y'''+3y''+3y'+y=0
Given the differential equation: "y'''+3y''+3y'+y=0"
Require to find the general solution of the given differential equation.
An Auxiliary Equation corresponding to the given homogeneous linear differential equation is
"m^3+3m^2+3m+1=0"
"\\Rightarrow (m+1)^3=0"
"\\Rightarrow m=-1,-1,-1"
The general solution to the given differential equation is
"y(x)=(c_1+c_2x+c_3x^2)e^{-x}"
Therefore, general solution to the given differential equation is
"y(x)=(c_1+c_2x+c_3x^2)e^{-x}"
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