Question

The differential equation of the family $y=a{e}^x+bx{e}^x+c{x}^2-e^x$ of curves, where $a,b,c$ are arbitary constants is :

• A. $y^{\prime }"+3y"+3/+y=0$
• B. $y"+3y"-3y^{\prime }-y=0$
• C. $y^{\prime \prime \prime }-3y"-3y^{\prime }+y$ = 0
• D. $y"+3y"+3y^{\prime }-y=0.$

Answer 1

Answer: (D)

$y=a{e}^x+bx{e}^x+c{x}^2{e}^x$
$y^{\prime }=a{e}^x+bx{e}^x+b{e}^x+c\left(x^2{e}^x+2x{e}^x\right)$
$=a{e}^x+6\left(1+x\right)e^x+cx{e}^x\left(x+2\right)$
$y^{\prime \prime }=a{e}^x+b\left(1+x\right)e^x+b{e}^x+cx{e}^{x}cx$
$\left(x+2\right)e^x+c\left(x+2\right)e^x=a{e}^x+b\left(2+x\right)e^x$
$+cx{e}^x+c\left(x+2\right)\left(x+1\right)e^x+c{e}^x\left(x+2+x+1\right)$
$+c\left(x+2\right)\left(x+1\right)s{e}^x$
$=a{e}^x+6\left(3+x\right)+c\left(x+1\right)e^x+c{e}^x\left(x+2\right)\left(x+1\right)\left(2x+3\right)$
$y^{\prime \prime \prime }-3y^{\prime \prime }-3y^{\prime }-y=a{e}^x+36+bx{e}^x$
$+c\left(x+1\right)e^x+c{e}^x\left(x^2+5x+5\right)-3a{e}^x$
$-3b\left(2+x\right)e^x-3bx{e}^x-3c\left(x+2\right)\left(x+1\right)e^x$
$+3a{e}^x+3b\left(1+x\right)e^x+3cx{e}^x\left(x+2\right)-a{e}^x-bx{e}^x-c{x}^2{e}^x=0$.