Suppose that the characteristic equation for a differential equation is (r-2)^2(

Question

Suppose that the characteristic equation for a differential equation is (r-2)^2(r-3)^2=0

(a) Find such a differential equation, assuming it is homogeneous and has constant coefficients. Enter your answer using y, y', y'', y''', y'''' for the dependent variable and its derivatives.

(b) Find the general solution to this differential equation. In your answer, use and to denote arbitrary constants, use for the dependent variable, and use for the independent variable. Enter as c1, as c2, etc.

a)What I tried was unfactoring the equation getting r^4-10r^3+37r^2-60r+36.
I then took the derivative four times and adding like terms each time.
y=r^4-10r^3+37r^2-60r+36.
y'=4r^3-30r^2+74r-60
y''=12r^2-60r+74
y'''=24r-60
y''''=24
Adding like terms I got r^4-6r^3+19r^2-22r+14
giving me
y''''-6y'''+19y''-22y'+14y=0
But it's saying that I got it incorrect?

b) can't do b without a

Explanation / Answer

(r-2)^2(r-3)^2=0 so here expand=(r^2-4r+4)(r^2-6r+9)=0 (r^4-6r^3+9r^2-4r^3+24r^2-36r+4r^2-24r+36)=0; (r^4-10r^3+37r^2-60r+36)=0 put y^4 for r^4 and subsequently this will be the eqaution; y^4-10y^3+37y^2-60y^1+36y=0 u got the whole concept of characteristic equation wrong the equation represents the characteristic eqn, and solution will be; y=Ae^(2x)+Bx(e^2x)+Ce^(3x)+Dxe^(3x)(because the roots are equal and it is 2)

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