Question

The solution of primitive equation $-y\frac{dy}{dx}=x\sqrt{1-y^2}$ is $y=y\left(x\right)$, where $y\left(x\right)$ is non-constant. If $y\left(0\right)=1$, then which of the following is/are correct

• A. $y\left(\sqrt{2}\right)=0$
• B. $y\left(3^{\frac{1}{4}}\right)=\pm \frac{1}{2}$
• C. $y\left(\sqrt{2\sin \theta }\right)=\pm \cos \theta \forall \theta \in (0,\pi )$
• D. $y\left(1\right)=1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $y\left(\sqrt{2}\right)=0$
B $y\left(3^{\frac{1}{4}}\right)=\pm \frac{1}{2}$
C $y\left(\sqrt{2\sin \theta }\right)=\pm \cos \theta \forall \theta \in (0,\pi )$

$\frac{-ydy}{\sqrt{1-y^2}}=xdxd\left(\sqrt{1-y^2}\right)=xdx\sqrt{1-y^2}=\frac{x^2}{2}+c$
Put $x=0,y=1\Rightarrow c=0$
$\Rightarrow 1-y^2=\frac{x^4}{4}\Rightarrow \frac{x^4}{4}+y^2=1$