diffeqtest2review - Section 3.4 Reduction of Order...

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Unformatted text preview: Section 3.4 Reduction of Order Solve for the characteristic equation. If the roots are r1,r2 y = C1er1t + C2er2t R1 and r2 complex roots, \+iu y = C1e\tcos(ut) + C2e\tsin(ut) R1=R2 y = C1er1t + C2ter2t Let y = v(t)*(solution already given) Take derivatives y’ and y’’. Plugging into the original equation. Using that t > 0 we use the equation: V’’ = 0 integrating twice we get: V = c1t + c2 Use c1 = 1, c2 = 0 for best C’s possible Plug v(t) into y(t) Second solution is found to be y(t) Section 3.5 Undetermined Coefficients The characteristic polynomial has roots # and # The solutions to the homogenous/complimentary problem are: Yh = c1e3t + C2e ­t To find a particular solution we try the form: Y(t) = Acos(2t) + Bsin(2t) And take the first and second derivatives Plug into the original equation. Solve for A and B, getting rid of “ghosts”. Once we solve for A and B, we add the particular and the homogenous/complimentary equations. For Equations with a g(x) =  ­3te ­t Try Ate ­t, then Ate ­t + Bt2e ­t Section 3.6 Variations of Parameters First we consider the homogenous equation: Y’’ + 4y’ + 4y = 0 Using the characteristic equation: R2 + 4R + 4 = 0 Solve for R. Then its complimentary solution is Yc(t) = C1e ­2t + C2te ­2t Thus we let y (t) = U1(t) e ­2t + U2 te ­2t Be a solution of Y’’ + 4y’ + 4y = t ­2e ­2t Next solve the wronskian: using U1, U2, U’1, U’2 Then integrate U1 =  ­ [y2g(x) / W(t)] U2 = [y1g(x)/ W(t)] Hence the particular solution becomes: Plug in U1 and U2 The general solution is the particular + complimentary Reminders: 1/(t2 + 16) = 1/4arctan(t/4) 1/(t2 ­25 ) = 1/10 int (1/t ­5) – 1/10 int(1/t+5) * separated w/ partial fractions Section 3.7 Mechanical and Electrical Vibrations No Damping The mass m = 100 g Gravity = 9.8 Stretch L = 5 cm = 5E ­2 m The spring constant k = mg/L = 100 * 196 g/s^2 Let u(t) be the position of the mass at time t. Thus we have: This is a second order homogeneous equation. Mu’’ + ku = 0 U(0) = 0, u’(0) = 10 cm/s Substituting our known information into the equation we have: “Equation” Its characteristic equation is : “equation” Solving for its roots we find r = + ­ 14 I The general solution of the equation is U(t) = Acos14t + Bsin14t We have the initial conditions and we can take the derivative twice. Setting the equations equal to the initials, we solve for A and B. The position of the mass at any time t is given by U(t) = equation U=Rsin(wt) = Rcos(wt ­pi/2) Period = 2pi/w Will return to equilibrium position at .5t Damping: Define mass, spring constant, damping coeff., and external force acting on mass The equation of motion is given by: Mu’’+ gamma*u + ku = f(t) Where u is measured in m, t in sec. LQ’’ + RQ’ + 1/C(Q) = 0 If weight given in lb. Mass = 8 lb/32 ft/sec^2 = ¼ lb sec^2 / ft K = 8/ (1.5/12) lb/ft Gamma = 2 (m*k)^1/2 R=2(L/C)^1/2 4.1 Higher Order Linear Equations To verify that a function is a solution of the differential equation just take relative derivatives and ensure that when you plug in you get 0=0 or a true statement. 4.2 Homogeneous Equations with Constant Coefficients Finding the general solution: The characteristic equation is: Then r1 = r2 = 1, r3 =  ­1 Therefore the general solution of the equation “ “ is Y (t) = c1et + c2t et + c3 e ­t Where c1 ­c3 are any constants 4.3 Undetermined Coefficients for Higher Order Lin. Equations same as 4.1. Therefore the general solution of the equation “ “ is Y (t) = c1et + c2t et + c3 e ­t Next, we can write a particular solution of “ “ as the sum of particular solutions of the differential equations y’’’ ­y’’ ­y’+y = 2e ­t and y’’’ ­y’’ ­y’+y = 3 For the first equation, since e ­t is a solution of the homogeneous equation, then we can assume that Y1(t) = Ate ­t Find y’,y’’,y’’’and plug in the values into the original equation and solve for A. Y1(t) is a particular solution of “ “ For the second one, y2(t) = 3 is a particular solution, so the general solution of “” is: Y(t) = complimentary + Y1 + Y2 *** if it is e ­t then just add a t in front of it and try it if it is a 4t then make At + A2 your trying solution If it is t2 then make it t2(At2 + At + A ) Section 6.1 & 2 The La Place Transformation Don’t forget partial fractions: 2/(s ­1)(s+4) = A/(s ­1) + B (s+4) 2 = A(s+4) + B(s ­1) solve for A and B. completing the square: 1/(s2 ­2s+2) = 1/(s2 ­2s+1)+(2 ­1) The two is what you had and you keep it to the right The +1 on the left is what you want, so put it there The  ­1 is subtracting out the opposite of what you want Consider the differential equation:” “ And the initial conditions “ “ The characteristic equation is “ “ The homogeneous/complimentary solution is “ “ Taking the La Place Transformation of the equation we obtain L{y’’} ­L{y’} ­2L{y} = 0 Where we have used the linearity of the transform to write the transform of a sum as the sum of the separate transforms. Using the corollary to express L{y’’} and L{y’} in terms of L{y} L(y’’) = s2L(y) –sy(0) – y’(0) L(y’) = sL(y) –y(0) Solve for L(y). Use partial fractions, if necessary. Transform the equation back by using the Corollary from the table. Hence, by the linearity of the Laplace transform, we have “ “ and is therefore the solution of the initial value problem. ...
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This note was uploaded on 02/01/2011 for the course MATH 427K taught by Professor Delallave during the Spring '11 term at University of Texas.

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