IV. Find the general solution of the given differential equation.
11/234) y‴ − y″ − y′ + y = 0
14/234) y^(4) − 4y‴ + 4y″ = 0
11/234)y′′′−y′′−y′+y=0λ3−λ2−λ+1=0λ2(λ−1)−(λ−1)=0(λ−1)(λ2−1)=0(λ−1)2(λ+1)=0λ1=λ2=1,λ3=−1y=c1ex+c2xex+c3e−x.14/234)y(4)−4y′′′+4y′′=0λ4−4λ3+4λ2=0λ2(λ2−4λ+4)=0λ2(λ−2)2=0λ1=λ2=0,λ3=λ4=2y=c1+c2x+c3e2x+c4xe2x11/234) y'''-y''-y'+y=0\\ \lambda^3-\lambda^2-\lambda+1=0\\ \lambda^2(\lambda-1)-(\lambda-1)=0\\ (\lambda-1)(\lambda^2-1)=0\\ (\lambda-1)^2(\lambda+1)=0\\ \lambda_1=\lambda_2=1, \lambda_3=-1\\ y=c_1e^x+c_2xe^x+c_3e^{-x}.\\ \\ 14/234) y^{(4)}-4y'''+4y''=0\\ \lambda^4-4\lambda^3+4\lambda^2=0\\ \lambda^2(\lambda^2-4\lambda+4)=0\\ \lambda^2(\lambda-2)^2=0\\ \lambda_1=\lambda_2=0, \lambda_3=\lambda_4=2\\ y=c_1+c_2x+c_3e^{2x}+c_4xe^{2x}11/234)y′′′−y′′−y′+y=0λ3−λ2−λ+1=0λ2(λ−1)−(λ−1)=0(λ−1)(λ2−1)=0(λ−1)2(λ+1)=0λ1=λ2=1,λ3=−1y=c1ex+c2xex+c3e−x.14/234)y(4)−4y′′′+4y′′=0λ4−4λ3+4λ2=0λ2(λ2−4λ+4)=0λ2(λ−2)2=0λ1=λ2=0,λ3=λ4=2y=c1+c2x+c3e2x+c4xe2x
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