \sqrt{a^{^{2}}-4b})} $$

Therefore there are 3 cases when it comes to finding the solution to Homogeneous Linear ODE's with CC:

Case 1:    Two real roots if  $$ \displaystyle {a^{^{2}}-4b>0} $$

Case 2:    A real double root if $$\displaystyle {a^{^{2}}-4b=0} $$

Case 3: Complex conjugate roots if $$\displaystyle {a^{^{2}}-4b=0} $$

-Problem 3 page 59 :
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$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$ (1) The discriminant for this equation (1) is:
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$$ \displaystyle { \sqrt{6^{^{2}}-4(8.96)}=.16} $$ which is >0 and therefore a case 1 (2)
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$$ \displaystyle {\lambda _{1}=-2.8, \lambda _{2}=-3.2} $$ and $$ \displaystyle {y _{1}=e^{-2.8x}, y _{2}=e^{-3.2x}} $$ (3)
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The general solution to (1) is


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$$ \displaystyle {y=c_{1}e^{-2.8x}+c_{2}e^{^{-3.2x}}} $$ (4) Check this solution by plugging in
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$$ \displaystyle {y _{1}=e^{-2.8x}} $$ and $$ \displaystyle {y _{2}=e^{-3.2x}} $$ (5) into (1). The results for plugging in the above equations(5) and their derivatives:
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$$ \displaystyle {y_{1}=e^{-2.8x},y_{1}{}'=-2.8e^{-2.8x},y_{1}{}''=7.84e^{-2.8x}} $$ and $$ \displaystyle {y_{2}=e^{-3.2x},y_{2}{}'=-3.2e^{-3.2x},y_{2}{}''=10.24e^{-3.2x}} $$ (6) into equation (1) respectively are:
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$$ \displaystyle {e^{-2.8x}(7.84-(6)(2.8)+8.96)=0} $$ and $$ \displaystyle {e^{-3.2x}(10.24-(6)(3.2)+8.96)=0} $$
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(7)
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Both of these reduce to
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$$ \displaystyle {0=0} $$ (8)
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Problem 4 page 59:
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$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$ (9)
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The discriminant for this equation is:


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$$ \displaystyle {4^{2}-4(\pi ^{2}+4)=-4\pi^{2}} $$ (10) which is a case 3 and solutions to the characteristic equation are:
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}a+i\omega, \lambda _{2}=-\frac{1}{2}a-i\omega} $$ (11) and our particular solutions for characteristic equation for this problem are:
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}(4)+i\pi, \lambda _{2}=-\frac{1}{2}4-i\pi} $$ (12)
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The general solution to (9) is


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$$ \displaystyle {y=e^{-\frac{4}{2}x}(A\cos \pi x+Bsin\pi x)} $$ (13) Check this solution by plugging in
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$$ \displaystyle {y _{1}=e^{-\frac{4}{2}x}\cos \pi x} $$ and $$ \displaystyle {y _{2}=e^{-\frac{4}{2}x}\sin \pi x} $$ (14) into (1). The results for plugging in the equations (14) and their derivatives:
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$$ \displaystyle {y_{1}=e^{-2x}\cos \pi x, y_{1}{}'=-\pi e^{-2x}-2e^{-2x}\cos \pi x, y_{1}{}''=e^{-2x}((4-\pi ^{2})\cos \pi x+4\pi \sin \pi x)} $$ (15)
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and
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$$ \displaystyle {y_{2}=e^{-2x}\sin \pi x, y_{2}{}'=e^{-2x}(\pi \cos \pi x-2\sin \pi x), y_{2}{}''=e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)} $$ (16)  into equation (1) respectively are:
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$$ \displaystyle {e^{-2x}((4-\pi ^{2})\cos \pi x+4\sin \pi x+4(-\pi \sin \pi x-2\cos \pi x)+(\pi ^{2}+4\cos \pi x)=0} $$ (17)
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$$ \displaystyle {e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)+4e^{-2x}(\pi \cos \pi x-2\sin \pi x)+(\pi ^{2}+4)e^{-2x}\sin \pi x=0} $$ (18) Both of these reduce to
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$$ \displaystyle {0=0} $$ (19) -
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Author
Solved and typed by - Egm4313.s12.team18.windham 22:41, 7 February 2012 (UTC)

Problem
K 2011 p.59 pb 5. Find a general solution. Check your answer by substation.


