Solve the initial value problem (x^2y^2+1)y + (xy-1)^2xy' =…

Solve the initial value problem $(x^2y^2+1)y + (xy-1)^2xy' = 0, \qquad y(a) = 1$.

Write in differential form

\[(x^2y^2+1)y\,dx + x(xy-1)^2\,dy = 0.\]

Substitute \(u = xy\)

With \(u = xy\), \(y = u/x\) and \(x\,dy/dx = du/dx - u/x\). Substituting into the equation:

\[\frac{u(u^2+1) - u(u-1)^2}{x} + (u-1)^2\frac{du}{dx} = 0.\]

Simplify the numerator

\[u(u^2+1) - u(u-1)^2 = u\left[(u^2+1)-(u^2-2u+1)\right] = 2u^2,\]

so the equation becomes

\[(u-1)^2\frac{du}{dx} = -\frac{2u^2}{x}.\]

Separate variables and integrate

\[\left(1 - \frac{2}{u} + \frac{1}{u^2}\right)du = -\frac{2\,dx}{x}\]

Integrating both sides:

\[u - 2\ln|u| - \frac{1}{u} = -2\ln|x| + C_1 \quad\Longrightarrow\quad u - \frac{1}{u} + 2\ln\frac{|x|}{|u|} = C_1.\]

Back-substitute \(u = xy\)

Since \(|x|/|u| = 1/|y|\), this becomes

\[xy - \frac{1}{xy} - 2\ln|y| = C.\]

Apply the initial condition \(y(a)=1\)

\[a(1) - \frac{1}{a(1)} - 2\ln|1| = C \quad\Longrightarrow\quad C = a - \frac{1}{a}.\]

Verify independently

The source never checks its own result, so this was verified from scratch. Implicit differentiation of \(F(x,y) = xy - \dfrac{1}{xy} - 2\ln y = C\) gives \(F_x = y + \dfrac{1}{x^2y}\) and \(F_y = x - \dfrac{2}{y} + \dfrac{1}{xy^2}\); computing \(y' = -F_x/F_y\) symbolically and comparing with \(y' = -\dfrac{(x^2y^2+1)y}{x(xy-1)^2}\) from the original equation gives an identical ratio of exactly \(1\) — the two expressions agree exactly, confirming the general solution. Separately, substituting \(x=a,\,y=1\) into \(xy-\frac1{xy}-2\ln y\) symbolically (with sympy, \(a\) real) gives \(a - 1/a\) exactly, confirming the constant. The derivation and final answer are correct as given.