Question

Step (1) "Rules for Guessing": 1. If g(x) = ekr, guess that yp(x) = Aekx then find A 2. If g(x) = sin(kx), guess that yp(x) = A cos(kt) + B sin(kt) then find A and B 3. If g(x) = cos(kx), guess that yp(x) = A cos(kt) + B sin(kt) then find A and B 4. If g(x) is a polyonomial of degree n, guess that yp(a) = A + AUX + A2r2 + ... + A," then find Ao, ...,A. 5. If g(x) = 91(2) + 92(x) is a sum of two functions, guess that yp(a) = guess for g(x) + guess for 92(x) 6. If g(x) = g(x)g() is a product of two functions, guess that Up(s) = (guess for gi(2)) (guess for 92(x)) 7. If you make a proper guess but it doesn't work out, then set new guess = v(old guess) 1. Find the general solution for each of the following differential equations (10 points each): (a) (b) y" - 2y - 3y = 3e21 y"-W'- 2y = -2x + 4x2 y"+y' - y = 12e3x + 12e-20 y" - 24 -3y = 3xe y" + 2y' + y = 2* [Hint: you'll use Rule 7. at least once (d)

solve all number 1 for me pls .

Answer 1

d² (ex) - 2d (&) - 3 0 - 0 2 x2 dx² dx an 22 substitute d² => e dx² xx and da xe - 2xe - 3 é (ex- 21 factor out ex (x-2x - 3) et -

xx e * O finite x, Since for any the zeros must come from the polynomial: x²–2x-3=0 factor: (x-3) (x+1)=0 solve for a 1 = -1 or x=3

-2 as a 37 The root x=-1 gives y(x)=c, e solution where G1 ; arbitrary constant The root x=3 gives 9₂(x) = Cze as a solution, where C2 ; arbitrary constant The general solution is the sum of above solutions yx) = 4, (x) + 4(x) = C, e + C₂ e 32

d 2x dy p(x) dy p(x) d (a, ex dx ax 2age d x2 d²4p(x) ď (a, en dx² da² =4a,e 2 2x

ticular Substitute the part solution Y, (2) into the differential equation: dyp (x) - 2 dy p(x) -39, (x) = 3 * da? da 49, 6* _ 2 (20, 0) – 3(a,e) = 302* Simplify: -3a, Зе 2x 2x Equate the coefficients of ex on both sides of the equation, -39, = 3

2x 2x е e² Solve the equation: 9, = -1 Substitute a, into 9p(x) = a, e Yp(x) The general solution is: 4x)= x (x) + (x) = - et cie + C₂ e Yp 20 32

2. 2 x2 and Substitute d² dx² lee)- ne d (ex) = x x d j de

t² AX x x té XX 2 e o Factor out e λα =.0 xa (x²-x-2) et Since et to for any finite the zeros must come from the polynomials 1²-9-2=0 factor: (4-2) (x+1)=0 Solve for ti i = -1 or x = 2

х 22 The root >= 1, y, (x) = Cie The root x=2, 4₂ (%) = Cze y (x)=.y, (x) + ₂(x) -x 2x cie + Cze 2 Determine the particular solution to dy (2) - dyln) - 256%) -26%) die dx² dx 4x² - 2x by the method of andetermined coefficients:

The particular solution to d²y (x) - dj(x) – 2yex) dx² da = 4x² - 2x is of the form: 9p(x) = a + ax + az x 2 Solve for the unknown constants ai, az and azi compute dy p(x): da dyp(x) dx da d ca, tazxto a3x²) - azt zaza

The particular solution to d²y (x) - dj(x) – 2yex) dx² da = 4x² - 2x is of the form: 9p(x) = a + ax + az x 2 Solve for the unknown constants ai, az and azi compute dy p(x): da dyp(x) dx da d ca, tazxto a3x²) - azt zaza

Compute d² up (2): dx² 9 p(x) da? dx² 2 da cari + a2x + Az x²) 2 = 223 Substitute the particular solution 4p(x) into the differential equation: Con

42²_22 2 both sides of the equation: dyp (x) - cup (x) - 2 0p(x) = 4x²2x da 2az - (A2+ 29₂x)-2(a,+Q₂x + Az x²) Simplify: - Q2 +2a3+ (-292-293) * -zaz -2x + 4x² azaz-zaz = -2 x zal on Equate the coefficients of a

al Substitute ai, az and az into yp(a) Equate the coefficients of x on both Sides of the equation : - 2az = 4 solve the system: -72 A2=3 Az = -2 =altxa2 + x²az: Sp (x) = -2x² + 3x - 72 yca)= YEx) + Up (x) = -2x² + 3x + cie" +CaC -7 2x 2