The particular solution of the differential equation \(e^x\sqrt{1-y^2}\,dx +\displaystyle\frac{y}{x}\, dy = 0\), given that \(y = 1\), when \(x = 0\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sqrt{1-y^2}=(x-1)e^x +1\)
The equation is variables separable. Moving the dx term to the other side and multiplying by \(\frac{x}{\sqrt{1-y^2}}\):
\( \displaystyle \frac{y}{\sqrt{1 - y^2}}\,dy = -xe^x\,dx \)
On the left, put \(1 - y^2 = t\), so \(y\,dy = -\frac{1}{2}dt\):
\( \displaystyle \int \frac{y}{\sqrt{1 - y^2}}\,dy = -\frac{1}{2}\int t^{-1/2}\,dt = -\sqrt{1 - y^2} \)
On the right, by integration by parts:
\( \displaystyle \int xe^x\,dx = xe^x - \int e^x\,dx = (x - 1)e^x \)
So, with a constant C,
\( \displaystyle -\sqrt{1 - y^2} = -(x - 1)e^x - C\)
\( \displaystyle \sqrt{1 - y^2} = (x - 1)e^x + C\)
Putting x = 0, y = 1:
\(0 = (0 - 1)e^0 + C = -1 + C,\quad C = 1\)
\( \displaystyle \sqrt{1 - y^2} = (x - 1)e^x + 1\)
Why the other options are incorrect
Option (2) \(\sqrt{1-y^2}=(x+1)e^x -1\) — Incorrect. It also gives 0 = 0 at x = 0, y = 1, but \((x + 1)e^x\) is not \(\int xe^x\,dx\): its derivative is \((x + 2)e^x\), not \(xe^x\).
Option (3) \(\sqrt{y-y^2}=(x+1)e^x -1\) — Incorrect. It also gives 0 = 0 at x = 0, y = 1, but the left side must be \(\sqrt{1 - y^2}\), from the integral of \(\frac{y}{\sqrt{1 - y^2}}\).
Option (4) \(\sqrt{y}=(x-1)e^x -1\) — Incorrect. At x = 0, y = 1 it gives 1 = −2, so it fails the initial condition.