For the differential equation y\" + 12y\' + 36y = 0 Find characteristic polynomi

Question

For the differential equation y" + 12y' + 36y = 0 Find characteristic polynomial & its zeroes. Find a solution of the form y = e^rt Use the substitution w = dy/dt + 6y to find the general solution of this differential equation. Find the solution to the differential equation with initial conditions y(0) = 0 and y'(0) = 1. Find the t-value at which the solution of part (d) hits its maximum value on the t-interval [0, infinity]. For the differential equation t^2y" + 14ty' + 36y = 0 Find two linearly independent solutions of the form y = t^r

Explanation / Answer

a)

For characteristic polynomial we assume solution of the form: e^{rt}

Substituting gives

r^2+12r+36=0   , THis is the characteristic polynomial

(r+6)^2=0

r=-6

So repeated roots

b)

e^{-6t} is one solution as we found r=-6 is a solution to the characteristic equation

c)

w=y'+6y

y'=w-6y

w'=y''+6y'=y''+6(w-6y)

y''=w'-6w+36y

Substituting gives

w'-6w+36y+12(w-6y)+36y=0

w'-6w+12w=0

w'+6w=0

w'=-6w

Solution to this equation is

w=Ae^{-6t}

y'+6y=Ae^{-6t}

(y'+6y)e^{6t}=A

(ye^{6t})'=A

Integrating gives

ye^{6t}=At+B

y=e^{-6t}(At+B) is the general solution

d)

y(0)=0

0=A*0+B

HEnce B=0

y(t)=Ate^{-6t}

y'(t)=e^{-6t}(A-6At)

y'(0)=A=1

y(t)=te^{-6t}

e)

y'(t)=e^{-6t}(1-6t)

Maximum is obtained at y'=0

y'=0 gives

1-6t=0

t=1/6

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