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. \(y=C_1e^{\frac{1}{3x}}+C_2e^{-x}-(x^2+4x+15)\)

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. \(y=C_1e^{\frac{1}{3x}}+C_2e^{-x}-(x^2+4x+15)\)