Question

Match the following

I. lf $y=a+b{\overline{e}}^{4x}$, then $y_2+4y_1=$	a) $0$
II. lf $y=ax+\frac{b}{x}$, then $x^2{y}_2+x{y}_1=$	b) $y$
III. lf $xy=a{e}^x+b{e}^{-x}$, then $x{y}_2+2y_1=$	c) $xy$
IV. lf $y=x{e}^{-\frac{1}{x}}$, then $x^3{y}_2-x{y}_1=$	d) $-y$

Which of the following is correct, if $y_n$ denotes $n^{th}$ derivative of $y$?

• A. I-a, II- b, III- c, IV-d
• B. I-b, II- d, III- a, IV-c
• C. I-a, II- d, III- b, IV-c
• D. I-d, II- c, III- b, IV-a

Answer 1

The correct option is A I-a, II- b, III- c, IV-d

$I$
$y_1=-4b{e}^{-4x}$
$y_2=16b{e}^{-4x}=-4y_1$
$y_2+4y_1=0\to$(a)
$II$
$y_1=a-\frac{b}{x^2}$
$y_2=\frac{2b}{x^3}$
$x^2{y}_2+x{y}_1=ax+\frac{b}{x}=y\to$(b)
$III$
$y+x{y}_1=a{e}^x-b{e}^{-x}$
$2y_1+x{y}_2=a{e}^x+b{e}^{-x}=xy\to$(c)
$IV$
$y=x{e}^{-1/x}$
$y_1=e^{-1/x}+x\times \frac{e^{-1/x}}{x^2}$
$y_1=e^{-1/x}(1+\frac{1}{x})$
$y_2=-\frac{e^{-1/x}}{x^2}+(1+\frac{1}{x})\frac{e^{-1/x}}{x^2}$
$y_2=\frac{e^{-1/x}}{x^3}$
$x^3{y}_2-x{y}_1=e^{-1/x}-x{e}^{-1/x}-e^{-1/x}$
$x^3{y}_2-x{y}_1=-y\to$(d)
Hence, option A.