Find the general solution of the given differential equation

Find the general solution of the given differential equation y\" + 9y\" + 27y\' + 27y = 0 y\" + 4y\" - 3y\' - 18y = 0 y^(4) + 10y\" + 25y\" = 0

Solution

a)

y\'\'\'+9y\'\'+27y\'+27y=0

Linear, constant-coefficient. Assume solution of form y = e^mx. This gives us the equation

m³+9m²+27m+27=0

(m+3)³=0 [(a+b)³=a³+b³+3ab(a+b)]

so

m=-3,-3,-3

So the general solution is

y=C₁e^m₁x+C₂e^m₂x+C₃e^m₃x

by substitutiion we get

y=C₁e^-3x+C₂e^-3x+C₃e^-3x

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b)

y\'\'\'+4y\'\'-3y\'-18y=0

Linear, constant-coefficient. Assume solution of form y = e^mx. This gives us the equation

m³+4m²-3m-18=0

(m-2)(m+3)(m+3)=0

So

m=2 ,-3,-3

So the general solution is

y=C₁e^m₁x+C₂e^m₂x+C₃e^m₃x

by substitutiion we get

y=C₁e^2x+C₂e^-3x+C₃e^-3x

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^c)

^{y\'\'\'\'+10y\'\'\'+25y\'\'=0}

Linear, constant-coefficient. Assume solution of form y = e^mx. This gives us the equation

m⁴+10m³+25m²=0

m²(m+5)(m+5)=0

So

m=0,0,-5 & -5

So the general solution is

y=C₁e^m₁x+C₂e^m₂x+C₃e^m₃x+C₄e^m₄x

by substitutiion we get

y=C₁e^(0)x+C₂e^(0)x+C₃e^-5x+C₄e^-5x

y=C₁+C₂+C₃e^-5x+C₄e^-5x