• # question_answer What is the order of the differential equation$\frac{dx}{dy}+\int{y\,dx={{x}^{3}}}$? A) 1 B) 2 C) 3 D) Cannot be determined

 [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.