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  • If y = 3 cos (log x) + 4 sin (log x), show that x ^ 2 y2 + xy1 + y = 0

Q13  If  y = 3 \cos (\log x) + 4 \sin (\log x), show that x^2 y_2 + xy_1 + y = 0

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Given function is
y = 3 \cos (\log x) + 4 \sin (\log x)
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}
                                                                           =\frac{4\cos (\log x)-3\sin(\log x)}{x}                       -(i)
Now, second order derivative is 
By using the Quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}
                                                               =\frac{-\sin(\log x)+7\cos (\log x)}{x^2}                                 -(ii)
Now, from equation (i) and (ii) we will get y_1 \ and \ y_2 
Now, we need to show
x^2 y_2 + xy_1 + y = 0
Put the value of y_1 \ and \ y_2  from equation (i) and (ii)
x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x) +4\sin(\log x)
-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\ cos (\log x) +4\sin(\log x)
=0
Hence proved


 

Posted by

Gautam harsolia

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