A) 1
B) 2
C) 3
D) Cannot be determined
Correct Answer: B
Solution :
| [b] \[\frac{dx}{dy}+\int{y.dx={{x}^{3}}\Rightarrow \int{y.dx={{x}^{3}}-\frac{dx}{dy}}}\] |
| \[\Rightarrow 1+\frac{dy}{dx}\left( \int{y.dx} \right)={{x}^{3}}.\frac{dy}{dx}\] |
| Differentiate both sides w.e.t.x |
| \[\Rightarrow 0+\frac{dy}{dx}(y)+\left( \int{y.dx} \right)\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)={{x}^{3}}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}(2{{x}^{2}})\]\[\Rightarrow y.\frac{dy}{dx}+\frac{{{d}^{2}}y}{d{{x}^{2}}}\left[ {{x}^{3}}-\frac{dx}{dy} \right]={{x}^{3}}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+2{{x}^{2}}\frac{dy}{dx}\] |
| \[\Rightarrow y\frac{dy}{dx}+{{x}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-\left( \frac{dx}{dy} \right)\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)={{x}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+2{{x}^{2}}\frac{dy}{dx}\]\[\Rightarrow y\frac{dy}{dx}-\frac{dx}{dy}.\frac{{{d}^{2}}y}{d{{x}^{2}}}=2{{x}^{2}}.\frac{dy}{dx}\] |
| Multiplying both side by \[\frac{dy}{dx}\] |
| \[y{{\left( \frac{dy}{dx} \right)}^{2}}-\frac{{{d}^{2}}y}{d{{x}^{2}}}=2{{x}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}\] |
| \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}+(2{{x}^{2}}-y){{\left( \frac{dy}{dx} \right)}^{2}}=0\] |
| Order = 2, degree = 1. |
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