A) \[{{y}^{2}}=x\]
B) \[{{x}^{2}}=2y-1\]
C) \[3{{x}^{2}}-2{{y}^{2}}=1\]
D) \[2{{x}^{2}}-{{y}^{2}}=1\]
Correct Answer: D
Solution :
\[2\left( x-\frac{y}{\frac{dy}{dx}} \right)\,=\frac{1}{x}\] \[x-\frac{1}{2x}=\frac{y}{\frac{dy}{dx}}\Rightarrow \,\frac{(2{{x}^{2}}-1)}{2x}=\frac{y}{\frac{dy}{dx}}\] \[\int_{{}}^{{}}{\frac{dy}{y}\,=\int_{{}}^{{}}{\frac{2x}{(2{{x}^{2}}-1)}dx}}\] so \[y=k{{(2{{x}^{2}}-1)}^{1/2}}\] Here, k = 1 thus \[{{y}^{2}}=2{{x}^{2}}-1\]\[2{{x}^{2}}-{{y}^{2}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
