Question

7. Given the initial-value problem y" + 3y' + 2y = 4x2, y(0) = 3, y'0) = 1, a. Find its homogeneous solution using the Constant Coefficient approach (10pts) b. Find is particular solution using the Annihilator method. (10pts) c. Find the general solution that satisfies the initial conditions. (5pts)

Answer 1

7 . using the Given the initial - Value problem y'+3y'+2y=4m" Y(0) = 3, Y'(0) = 1 (a) find its homogeneous solution constant coefficient approach 6) find it particular solution using the Annihitator method. (@ find the general solution that satisfies the initial condition: A la ® Given differential equation Y"+34'424 = 402 thus the homogeous equation of constant Y"+By '+ 2y = 0 (62 Auxiliary equation 2 +3D+297 +38+2 =0 rn +28+8+2 = 0 solving 3) (+22+1(°+2) = (3+2)(ru) = 0 7= -2, 17 S-1

AtB n tear therefore the complem mentary Ye (n = qe solution (funetion) -an teek b particular solution of Annihilator method B taca ac Yo (n) and Ycm) : From the equation (1) Y" + 3y + 2y = 402 then is ( T 2c +31 B+2CX) + 2 ( A+But cuy=442 20+ 3B+ 6ck t2 A+2B2+2 on=402 =) (2C+36 +2A): + (60+23) * +26nº=482 comparin 20=4 a) le=2 eing 2e + 3B+2 Aco and 6c+2B=0 5 2) 2B = ä-be B -6X Z 2 TB -6 therefore 11 7 2 A E20-3B A- 282-3X-6 2 - 4 + 18 2 A=7 А - 쁠 2 ㅖ Heree Yplay 2 7 -62 +24? therefore the particular solution is Yp(") = 7-6x tan (Ans):

General solution of the equation is yeu); Ye (M) + Yp (m) Ciec. F +P. I) y you) = (a een tegen + #-64+24²) use the initial condition YO) 23 and y'(o) - 1 thus y'(") = -za eza eza_qē"_6+40 - ' . = ) y (o)=3 atq+7-6.0+0=3 >) =) fi) T =) Ct2 = 3-7 =-4 (4 +2=-4 And yl (0)=1 -29 -62 - 6 = 1 -29-C=7 5) 26 +22=-7] solving (i) and (ii) we get, . ((W) subtraction o0) 201 + 4 =-7 a +42=-4 -7+4 =) TG = -3 (11) G G = - 4-4 =-4+3 = = -1 = G = -1 therefore the required general solution sy(") [-3 e 2x en +7-6x + 2x² (Ans).