A) \[y.y''-2{{\left( y' \right)}^{2}}+1=0\]
B) \[y.y''+{{(y')}^{2}}+1=0\]
C) \[y.y''+{{\left( y' \right)}^{2}}-1=0\]
D) \[y.y''+2{{\left( y' \right)}^{2}}+1=0\]
Correct Answer: B
Solution :
[b] \[\because {{x}^{2}}+y=1\] \[2x+2y.{{y}_{1}}=0\] \[{{y}_{1}}=-\frac{x}{y}\] ..............(1) Again differentiating w.r.t. x, we have \[1+y_{1}^{2}+y.{{y}_{2}}=0\Rightarrow y.{{y}_{2}}+y_{1}^{2}+1=0\] \[\because {{y}_{1}}=\frac{dy}{dx}\,\,{{y}_{2}}=\frac{{{d}^{2}}y}{d{{x}^{2}}}\] Hence, option [b]is correct.
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