Question

part b please

1. (15 Points Each) Find the solution to the given differential equations satisfying the initial conditions using the method of undetermined coefficients. 1'(0) = 6 (a) (Problem 3.7.28) y" + 3y + 2y = 20 cos 2. y(0) = -1 (b) y" - V - 2y = 4e3+ +6 (0)=0 7(0) = 1

Answer 1

Ques y"y'+62) ye 4e3" 76 --- y(o)= 0 4 4' () = 1 sol: The characteristic equation corresponding to the associated Homogeneous equation as r² &-200 => &=-(-1) $ JT-4X(-2) 2 » r = 1 # 3 Now, the non *8=2 or -1 ou The complementary solution is S dc (x)= q extG ex erhere, a and ca are arbitrary . constants The non- Homogeneous Past is : F(x)= 483% +6. Let yp (x) = ax + b be a solution of 1. Thens Čř (2) = 34 e3x Yp" () = 94 est Setting these equations in , We get : 1. q A ex _ ( 3A 63x) - 2 (A 3* + B) = 40 +6 >> (94–34–2A) e – 2B = 4e3x + 6 7. 4A eBX – 2B = 4634 +6.

Comparing the respective components, We get: - 4A=4 f - 2B=6 ? - -> A=1 and B=-3 So that: yo (2)= é 2-3 oo The general solution is : - y (x) = ciet + cq et + (211_3) y (x) = -ger +2 et + 3e 3X Since, y (o)=0 7 0=9+62 +(1-3) act Cg = 2 - ④ AND, y (0) oo -a +262 +3 = 1 → 9.-26=4 ---O Subtracting 0 € 0, ble get: (4-4).+ (6 +2 (2) = 2-2 7 [C2=0 ] And, a ci +c=2 7/4=25 y (x) = q ex + 8x_3 - (Auswer)