A) \[x+y=a\]
B) \[\sqrt{x}-\sqrt{y}=\sqrt{a}\]
C) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
D) \[\sqrt{x}+\sqrt{y}=\sqrt{a}\]
E) \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]
Correct Answer: D
Solution :
\[x{{\left( \frac{dy}{dx} \right)}^{2}}+2\sqrt{xy}\frac{dy}{dx}+y=0\] \[\Rightarrow \] \[{{\left( \sqrt{x}\frac{dy}{dx}+\sqrt{y} \right)}^{2}}=0\] On integrating both sides \[\int{\frac{1}{\sqrt{y}}}dy+\int{\frac{1}{\sqrt{x}}}dx=0\] \[\Rightarrow \] \[2\sqrt{y}+2\sqrt{x}={{c}_{1}}\] \[\Rightarrow \] \[\sqrt{x}+\sqrt{y}=\frac{{{c}_{1}}}{2}\] which is similar to\[\sqrt{x}+\sqrt{y}=\sqrt{a}.\] \[\therefore \]Solution of given differential equation is \[\sqrt{x}+\sqrt{y}=\sqrt{a}\]You need to login to perform this action.
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