Question

If $y=3cos(logx)+4sin(logx)$,show that $x^2y_2+xy_1+y=0$

Answer 1

It is given that,$y=3cos(logx)+4sin(logx)$
Then,
$y_1=3.\frac{d}{dx}[cos(logx)]+4.\frac{d}{dx}[sin(logx)]$
$=3.[-sin(logx).\frac{d}{dx}(logx)]+4.[cos(logx).\frac{d}{dx}(logx)]$
$∴y_1=\frac{-3sin(logx)}{x}+\frac{4cos(logx)}{x}=\frac{4cos(logx)-3sin(logx)}{x}$
$∴y_2=\frac{d}{dx}(\frac{4cos(logx)-3sin(logx)}{x})$
$=x\frac{[4cos(logx)-3sin(logx)']-[{4cos(logx)-3sin(logx)}(x)']}{x2}$
$=\frac{x[4{cos(logx)}'-3{sin(logx)}']-{4cos(logx)-3sin(logx)}.1}{x^2}$
$=\frac{x[-4sin(logx).(logx)'-3cos(logx).(logx)']-4cos(logx)+3sin(logx)}{x^2}$
$=\frac{x[-4sin(logx).\frac{1}{x}-3cos(logx).\frac{1}{x}]-4cos(logx)+3sin(logx)}{x^2}$
$=\frac{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}{x^2}$
$=\frac{-sin(logx)-7cos(logx)}{x^2}$
$∴x^2y_2+xy_1+y$
$=x^2\frac{(-sin(logx)-7cos(logx))}{x^2}+x\frac{(4cos(logx)-3sin(logx))}{x}+3cos(logx)+4sin(logx)$
$=-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+3cos(logx)+4sin(logx)$
$=0$
Hence,proved.