Let \( y(x) \) be the solution curve of the differential equation \( x(1+y^2)dx-y(1+x^2)dy=0 \) and it passes through the point (0, 1), then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y^2=2x^2+1 \)
The variables separate:
\( \displaystyle \frac{y}{1+y^2}\,dy=\frac{x}{1+x^2}\,dx \)
\( \displaystyle \frac{1}{2}\log(1+y^2)=\frac{1}{2}\log(1+x^2)+\frac{1}{2}\log\mathrm{C} \)
\( \displaystyle 1+y^2=\mathrm{C}(1+x^2) \)
At (0, 1): \( 1+1=\mathrm{C}(1+0) \), so C = 2.
\( \displaystyle 1+y^2=2+2x^2 \)
\( \displaystyle y^2=2x^2+1 \)
Shortcut: All four options pass through (0, 1), so test the equation instead. From Option (1), \( 2y\frac{dy}{dx}=4x \), so \( y\frac{dy}{dx}=2x \). The equation needs \( y\frac{dy}{dx}=\frac{x(1+y^2)}{1+x^2}=\frac{x(2x^2+2)}{1+x^2}=2x \). It matches.