91Ó°ÊÓ

We have \(M(x, y) = -\left(\frac{y}{x^2 + y^2} - 1\right)\) and \(N(x, y) = \frac{x}{x^2 + y^2}\). Compute the partial derivatives: \(\frac{\partial M}{\partial y} = \frac{x^2-y^2}{(x^2+y^2)^2}\) and \(\frac{\partial N}{\partial x} = \frac{x^2-y^2}{(x^2+y^2)^2}\). Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.

Since the equation is exact, we can find a function \(F(x, y)\) such that \(\frac{\partial F}{\partial x} = M(x, y)\) and \(\frac{\partial F}{\partial y} = N(x, y)\). Integrate \(\frac{\partial F}{\partial x} = M(x, y)\) with respect to \(x\): \(F(x, y) = \int M(x, y) dx = \int -\left(\frac{y}{x^2 + y^2} - 1\right) dx = -y\arctan\frac{x}{y} + x + C(y)\). To find \(C(y)\), differentiate \(F(x, y)\) with respect to \(y\) and compare it with \(N(x, y)\): \(\frac{\partial F}{\partial y} = -\arctan\frac{x}{y} - \frac{xy}{x^2+y^2} + C'(y) = \frac{x}{x^2+y^2}\). This implies \(C'(y) = 0\), so \(C(y) = C\). The solution is \(F(x, y) = C\), i.e., \(-y\arctan\frac{x}{y} + x = C\). (ii) \(\frac{x dx + y dy}{\sqrt{x^2 + y^2}} = \frac{y dx - x dy}{x^2}\)

Write \(y=vx\), so \(dy = vdx + xdv\). Substitute into the given equation: \(\frac{x(vdx + xdv)}{\sqrt{x^2 + x^2v^2}} = \frac{x(vdx + xdv) - x^2dv}{x^2}\).

Simplify the equation: \(\frac{1+v^2}{\sqrt{1+v^2}} = v - \frac{1-v^2}{(1+v^2)v}\). Separate variables: \(\frac{2v^2}{(1 - v^2)^{3/2}}dv = dx\).

Integrate both sides: \(\int\frac{2v^2}{(1 - v^2)^{3/2}}dv = \int dx\). After the integration, we have: \(-\frac{2v}{\sqrt{1 - v^2}} = x + C\).

Write the solution in terms of \(x\) and \(y\): \(-\frac{2(y/x)}{\sqrt{1 - (y/x)^2}} = x + C\). Solve for \(y\): \(y = -\frac{2x^2}{\sqrt{x^2 - y^2}} - Cx\).