A) \[{{y}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( b-\frac{dy}{dx}+x \right)\]
B) \[xy\frac{{{d}^{2}}y}{d{{x}^{2}}}+x{{\left( \frac{dy}{dx} \right)}^{2}}-y\frac{dy}{dx}=0\]
C) \[y{{\left( \frac{dy}{dx} \right)}^{2}}-x\frac{dy}{dx}+1=0\]
D) None of the above
Correct Answer: B
Solution :
Given, \[{{y}^{2}}=a(b-{{x}^{2}})\] On differentiation w. r. t. to x, we get \[2y\frac{dy}{dx}=-2ax\,\,\,\,\Rightarrow a=-\frac{y}{x}\,\,\frac{dy}{dx}\] From Eq. (i), \[{{y}^{2}}=-\frac{y}{x}\frac{dy}{dx}\,(b-{{x}^{2}})\] \[\Rightarrow \] \[-xy{{\left( \frac{dy}{dx} \right)}^{-1}}=b-{{x}^{2}}\] On differentiation w. r. t. x, we get \[-\left\{ -xy{{\left( \frac{dy}{dx} \right)}^{-2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{-1}}\left( x\frac{dy}{dx}+y \right) \right\}=-2x\]\[\Rightarrow \] \[-\left\{ -xy\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\left( x\frac{dy}{dx}+y \right) \right\}=-2x{{\left( \frac{dy}{dx} \right)}^{2}}\] \[\Rightarrow \] \[xy\frac{{{d}^{2}}y}{d{{x}^{2}}}-x{{\left( \frac{dy}{dx} \right)}^{2}}-y\left( \frac{dy}{dx} \right)=-2x{{\left( \frac{dy}{dx} \right)}^{2}}\] \[\Rightarrow \] \[xy\frac{{{d}^{2}}y}{d{{x}^{2}}}+x{{\left( \frac{dy}{dx} \right)}^{2}}-y\frac{dy}{dx}=0\] Which is the required differential equation solution.
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