If you solve your ode without initial conditions
ode = y'[x] == (y[x]^2 - x^2 - 2*x*y[x])/(y[x]^2 + 2*x*y[x] - x^2)
ic = y[1] == -1
{y1, y2} = (Values@DSolve[{ode }, y, x ] // Flatten) /. C[1] -> c1
you get two solutions.
Only the secondfirst solution may fullfill y[1]==1y[1]==-1
Plot[{y1[1], y2[1]}, {c1, -5, 5}, PlotRange -> {-2, 2},PlotStyle -> {Blue, Red}]
Asymptotic[y1[x], c1 -> Infinity]
(*ConditionalExpression[-x, x \[Element] Reals]*)
If you solve your ode without initial conditions
ode = y'[x] == (y[x]^2 - x^2 - 2*x*y[x])/(y[x]^2 + 2*x*y[x] - x^2)
ic = y[1] == -1
{y1, y2} = (Values@DSolve[{ode }, y, x ] // Flatten) /. C[1] -> c1
you get two solutions.
Only the second solution may fullfill y[1]==1
Plot[{y1[1], y2[1]}, {c1, -5, 5}, PlotRange -> {-2, 2},PlotStyle -> {Blue, Red}]
If you solve your ode without initial conditions
ode = y'[x] == (y[x]^2 - x^2 - 2*x*y[x])/(y[x]^2 + 2*x*y[x] - x^2)
ic = y[1] == -1
{y1, y2} = (Values@DSolve[{ode }, y, x ] // Flatten) /. C[1] -> c1
you get two solutions.
Only the first solution may fullfill y[1]==-1
Plot[{y1[1], y2[1]}, {c1, -5, 5}, PlotRange -> {-2, 2},PlotStyle -> {Blue, Red}]
Asymptotic[y1[x], c1 -> Infinity]
(*ConditionalExpression[-x, x \[Element] Reals]*)
If you solve your ode without initial conditions
ode = y'[x] == (y[x]^2 - x^2 - 2*x*y[x])/(y[x]^2 + 2*x*y[x] - x^2)
ic = y[1] == -1
{y1, y2} = (Values@DSolve[{ode }, y, x ] // Flatten) /. C[1] -> c1
you get two solutions.
Only the second solution may fullfill y[1]==1
Plot[{y1[1], y2[1]}, {c1, -5, 5}, PlotRange -> {-2, 2},PlotStyle -> {Blue, Red}]