Find the general solution of the higher order and solve the

Find the general solution of the higher order and solve the given initial value. Please show work, thank you.

m^3+12m^2+36m=0    y(0)= 0 , y\'(0)= 1 , y\"(0)= -7

Solution

WE are given the characteristic equation

m^3+12m^2+36m=0

m(m+6)^2=0

SO general solution is

y(x)=A +exp(-6x)(B+Cx)

y(0)=A+B=0 ie A=-B

y(x)=B(-1+exp(-6x))+Cx exp(-6x)

y\'(x)=-6B exp(-6x)+C exp(-6x)(-6x+1)

y\'(0)=-6B+C=1

Hence, C=1+6B

y(x)=B(-1+exp(-6x))+(1+6B) x exp(-6x)

y\'\'(x)=12e^(-6 x) (18 B x-3 B+3 x-1)

y\'\'(0)=12(-3B-1)=-7

3B+1=7/12

3B=-5/12

B=-5/36

y(x)=-5(-1+exp(-6x))/36+(1-5/6) x exp(-6x)

y(x)=-5(-1+exp(-6x))/36+x exp(-6x)/6