Answer to Question #275853 in Differential Equations for mary

Question

1.    y''' - 9y''+ 27y'-27y=0        ans, y=c₁e^3x+c₂xe^3x+c₃x²e^3x

2.    y''' + 2y''- 13y' + 10y =0    ans, y= c₁e^-5x+c₂e^2x+c₃e^x

Answer 1

 Q1)

y''' - 9y'' + 27y' - 27y = 0

Let us substitute y = e^mx .

y' = me^mx , y'' = m²e^mx and y''' = m³e^mx

So the auxiliary equation is

m³ - 9m² + 27m - 27 = 0

=> m³ - 3m².3 + 3m.3² - 3³=0

=> (m-3)³=0

=> m = 3,3,3

So the general solution of the differential equation is y = c₁e^3x + c₂xe^3x + c₃x²e^3x

Q2)

y''' + 2y''-13y'+10y = 0

Let us substitute y = e^mx .

y' = me^mx , y'' = m²e^mx and y''' = m³e^mx

So the auxiliary equation is

So m³ + 2m² - 13m + 10 = 0

=> m²(m-1)+3m(m-1)-10(m-1)=0

=> (m-1)(m²+3m-10)=0

=> (m-1)(m²+5m-2m-10)=0

=> (m-1){m(m+5)-2(m+5)}=0

=> (m-1)(m+5)(m-2)=0

=> m = -5, 2, 1

So the general solution is

y = c₁e^-5x + c₂e^2x + c₃e^x