Solve the second order linear differential equation using method of undetermined coefficients 3y''+2y'-y=x^2+1

Question
Solve the second order linear differential equation using method of undetermined coefficients
\(3y''+2y'-y=x^2+1\)

Answers (1)

2020-12-08
To solve:
The second order linear differential equation, \(3y''+2y'-y=x^2+1\) using method of undetermined coefficients.
\(3y''+2y"-y=x^2+1\)
Rewrite the avove equation as
\(3D^2y+2Dy-y=x^2+1\)
\((3D^2+2D-1)y=x^2+1\)
The auxiliary equation is
\(3m^2+2m-1=0\)
\(3m^2+3m-m-1=0\)
\(3m(m+1)-1(m+1)=0\)
\((3m-1)(m+1)=0\)
\(3m-1=0\) or \(m+1=0\)
\(m=\frac{1}{3}\) or \(m=-1\)
The complimentary function is
\(C.F.=C_1e^{\frac{1}{3x}}+C_2e^{-x}\)
To find the particular integral using the method of undetermined coefficients :
\(3y''-2y'-y=x^2+1\)
The most general linear combination of the functions in the family is
\(y_p=Ax^2+Bx+c\)
\(\frac{dy}{dx}=2Ax+B\)
\(\frac{d^2y}{dx^2}=2A\)
Plug \({dy}{dx}\) and \(\frac{d^2}{dx^2}\) in \(3y''+2y'-y=x^2+1\), we have
\(3(2A)+2(2Ax+B)-(Ax^2+Bx+C)=x^2+1\)
\(6A+4Ax+2B-Ax^2-Bx-C=x^2+1\)
\(-Ax^2+(4A-B)x+(6A+2B-C)=x^2+1\)
Equating the coefficients of \(x^2\),we have
\(-A=1\)
\(A=-1\)
Equating the coefficients of x, we have
\(4A-B=0\)
\(4(-1)-B=0\)
\(B=-4\)
Equating the constant term, we have
\(6A+2B-C=1\)
\(6(-1)+2(-4)-C=1\)
\(-6-8-C=1\)
\(-14-C=1\)
\(C=-14-1\)
\(C=-15\)
Therefore, the particular solution is
\(P.I.=Ax^2+Bx+C\)
\(=-x^2-4x-15\)
\(=-(x^2+4x+15)\)
The general solution for the given differential equation is
y=C.F.+P.I. \(y=C_1e^{\frac{1}{3x}}+C_2e^{-x}-(x^2+4x+15)\)
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