# 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
$$3y''+2y'-y=x^2+1$$

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

### Relevant Questions

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(d) $$y''+(3-a)y'+3ay=0$$, a any real number.
(e) Cannot be determined.
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