A) order 1, degree 2
B) order 1, degree 3
C) order 2, degree 3
D) order 2, degree 2
E) order 1, degree 1
Correct Answer: B
Solution :
\[{{y}^{2}}=2c(x+\sqrt{c})\] ?.. (i) Differentiating w.r.t.\[x,\] \[2y\frac{dy}{dx}=2c\] \[\Rightarrow \] \[c=y\frac{dy}{dx}\] On putting this value in Eq. (i), \[{{y}^{2}}=2x.y\frac{dy}{dx}+2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}\] \[2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}={{y}^{2}}-2xy\frac{dy}{dx}\] Squaring on both sides, \[4{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}\left( y.\frac{dy}{dx} \right)={{\left\{ {{y}^{2}}-2xy\frac{dy}{dx} \right\}}^{2}}\] \[4{{y}^{3}}{{\left( \frac{dy}{dx} \right)}^{3}}={{y}^{2}}+4{{x}^{2}}{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}-2x{{y}^{3}}\left( \frac{dy}{dx} \right)\] \[4y{{\left( \frac{dy}{dx} \right)}^{3}}=-4{{x}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}+2xy\left( \frac{dy}{dx} \right)-{{y}^{2}}=0\] Hence, order\[\to 1,\]degree\[\to 3\]
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