Question

The particular integral of $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=x^2+2x+4$ is

• A. $\frac{x^3}{3}+x^2+4x+C$
• B. $\frac{x^3}{3}+x^2+C$
• C. $\frac{x^3}{3}+4x+C$
• D. $x^2+4x+C$

Answer 1

The correct option is C $\frac{x^3}{3}+4x+C$

Given equation is of the form

$(D^2+D)y=x^2+2x+4$

$\therefore$ $P.I.=\frac{1}{D(D+1)}(x^2+2x+4)$

$=\frac{1}{D}(1+D{)}^{-1}(x^2+2x+4)$

$=\frac{1}{D}(1-D+D^2-...)(x^2+2x+4)$

$=\frac{1}{D}[x^2+2x+4-(2x+2)+2]$

$=\int (x^2+4)dx=\frac{x^3}{3}+4x+C$