A) \[x+y=a\]
B) \[\sqrt{x}-\sqrt{y}=\sqrt{a}\]
C) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
D) \[\sqrt{x}+\sqrt{y}=\sqrt{a}\]
Correct Answer: D
Solution :
Given differential equation can be rewritten as \[{{\left( \sqrt{x}\frac{dy}{dx}+\sqrt{y} \right)}^{2}}=0\] \[\Rightarrow \] \[\frac{dy}{\sqrt{y}}+\frac{dx}{\sqrt{x}}=0\] On integrating, we get \[2\sqrt{y}+2\sqrt{x}={{c}_{1}}\] \[\Rightarrow \] \[\sqrt{x}+\sqrt{y}=\frac{{{c}_{1}}}{2}\] \[\Rightarrow \] \[\sqrt{x}+\sqrt{y}=\sqrt{a}\] \[\left( \text{where}\,\,a=\frac{{{c}_{1}}}{2} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
