diffeqtest2review

# diffeqtest2review - Section 3.4 Reduction of Order...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

## 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.

Ask a homework question - tutors are online