Question

If y = 3 cos(logx) + 4 sin(logx), show that $ {{x}^{2}}{{y}_{2}}+x{{y}_{1}}+y=0 $ .

Answer 1

Hint: To prove that the given equation is true, we will find the first derivative of y and then second derivative of y. As we can see, y is a composite function, we need to use the chain rule to find its derivative. We will manipulate the first derivative in such a way that on second derivation, we will find that $ {{x}^{2}}{{y}_{2}}+x{{y}_{1}}+y=0 $ . We will also require the product rule of derivation.

Complete step-by-step answer:
It is given to us that y = 3 cos(logx) + 4 sin(logx).
We will be required to use chain rule to derivate the function y, as it is a composite function.
According to the chain rule, if f and g are two functions, then $ \dfrac{df\left( g\left( x \right) \right)}{dx}=f'\left( g\left( x \right) \right)\times g'\left( x \right) $ , where f’ and g’ are the derivatives of f and g with respect to x, respectively.
We need to keep in mind that derivative of cos(x) = ─sin(x), derivative of sin(x) = cos(x) and derivative of logx = $ \dfrac{1}{x} $
Thus, the first derivative of y with respect to x is $ {{y}_{1}}=\dfrac{dy}{dx}=-3\sin \left( \log x \right)\times \dfrac{1}{x}+4\cos \left( \log x \right)\times \dfrac{1}{x} $ .
To find the second derivative, we need to derivate the first derivative of y. To make the derivation easy, we will cross multiply x to the other side.
 $ \Rightarrow x\dfrac{dy}{dx}=-3\sin \left( \log x \right)+4\cos \left( \log x \right) $
Thus, we will derivate $ x\dfrac{dy}{dx}=-3\sin \left( \log x \right)+4\cos \left( \log x \right) $ with respect to x.
We will need to use the product rule of derivation. If f and g are two functions, then according to product rule of differentiation $ \dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right) $ where f’ and g’ are the derivatives of f and g with respect to x, respectively.
\[\begin{align}
  & \Rightarrow \dfrac{d\left( x\dfrac{dy}{dx} \right)}{dx}=-3\cos \left( \log x \right)\times \dfrac{1}{x}-4\sin \left( \log x \right)\times \dfrac{1}{x} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}=-3\cos \left( \log x \right)\times \dfrac{1}{x}-4\sin \left( \log x \right)\times \dfrac{1}{x} \\
 & \Rightarrow {{x}^{2}}{{y}_{2}}+x{{y}_{1}}=-3\cos \left( \log x \right)-4\sin \left( \log x \right) \\
 & \Rightarrow {{x}^{2}}{{y}_{2}}+x{{y}_{1}}=-\left[ 3\cos \left( \log x \right)+4\sin \left( \log x \right) \right] \\
\end{align}\]
But it is given to us that y = 3 cos(logx) + 4 sin(logx).
\[\Rightarrow {{x}^{2}}{{y}_{2}}+x{{y}_{1}}=-y\]
Thus, we have shown that \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}+y=0\]


Note: Derivation of $ {{y}_{1}}=\dfrac{dy}{dx}=-3\sin \left( \log x \right)\times \dfrac{1}{x}+4\cos \left( \log x \right)\times \dfrac{1}{x} $ directly without taking x on the other side is possible, but will be very tedious, as now for product rule, there will be 3 terms. Thus, to make our differentiation easier, we cross multiply x to the other side.