Question

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

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

Answer 1

Answer: (D)

$y^{\prime \prime \prime }-3y^{\prime \prime }+3y^{\prime }-y=0$

$y=a{e}^x+bx{e}^x+c{x}^2{e}^x$ ...(1)

On differentiating w.r.t $x$, we get

$y^{\prime }=a{e}^x+b(x{e}^x+e^x)+c(x^2{e}^x+2x{e}^x)$

$\Rightarrow y^{\prime }=a{e}^{x}bx{e}^x+c{x}^2{e}^x+b{e}^x+2cx{e}^x$

$\Rightarrow y^{\prime }=y+b{e}^x+2cx{e}^x$ ...(2)

Again differentiating w.r.t $x$, we get

$y^{\prime \prime }=y^{\prime }+b{e}^x+2c(x{e}^x+e^x)$

$\Rightarrow y^{\prime \prime }=y^{\prime }+b{e}^x+2cx{e}^x+2c{e}^x$

$\Rightarrow y^{\prime \prime }=2y^{\prime }-y+2c{e}^x$ ...(3) (from (2)

Again differentiating w.r.t $x$4, we get

$y^{\prime \prime \prime }=2y^{\prime \prime }-y^{\prime }+2c{e}^x$

$\Rightarrow y^{\prime \prime \prime }=2y^{\prime \prime }-y^{\prime }+(y^{\prime \prime }-2y^{\prime }+y)$ (from (3)

$\Rightarrow y^{\prime \prime \prime }-3y^{\prime \prime }+3y^{\prime }-y=0$