Question

The equation of the curve passing through $(\frac{{\pi }^2}{4},1),$ which has a solution of the equation as $y^{2}cos\sqrt{x}dx-2\sqrt{x}{e}^{\frac{1}{y}}dy=0$

• A. $sin\sqrt{x}+e^{\frac{1}{y}}dy=e$
• B. $sin\sqrt{x}+e^{\frac{1}{y}}=1-e$
• C. $sin\sqrt{x}=1+e+e^{\frac{1}{y}}$
• D. $sin\sqrt{x}+e^{\frac{1}{y}}=1+e$

Answer 1

The correct option is D $sin\sqrt{x}+e^{\frac{1}{y}}=1+e$

$y^{2}cos\sqrt{x}dx-2\sqrt{x}{e}^{\frac{1}{y}}dy=0\Rightarrow \frac{cos\sqrt{x}}{2\sqrt{x}}dx-\frac{e^{\frac{1}{y}}}{y^2}dy=0\Rightarrow sin\sqrt{x}+e^{\frac{1}{y}}+c=0$
Curve passing through $(\frac{{\pi }^2}{4},1)\Rightarrow 1+e+c=0\Rightarrow c=-1-e.$
$\therefore$ Required curve is $sin\sqrt{x}+e^{\frac{1}{y}}=1+e.$