Solve the differential equation (x^2 - 1)y' + y^2 - 2xy + 1…

Solve the differential equation $(x^2 - 1)y' + y^2 - 2xy + 1 = 0$.

Explanation of Method

The nonlinear terms $y^2-2xy$ suggest completing the square around $y-x$. After rewriting the right-hand side as $(x^2-1)-(y-x)^2$, the substitution $u=y-x$ turns the equation into a separable equation in $u$ and $x$, which is then integrated using partial fractions.

Isolate the derivative

Rearranging gives \[\frac{dy}{dx}=\frac{-y^2+2xy-1}{x^2-1}.\]

Complete the square in the numerator

Since $-(y-x)^2=-y^2+2xy-x^2$, we can write \[-y^2+2xy-1=-(y-x)^2+(x^2-1),\] so \[\frac{dy}{dx}=1-\frac{(y-x)^2}{x^2-1}.\]

Substitute $u=y-x$

With $u=y-x$, $u'=y'-1$, so $u'+1=1-\dfrac{u^2}{x^2-1}$, giving the separable equation \[\frac{du}{dx}=-\frac{u^2}{x^2-1}.\]

Separate and integrate

Separating variables, $\dfrac{du}{u^2}=-\dfrac{dx}{x^2-1}$. The left side integrates to $-1/u$. For the right side, partial fractions give \[\frac{1}{x^2-1}=\frac{1}{2(x-1)}-\frac{1}{2(x+1)},\] so \[\int\frac{dx}{x^2-1}=\frac12\ln\left|\frac{x-1}{x+1}\right|+C_2.\]

Combine and solve for $u$

Combining, $-\dfrac1u=-\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C$, so \[u=\frac{2}{\ln\left|\dfrac{x-1}{x+1}\right|+C}\] (relabeling the constant).

Back-substitute and verify

Since $u=y-x$, \[y=x+\frac{2}{\ln\left|\dfrac{x-1}{x+1}\right|+C}.\] Differentiating this and substituting back into the original ODE reproduces $(x^2-1)y'+y^2-2xy+1\equiv 0$ identically (confirmed symbolically), so the solution is correct.

Result

Final answer:

Verified as correct. The general solution is:

\[\boxed{\,y=x+\dfrac{2}{\ln\left|\dfrac{x-1}{x+1}\right|+C}\,}\]

equivalently $(y-x)\left[\ln\left|\frac{x-1}{x+1}\right|+C\right]=2$, valid for $x\neq\pm1$, with $C$ an arbitrary constant.