A) \[{{y}^{2}}={{x}^{2}}(ln\,\,{{x}^{2}}-1)+C\]
B) \[y={{x}^{2}}(ln\,\,x-1)+C\]
C) \[{{y}^{2}}=x(ln\,\,x-1)+C\]
D) \[y={{x}^{2}}{{e}^{{{x}^{2}}}}+C\]
Correct Answer: A
Solution :
[a] \[{{x}^{2}}={{e}^{{{\left( \frac{x}{y} \right)}^{-1}}\left( \frac{dy}{dx} \right)}}\Rightarrow {{x}^{2}}={{e}^{\left( \frac{y}{x} \right)\left( \frac{dy}{dx} \right)}}\] \[\Rightarrow \] ln \[{{x}^{2}}=\frac{y}{x}\frac{dy}{dx}\] or \[\int{x\,ln\,{{x}^{2}}dx=\int{ydy}}\] Put \[{{x}^{2}}=t\Rightarrow 2xdx=dt\therefore \frac{1}{2}\int{\,ln\,t\,dt=\frac{{{y}^{2}}}{2}}\] \[C+t\,\,\ln \,\,or\,\,t-t={{y}^{2}}\,\,or\,\,{{y}^{2}}={{x}^{2}}(ln\,\,{{x}^{2}}-1)+C\]
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