a)Find the general solution of the following problem. y\'\'-8y\'+12y=24t^2-32t+2
ID: 1946545 • Letter: A
Question
a)Find the general solution of the following problem.y''-8y'+12y=24t^2-32t+28
b)Find the general solution of the following problem.
y''+18y'+81y=6e^-9t
Explanation / Answer
a)Find the general solution of the following problem. y''-8y'+12y=24t^2-32t+28 homogeneous solution y''-8y'+12y = 0 ==> m^2- 8 m+12= 0 ==> (m-2)(m-6)=0 ==> m= 2 AND m=6 yh= C1 e^(2t) + C2 e^(6t) paricular solution yp= At^2 +Bt +C y' = 2At+B y'' = 2A y''-8y'+12y = 24t^2-32t+28 2A -8(2At+B ) + 24 (At^2 +Bt +C ) = 24t^2-32t+28 24 A t^2 = 24 t^2 ==> A= 1 t( -16A + 24 B) = -32 t -16 +24 B=-32 ==> 24 B= -16 ==> B= -16 / 24 = - 2/3 2A -16B +24 C = 28 2 + 32/3 +24 C = 28 24 C= 28 - 38/3 = 46/3 C= 23 /36 yp= At^2 +Bt +C = t^2 -(2/3)t + 46/3 general solution: y= yh+ yp = C1 e^(2t) + C2 e^(6t) +t^2 -(2/3)t + 46/3 b)Find the general solution of the following problem. y''+18y'+81y=6e^-9t homogeneous solution y''+ 18y'+81y=0 ==> m^2+18 m+81= 0 ==> (m+9)^2=0 ==> m= -9 double root yh= C1 e^(-9t) + C2 .t . e^(-9t) paricular solution yp= A e^(-9t) y'= -9 A e^(-9t) y'= 81 A e^(-9t) y''+18y'+81y=6e^(-9t) 81 A e^(-9t) +18(-9 A e^(-9t) )+ A e^(-9t) = 6e^(-9t) 0 = 6e^(-9t) SO WE have to Change yp= A t e^(-9t) y' = A e^(-9t) - 9 A t e^(-9t) y'' = -9 A e^(-9t) - 9 A e^(-9t) + 81 t A e^(-9t) = -18 A e^(-9t) + 81 t A e^(-9t) y''+18y'+81y = 6e^(-9t) -18 A e^(-9t) + 81 t A e^(-9t) + 18 (y' = A e^(-9t) - 9 A t e^(-9t) ) +81A t e^(-9t) = 6e^(-9t) 0= 6e^(-9t) Hence WE have to Change yp= A t^2 e^(-9t) y' = 2 A t e^(-9t) - 9 A t^2 e^(-9t) y'' = A e^(-9t) (2-36 t + 81 t^2 ) y''+18y'+81y = 6e^(-9t) A e^(-9t) (2-36 t + 81 t^2 ) +18 ( 2 A t e^(-9t) - 9 A t^2 e^(-9t) ) + 81 (A t^2 e^(-9t) ) = 6 e^(-9t) 2A e^(-9t) = 6 e^(-9t) 2A = 6 A=3 yp= A t^2 e^(-9t) = 3 t^2 e^(-9t) general solution : y(t) = yh + yp = C1 e^(-9t) + C2 .t . e^(-9t) + 3 t^2 e^(-9t)
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