a)
y ′ ′ − 8 y ′ + 12 y = 0 y''-8y'+12y=0 y ′′ − 8 y ′ + 12 y = 0 Auxiliary (characteristic) equation
r 2 − 8 r + 12 = 0 r^2-8r+12=0 r 2 − 8 r + 12 = 0
( r − 2 ) ( r − 6 ) = 0 (r-2)(r-6)=0 ( r − 2 ) ( r − 6 ) = 0
r 1 = 2 , r 2 = 6 r_1=2, r_2=6 r 1 = 2 , r 2 = 6 The family of solutions of the given differemtial equation is
y = C 1 e 2 t + C 2 e 6 t y=C_1e^{2t}+C_2e^{6t} y = C 1 e 2 t + C 2 e 6 t
b)
y ′ ′ + 10 y ′ + 25 y = 0 y''+10y'+25y=0 y ′′ + 10 y ′ + 25 y = 0 Auxiliary (characteristic) equation
r 2 + 10 r + 25 = 0 r^2+10r+25=0 r 2 + 10 r + 25 = 0
( r + 5 ) 2 = 0 (r+5)^2=0 ( r + 5 ) 2 = 0
r 1 = r 2 = − 5 r_1=r_2=-5 r 1 = r 2 = − 5 The family of solutions of the given differemtial equation is
y = C 1 e − 5 t + C 2 t e − 5 t y=C_1e^{-5t}+C_2te^{-5t} y = C 1 e − 5 t + C 2 t e − 5 t
c)
y ′ ′ − 8 y ′ + 65 y = 0 y''-8y'+65y=0 y ′′ − 8 y ′ + 65 y = 0 Auxiliary (characteristic) equation
r 2 − 8 r + 65 = 0 r^2-8r+65=0 r 2 − 8 r + 65 = 0
D = ( − 8 ) 2 − 4 ( 1 ) ( 65 ) = − 196 < 0 D=(-8)^2-4(1)(65)=-196<0 D = ( − 8 ) 2 − 4 ( 1 ) ( 65 ) = − 196 < 0
r 1 , 2 = 8 ± − 196 2 ( 1 ) = 4 ± 7 i r_{1,2}=\dfrac{8\pm\sqrt{-196}}{2(1)}=4\pm7i r 1 , 2 = 2 ( 1 ) 8 ± − 196 = 4 ± 7 i The family of solutions of the given differemtial equation is
y = C 1 e 4 t cos ( 7 t ) + C 2 e 4 t sin ( 7 t ) y=C_1e^{4t}\cos(7t)+C_2e^{4t}\sin(7t) y = C 1 e 4 t cos ( 7 t ) + C 2 e 4 t sin ( 7 t )
Comments