(i) Express \(\frac{1}{4-y^2}\) as partial fractions:

\(\frac{1}{4-y^2} = \frac{A}{2+y} + \frac{B}{2-y}\)

Multiply through by \((2+y)(2-y)\):

\(1 = A(2-y) + B(2+y)\)

Equating coefficients, solve for \(A\) and \(B\):

\(A = \frac{1}{4}, \; B = \frac{1}{4}\)

(ii) Solve the differential equation:

Separate variables:

\(\frac{dy}{4-y^2} = \frac{x}{4-y^2} dx\)

Integrate both sides:

\(\int \frac{1}{4-y^2} dy = \int x dx\)

\(\frac{1}{4} \ln |2+y| - \frac{1}{4} \ln |2-y| = \frac{1}{2}x^2 + C\)

Use initial condition \(x = 1, y = 1\):

\(\ln 3 = \frac{1}{2} + C\)

\(C = \ln 3 - \frac{1}{2}\)

Substitute back:

\(\ln x = \frac{1}{4} \ln(2+y) - \frac{1}{4} \ln(2-y) - \frac{1}{4} \ln 3\)

Rearrange to find \(y\):

\(y = \frac{2(x^4 - 1)}{3x^4 + 1}\)