The solution of the equation \[\frac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=\ln x,\] when \[x=1\], \[y=0\] and \[\frac{dy}{dx}=-1\] is [Orissa JEE 2003]

\[\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]

\[\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]

\[-\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]

\[-\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]

The solution of the equation \[\frac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=

1.	The solution of the equation \[\frac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=\ln x,\] when \[x=1\], \[y=0\] and \[\frac{dy}{dx}=-1\] is [Orissa JEE 2003]
A.	\[\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]
B.	\[\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]
C.	\[-\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]
D.	\[-\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]
Answer» E.

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The solution of the equation \[\frac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=\ln x,\] when \[x=1\], \[y=0\] and \[\frac{dy}{dx}=-1\] is [Orissa JEE 2003]

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