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.