Solve the initial value problem (x^2 - a^2) dx + 2xy dy = 0…

Solve the initial value problem $(x^2 - a^2)\,dx + 2xy\,dy = 0, \qquad y(2a) = 3a$.

Check exactness

With \(M = x^2 - a^2\) and \(N = 2xy\): \(M_y = 0\), \(N_x = 2y\). Since \(M_y \ne N_x\), the equation is not exact.

Find the integrating factor

\[\frac{M_y - N_x}{N} = \frac{0 - 2y}{2xy} = -\frac{1}{x},\]

which depends only on \(x\), so \(\mu(x) = e^{\int(-1/x)dx} = 1/x\).

Multiply through and re-check exactness

Multiplying by \(1/x\):

\[\left(x - \frac{a^2}{x}\right)dx + 2y\,dy = 0.\]

Now \(\tilde M_y = 0 = \tilde N_x\), so the equation is exact.

Find the potential function

Integrating \(\tilde M\) with respect to \(x\):

\[F(x,y) = \frac{x^2}{2} - a^2\ln|x| + g(y).\]

Since \(F_y = g'(y)\) must equal \(\tilde N = 2y\), \(g(y) = y^2\). Thus

\[F(x,y) = \frac{x^2}{2} - a^2\ln|x| + y^2 = C.\]

Apply the initial condition \(y(2a) = 3a\)

\[\frac{(2a)^2}{2} - a^2\ln(2a) + (3a)^2 = C \quad\Longrightarrow\quad 2a^2 + 9a^2 - a^2\ln(2a) = C,\]

so \(C = a^2\left[11 - \ln(2a)\right]\).

Verify independently

The source applies the initial condition without first checking the general solution, so this was verified independently. Confirmed with sympy: (1) after multiplying by \(1/x\), \(\tilde M_y = \tilde N_x = 0\) identically; (2) differentiating \(F(x,y) = \dfrac{x^2}{2} - a^2\ln x + y^2\) gives \(F_x = x - a^2/x\) and \(F_y = 2y\), which multiplied by \(dx\) and \(dy\) and then by \(x\) reproduces exactly \((x^2-a^2)\,dx + 2xy\,dy = 0\); (3) substituting \(x=2a,\,y=3a\) into \(F\) with \(a\) treated as a positive symbolic variable and solving for the constant gives \(C = a^2(11-\ln(2a))\) exactly, matching the source. No corrections were needed.