Separate variables: \(\frac{dy}{1-y^2} = \frac{dx}{x}\).

Express \(\frac{1}{1-y^2}\) as partial fractions: \(\frac{1}{1-y^2} = \frac{1}{2(1-y)} + \frac{1}{2(1+y)}\).

Integrate both sides: \(-\frac{1}{2} \ln(1-y) + \frac{1}{2} \ln(1+y) = \ln x + C\).

Combine logarithms: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x + C\).

Use initial condition \(x = 2, y = 0\) to find \(C\):

\(\frac{1}{2} \ln 1 = \ln 2 + C \Rightarrow C = -\ln 2\).

Substitute back: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x - \ln 2\).

Exponentiate to solve for \(y\):

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

Cross-multiply and solve for \(y\):

\(4(1+y) = x^2(1-y)\).

\(4 + 4y = x^2 - x^2y\).

\(4y + x^2y = x^2 - 4\).

\(y(4 + x^2) = x^2 - 4\).

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

(a) Start by separating variables: \(e^{-y} dy = x e^{-x} dx\).

Integrate both sides: \(\int e^{-y} dy = \int x e^{-x} dx\).

The left side integrates to \(-e^{-y}\).

For the right side, use integration by parts: let \(u = x\) and \(dv = e^{-x} dx\), then \(du = dx\) and \(v = -e^{-x}\).

Integration by parts gives: \(\int x e^{-x} dx = -xe^{-x} - \int -e^{-x} dx = -xe^{-x} - e^{-x} + C\).

Thus, \(-e^{-y} = -xe^{-x} - e^{-x} + C\).

Using the initial condition \(y = 0\) when \(x = 0\), we find \(-1 = -1 + C\), so \(C = 0\).

Therefore, \(-e^{-y} = -xe^{-x} - e^{-x}\).

Solve for \(y\): \(e^{-y} = (x+1)e^{-x}\), so \(y = -\ln((x+1)e^{-x})\).

(b) Substitute \(x = 1\) into the expression for \(y\):

\(y = -\ln((1+1)e^{-1}) = -\ln(2e^{-1}) = -\ln 2 + 1\).

Thus, \(y = 1 - \ln 2\).

First, separate the variables:

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

Integrate both sides:

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

The left side integrates to:

\(-\frac{1}{y}\).

For the right side, use integration by parts:

Let \(u = x\) and \(dv = 4e^{-2x} dx\).

Then \(du = dx\) and \(v = -2e^{-2x}\).

So, \(\int 4xe^{-2x} dx = -2xe^{-2x} - \int -2e^{-2x} dx\).

\(= -2xe^{-2x} + e^{-2x}\).

Thus, the integrated equation is:

\(-\frac{1}{y} = -2xe^{-2x} + e^{-2x} + C\).

Using the initial condition \(y = 1\) when \(x = 0\):

\(-1 = -2(0)e^{0} + e^{0} + C\).

\(-1 = 1 + C\).

\(C = -2\).

Substitute back:

\(-\frac{1}{y} = -2xe^{-2x} + e^{-2x} - 2\).

\(\frac{1}{y} = 2xe^{-2x} - e^{-2x} + 2\).

\(y = \frac{1}{2xe^{-2x} - e^{-2x} + 2}\).

Multiply numerator and denominator by \(e^{2x}\):

\(y = \frac{e^{2x}}{2x + 1}\).

(i) Separate variables: \(\frac{dy}{dx} = xe^{x+y}\) becomes \(e^{-y} dy = x e^x dx\).

Integrate both sides: \(\int e^{-y} dy = -e^{-y}\) and \(\int x e^x dx\) by parts gives \(x e^x - e^x\).

Combine: \(-e^{-y} = x e^x - e^x + C\).

Use initial condition \(y = 0\) when \(x = 0\): \(-1 = 0 - 1 + C\), so \(C = 0\).

Final solution: \(-e^{-y} = e^x(1-x)\), thus \(y = -\ln(e^x(1-x))\).

(ii) For \(y\) to be real, \(e^x(1-x) > 0\). Since \(e^x > 0\), it follows that \(1-x > 0\), so \(x < 1\).

(i) Separate variables and integrate: \(2y^{\frac{1}{2}} = \ldots\)

Integrate by parts for \(x \sin \left( \frac{1}{3}x \right)\):

\(-3x \cos \left( \frac{1}{3}x \right) + 9 \sin \left( \frac{1}{3}x \right)\)

General solution: \(y = \left( -\frac{3}{10}x \cos \left( \frac{1}{3}x \right) + \frac{9}{10} \sin \left( \frac{1}{3}x \right) + c \right)^2\)

(ii) Use \(x = 0\) and \(y = 100\) to find constant:

Substitute \(x = 25\) to find \(y\):

\(y = 203\)