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Question: The solution of the equation \(x^{2}\frac{d^{2}y}{dx^{2}} = \ln x\) when *x* = 1, *y* = 0 and \(\fra...

The solution of the equation x2d2ydx2=ln⁡xx^{2}\frac{d^{2}y}{dx^{2}} = \ln x when x = 1, y = 0 and dydx=−1\frac{dy}{dx} = - 1 is

A

12(ln⁡x)2+ln⁡x\frac{1}{2}(\ln x)^{2} + \ln x

B

12(ln⁡x)2−ln⁡x\frac{1}{2}(\ln x)^{2} - \ln x

C

12(ln⁡x)2+ln⁡x\frac{1}{2}(\ln x)^{2} + \ln x

D

−12(ln⁡x)2−ln⁡x- \frac{1}{2}(\ln x)^{2} - \ln x

Answer

−12(ln⁡x)2−ln⁡x- \frac{1}{2}(\ln x)^{2} - \ln x

Explanation

Solution

We haved2ydx2=ln⁡xx2\frac{d^{2}y}{dx^{2}} = \frac{\ln x}{x^{2}} ⇒ d(dydx)=ln⁡xx2dxd\left( \frac{dy}{dx} \right) = \frac{\ln x}{x^{2}}dx

Integrating, dydx=∫ln⁡xd(−1x)=−ln⁡xx+∫1x2dx=−ln⁡xx−1x+c\frac{dy}{dx} = \int_{}^{}{\ln x}d\left( - \frac{1}{x} \right) = - \frac{\ln x}{x} + \int_{}^{}{\frac{1}{x^{2}}dx = - \frac{\ln x}{x} - \frac{1}{x} + c}

⇒ dydx=−1+ln⁡xx+c\frac{dy}{dx} = - \frac{1 + \ln x}{x} + c

When x = 1, dydx=−1\frac{dy}{dx} = - 1

∴ – 1 = – 1 + c ⇒ c = 0

∴ dydx=−1+ln⁡xx\frac{dy}{dx} = - \frac{1 + \ln x}{x} ⇒ dy=−1+ln⁡xxdxdy = - \frac{1 + \ln x}{x}dx

⇒ −∫dy=+∫dxx+∫ln⁡x.1xdx- \int_{}^{}{dy} = + \int_{}^{}\frac{dx}{x} + \int_{}^{}{\ln x}.\frac{1}{x}dx ⇒ −y=ln⁡x+12(ln⁡x)2+λ- y = \ln x + \frac{1}{2}(\ln x)^{2} + \lambda

y = 0 when x = 1

∴ 0=0+02+λ0 = 0 + 0^{2} + \lambda ⇒ λ=0\lambda = 0 ⇒ −y=ln⁡x+12(ln⁡x)2- y = \ln x + \frac{1}{2}(\ln x)^{2}

∴ y=−12(ln⁡x)2−ln⁡xy = - \frac{1}{2}(\ln x)^{2} - \ln x