Yasmin

## Answered question

2020-12-07

Solve the second order linear differential equation using method of undetermined coefficients
$3{y}^{″}+2{y}^{\prime }-y={x}^{2}+1$

### Answer & Explanation

avortarF

Skilled2020-12-08Added 113 answers

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

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