Question

Let \(y=y(x)\) be the solution of the differential equation
\(e^x\sqrt{1-y^2}\,dx+\left(\frac{y}{x}\right)dy=0\)
, \(y(1)=-1\). Then the value of \((y(3))^2\) is equal to :

• A. \(1-4e^3\)
• B. \(1-4e^6\)
• C. \(1+4e^3\)
• D. \(1+4e^6\)

Hint:

Separate variables: \(\frac{y\,dy}{\sqrt{1-y^2}}=-xe^x\,dx\); integrate (by parts on the right) and use \(y(1)=-1\).

Answer 1

The Correct Option is B

\(\frac{y}{\sqrt{1-y^2}}dy=-xe^x\,dx\).
Left side: with \(u=1-y^2\), \(\int\frac{y\,dy}{\sqrt{1-y^2}}=-\sqrt{1-y^2}\). Right side: \(-\int xe^x\,dx=-(x-1)e^x+C\).
So \(\sqrt{1-y^2}=(x-1)e^x-C\). At \(x=1,\ y=-1\): \(0=0-C\Rightarrow C=0\).
At \(x=3\): \(\sqrt{1-y^2}=2e^3\Rightarrow1-y^2=4e^6\Rightarrow y^2=1-4e^6\).
Answer: (2) \(1-4e^6\)