Question

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

• A. $ y'" + 3y" + 3 / + y = 0 $
• B. $ y" + 3y" -3y' -y = 0 $
• C. $ y''' - 3y" - 3 y'+ y$ = 0
• D. $y" + 3y" + 3y' - y = 0. $

Answer 1

Answer: (D)

$y = ae^{x} + bx \,e^{x}+ cx^{2}\, e^{x}$
$y' = ae^{x}+bx\,e^{x}+be^{x}+c\left(x^{2}e^{x}+2xe^{x}\right)$
$= ae^{x} + 6\left( 1 +x\right)e^{x} + cxe^{x} \left(x + 2\right)$
$y'' = ae^{x}+b\left(1+x\right)e^{x}+be^{x}+cxe^{x}cx$
$\left(x+2\right)e^{x}+c\left(x+2\right)e^{x} = ae^{x}+ b\left(2+x\right)e^{x}$
$+ cxe^{x} +c\left(x+2\right)\left(x+1\right)e^{x}+ce^{x}\left(x+2+x+1\right)$
$+c \left(x+2\right)\left(x+1\right)se^{x}$
$= ae^{x} + 6\left(3 + x\right) + c\left(x + 1\right)e^{x} + ce^{x}\left(x + 2\right)\left(x+ 1\right) \left(2x + 3\right)$
$y'''-3y''-3y'-y = ae^{x} + 36 + bx\,e^{x}$
$+c\left(x+1\right)e^{x}+ce^{x}\left(x^{2}+5x+5\right)-3ae^{x}$
$-3b\left(2+x\right)e^{x}-3bxe^{x}-3c\left(x+2\right)\left(x+1\right)e^{x}$
$+3ae^{x}+3b\left(1+x\right)e^{x}+3cx\,e^{x}\left(x+2\right)-ae^{x}-bxe^{x}-cx^{2}e^{x} = 0$.