A) \[xyy'+{{y}^{2}}-9=0\]
B) \[x+y{{y}^{''}}=0\]
C) \[xyy''+x{{(y')}^{2}}-yy'=0\]
D) \[xyy'-{{y}^{2}}+9=0\]
Correct Answer: C
Solution :
\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] it passes through (0,3), so it will become \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{9}=1\] different w.r.t x, we get \[\frac{2x}{{{a}^{2}}}+\frac{2y}{9}\frac{dy}{dx}=0\] \[\frac{y}{x}\left( \frac{dy}{dx} \right)=\frac{-9}{{{a}^{2}}}\] diff w.r.t x, we get \[(\frac{y}{x}\frac{{{d}^{2}}y}{d{{x}^{2}}})+\frac{x\frac{dy}{dx}-y}{{{x}^{2}}}\frac{dy}{dx}=0\] \[xyy''+xy{{'}^{2}}-yy'=0\]
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