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$$  \displaystyle y'' + 2{\pi}y' +{\pi ^2}y = 0
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$$     (1)
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Solution
This is a homogeneous linear ODE with constant coefficients.
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$$  \displaystyle a=2\pi $$     (2)
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$$  \displaystyle b=\pi ^2 $$     (3)
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$$  \displaystyle a^2 -4b $$     (4)
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Substitution equations 2 and 3 into equation 4 we see that
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$$  \displaystyle a^2 -4b = (2 \pi) ^2 - 4( \pi ^2)= 4 \pi ^2 -4 \pi ^2 = 0 $$     (5)
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Because equation 5 is equal to 0, the solution is a real double root. Therefore the solution will be in the form
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$$  \displaystyle y = (c_1 + c_2 x)e^ {\lambda x}
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$$     (6)
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Where
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$$  \displaystyle \lambda = -\frac{a}{2}= -\frac{2 \pi}{2} = -\pi
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$$     (7)
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By combining equations 6 and 7 we see that the general solution is
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$     (8)
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In order to check the solution, we take the first and second derivatives of equation 8


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$$  \displaystyle y' = c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}
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$$     (9)
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$$  \displaystyle y'' = -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x}
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$$     (10)
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After taking the derivates, we substitute equations 8, 9, and 10 back into equation 1


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$$  \displaystyle -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x} + 2\pi (c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}) + {\pi} ^2 ( (c_1 + c_2 x)e^ {\lambda x}) = 0
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$$     (11)
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Equation 11 is true. Therefore,
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$
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Is a general solution to equation 1.

Solved and typed by - Egm4313.s12.team18.burgher.wb 14:28, 7 February 2012

Problem
K 2011 p.59 pb 6. Find a general solution. Check your answer by substation.


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$$  \displaystyle 10y'' - 32y' +25.6y = 0
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$$     (1)
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Solution
This is a homogeneous linear ODE with constant coefficients.
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$$  \displaystyle a=10 $$     (2)
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$$  \displaystyle b=-32 $$     (3)
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To find the value of the discriminant we use the equation
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$$  \displaystyle a^2 -4b $$     (4)
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By substitution equations 2 and 3 into equation 4 we see that
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$$  \displaystyle a^2 -4b = 10^2 - 4(-32)= 100 +128 = 228 $$     (5)
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Because equation 5 is greater than 0, the solution is has two real roots. Therefore the solution will be in the form
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$$  \displaystyle y = c_1 e ^ {\lambda_1 x} + c_2 e ^ {\lambda_2 x}
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$$     (6)
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Where
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$$  \displaystyle \lambda_1 = \frac{1}{2} (-a + \sqrt{a^2 -4b})
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$$
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$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + \sqrt{10^2 -4(-32})
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


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

$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + \sqrt{100 +128})
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


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

$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + 15.0997) = 2.55
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

And


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

$$  \displaystyle \lambda_2 = \frac{1}{2} (-a - \sqrt{a^2 -4b})
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


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

$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - \sqrt{10^2 -4(-32})
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


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

$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - \sqrt{100 +128})
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


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

$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - 15.0997) = -12.55
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

By substitution \lambda_1 and \lambda_2 into equation 6 we see that the general solution is
 * {| style="width:100%" border="0"

$$  \displaystyle y = c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (7)
 * <p style="text-align:right">
 * }

In order to check the solution, we take the first and second derivatives of equation 7


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

$$  \displaystyle y' = 2.55 c_1 e ^ {2.55 x} - 12.55 c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (8)
 * <p style="text-align:right">
 * }


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

$$  \displaystyle y'' = 6.5025 c_1 e ^ {2.55 x} + 157.5025 c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (9)
 * <p style="text-align:right">
 * }

After taking the derivates, we substitute equations 7, 8, and 9 back into equation 1


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

$$  \displaystyle 10 (6.5025 c_1 e ^ {2.55 x} + 157.5025 c_2 e ^ {-12.55 x}) - 32 (2.55 c_1 e ^ {2.55 x} - 12.55 c_2 e ^ {-12.55 x}) + 25.6 (c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}) = 0
 * style="width:95%" |
 * style="width:95%" |

$$     (10)
 * <p style="text-align:right">
 * }

Equation 10 is true. Therefore,
 * {| style="width:100%" border="1"

$$  \displaystyle y = c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Is a general solution to equation 1.

Solved and typed by - Egm4313.s12.team18.burgher.wb 14:28, 7 February 2012

Problem
K 2011 p.59 problems 16 and 17.

