find the general solution of 3yy0 a yc1ex3c1xex3 b ycex3 c y

find the general solution of 3y\"+y\'=0

a) y=c₁e^-x/3+c₁xe^-x/3

b) y=ce^-x/3

c) y=c₁+c₂e^-x/3

Solution

given equation : 3y\" + y\' =0

the auxiliary equation is 3r² +r =0

the roots for above equation is : r (3r +1) =0

whose roots are r =0 , r = -1/3

the general solution for differential equation is y = c₁ + c₂ e^(-x/3)

there fore, the option is c