User:Egm4313.s12.team18.roberts.jr

= Josh Roberts =

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Problem
Find a general solution. Check your answer by substitution.


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$$  \displaystyle y''+4y'+(\pi^2+4)y=0 $$     (1)
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Solution
Note This is a homogenous linear 2nd order ordinary differential equation (ODE) with constant coefficients. Trial Solution:
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$$  \displaystyle y=e^{\lambda x} $$ (2)
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$$  \displaystyle y'=\lambda e^{\lambda x} $$ (3)
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$$  \displaystyle y''=\lambda^2 e^{\lambda x} $$ (4) Plugging (2),(3), and (4) into (1)
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$$  \displaystyle (\lambda^2 + 4\lambda + (\pi^2+4))e^{\lambda x} = 0 $$     (5)
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The characteristic equation is
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$$  \displaystyle \lambda^2 + 4\lambda + (\pi^2+4) = 0 $$     (6)
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Note The coefficients of this quadratic equation are
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$$  \displaystyle a=1 $$
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$$  \displaystyle b=4 $$
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$$  \displaystyle c=\pi^2+4 $$
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The discriminant can be found as follows
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$$  \displaystyle \Delta = b^2 - 4ac $$     (7)
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$$  \displaystyle \Delta = 4^2 - 4(1)(\pi^2+4) $$
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Note The discriminant is negative. Therefore, Eq (6) has complex roots of the form


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$$  \displaystyle \lambda = -1/2a \pm i\omega $$     (8) where
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$$  \displaystyle \omega=\sqrt{c- b^2/4} $$     (9)  Now The general solution to the ODE is
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$$  \displaystyle y(x)=e^{-bx/2} (Acos\omega x + Bsin\omega x) $$ (10) Plugging known quantities into (9) and (10)
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$$  \displaystyle y(x)=e^{-2x} (Acos\sqrt{(\pi^2+4)-4}\,\,x + Bsin\sqrt{(\pi^2+4)-4}\,\,x) $$
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$$  \displaystyle y(x)=e^{-2x} (Acos\pi\,x + Bsin\pi\,x) $$
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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 in (3a) p.3-7 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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Exam 1 Material
Date: Tue, Feb 28, 2012

Study Guide
Sec 2.1 - Homogeneous L2-ODEs
 * A 2nd order ODE can be called linear if it can be written as
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$$  \displaystyle y'' + p(x)y'+q(x)y = r(x) $$     (1)
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 * Linear in y and its derivatives.
 * This equation is in standard form.

If $$r(x)\equiv 0$$, then (1) reduces to
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$$  \displaystyle y'' + p(x)y'+q(x)y = 0 $$     (2)
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 * which is called a homogeneous L2-ODE.

If $$r(x)\not\equiv 0 $$, then (1) is called non homogeneous.
 * Superposition Principle (Linearity Principle)
 * For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution
 * of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

Sec 2.2 - Homogeneous Linear ODEs with Constant Coefficients These equations take the form
 * Initial Value Problems
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$$  \displaystyle y'' + ay'+by = 0 $$     (3) Homogeneous solutions to these equations take the form
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$$  \displaystyle y_h'' + ay_h'+by_h = 0 $$     (4) and can be solved by the method of trial solution, also known as the method of undetermined coefficients. The trial solution is
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$$  \displaystyle y = e^{\lambda x} $$ (5) By inserting (5) and its 1st and 2nd derivatives into (4), one can obtain the characteristic equation
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$$  \displaystyle \lambda^2 + a\lambda + b = 0 $$     (6) Finding the roots to the characteristic equation will allow for the homogeneous solution to be found. We know that for a quadratic equation such as (6) that there are 3 kinds of roots, depending on the sign of the discriminant $$a^2 - 4b$$,namely.
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 * (Case 1)  Two real roots if $$a^2 - 4b > 0$$
 * (Case 2)  A real double root if $$a^2 - 4b = 0$$
 * (Case 3)  Complex conjugate roots if $$a^2 - 4b < 0$$

There will be two distinct roots $$\lambda_1 and  \lambda_2$$ Therefore, the solution to (4) will be
 * Case 1
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$$  \displaystyle y = C_1e^{\lambda_1 x} + C_2e^{\lambda_2 x} $$ (7) If the discriminant is zero it is easily seen that we only get one root (double root), hence only one solution.
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 * Case 2
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$$  \displaystyle y = (C_1 + C_2x)e^{-ax/2} $$     (8) If the discriminant is negative, we know that the roots will be complex and of the form
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 * Case 3
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$$  \displaystyle \lambda = -(1/2)a \pm i\omega $$ A basis of real solutions can be obtained that is
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$$  \displaystyle y_1 = e^{-ax/2}cos{\omega x} $$
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$$  \displaystyle y_2 = e^{-ax/2}sin{\omega x} $$ where
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$$  \displaystyle \omega^2 = b - (1/4)a^2 $$ Finally, the general solution is
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$$  \displaystyle y = e^{-ax/2}(Acos{\omega x} + Bsin{\omega x}) $$     (9)
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Exam 2 Material
Date: Tue, Apr 24, 2012