Solution
Find an ODA of form $$y''+ay'+by=0$$ for the given basis: #16
 * {| style="width:100%" border="0"

Given: $$ \displaystyle e^{2.6x}, e^{-4.3x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

From the given, we can see that this is a scenario of two distinct roots of form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Introducing a variable $$c$$ to simplify calculations:


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

$$  \displaystyle c=\sqrt{a^2-4b} $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting (3) into (1) and (2):
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=\frac{1}{2}(-a+c)=2.6 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \lambda_2=\frac{1}{2}(-a-c)=-4.3 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From (4) and (5) we have two equations and two unknowns. Begin by solving for $$a$$ via adding (4) and (5):


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

$$  \displaystyle -a=-1.7 $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \rightarrow a=1.7 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting the value from (6) into (4) and solving for $$c$$:


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

$$  \displaystyle \frac{1}{2}(-a+c)=2.6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \rightarrow c=6.9 $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting the value from (7) into (3) and solving for $$b$$:


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

$$  \displaystyle 6.9=\sqrt{1.7^2-4b} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle b=\frac{1.7^2-6.9^2}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \rightarrow b=-11.18 $$     (8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting the values from (6) and (8) into the general form yields:


 * {| style="width:100%" border="2"

$$  \displaystyle y''+1.6y'-11.18y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

#17
 * {| style="width:100%" border="0"

Given: $$ \displaystyle e^{-\sqrt{5}x}, xe^{-\sqrt{5}x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

From the given we can see that this is a scenario of a real double root:


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

$$  \displaystyle \lambda=-\frac{1}{2}a $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle a^2-4b=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From the given basis:


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

$$  \displaystyle e^{\frac{-ax}{2}} $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Setting (3) equal to (1) and solving for $$a$$:


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

$$  \displaystyle e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \frac{a}{2}=\sqrt{5} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \rightarrow a=2\sqrt{5} $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solving for $$b$$ by substituting (4) into (2):


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

$$  \displaystyle {2\sqrt{5}}^2-4b=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle b=\frac{{2\sqrt{5}}^2}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


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

$$  \displaystyle \rightarrow b=\sqrt{5} $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting the results from (4) and (5) into the general form yields:


 * {| style="width:100%" border="2"

$$  \displaystyle y''+2\sqrt{5}y'+5y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Problem
Realize spring-dashpot-mass systems in series as shown in Fig. p.1-4 with the similar characteristic as in (3) pp.5-5, but with double real root $$\lambda=-3$$, i.e., find the values for the parameters $$k,c,m$$.

Equation for Spring-Dashpot-Mass System:
 * {| style="width:100%" border="0"

$$ \displaystyle My_{k}''+\frac{Mk}{c}y_{k}'+ky=f(t) $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Given the real double-root$$\lambda=-3$$:
 * {| style="width:100%" border="0"

$$ \displaystyle (\lambda+3)^{2}=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \lambda^{2}+6\lambda+9=0 $$     (3) Therefore:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle M=1 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{Mk}{c}=6 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle k=9 $$     (6) Solving for all unknowns: M=1 k=9 c=1.5
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Author
Solved and typed by - Egm4313.s12.team18.alvinb 14:20, 6 February 2012 (UTC)

Problem
Develop the MacLaurin series (Taylor series at T=0) for: $$e^t \, \ \cos t \ , \ \sin t $$

Problem
K 2011 p. 60 problems 8 and 15. These problems give you an equation and ask for the general solution. Once the general solution has been found, use substitution to check the answer.

Problem 8) $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

Problem 15) $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
Problem 8)
 * $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:


 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Or
 * $$\displaystyle a^{2}-4b=-12 $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:


 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{3.25-\frac{1}{4}(1)^{2}}) \rightarrow \omega=\sqrt{3} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{3}$$
 * $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking the solution by using substitution:


 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$


 * $$\displaystyle y^{'}=\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{-2}+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$


 * $$\displaystyle y^{''}=-\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{4}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$


 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$


 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-x/2}Bsin(\sqrt{3}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Problem 15)
 * $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:


 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Or
 * $$\displaystyle a^{2}-4b=-4\pi $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:


 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{(0.0729-\pi)-\frac{1}{4}(0.54)^{2}}) =\sqrt{\pi} \rightarrow \omega=\sqrt{\pi} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{\pi}$$ $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking the solution by using substitution:


 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$


 * $$\displaystyle y^{'}=-0.27{e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))}+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$


 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$


 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$


 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-0.27}Bsin(\sqrt{\pi}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Author
Solved and typed by - Egm4313.s12.team18.gibson.mg 18:22, 7 February 2012 (UTC)

Problem
Initial conditions: $$y(0)=1 \, \ y'(0)=0$$ No excitation:$$r(x)=0$$ Find and plot the solution for L2-ODE-CC corresponding to (1). In another Fig. superpose 3 Figs.: (a) this Fig., (b) the Fig in R2.6 p..5-6, (c) the Fig. in R2.5 p.3-6.