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(A) Let $y^2 = t$. Then $2y \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$.Substituting this into the differential equation:$\left( {{e^{{x^2}}} + {e^t}} \rig

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${e^{{x^2}}} (y^2 - 1) + {e^{{y^2}}} = C$

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(A) Let $y^2 = t$. Then $2y \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$.Substituting this into the differential equation:$\left( {{e^{{x^2}}} + {e^t}} \right) \frac{1}{2} \frac{{dt}}{{dx}} + {e^{{x^2}}} x (t - 1) = 0$$\left( {{e^{{x^2}}} + {e^t}} \right) \frac{{dt}}{{dx}} + 2 x {e^{{x^2}}} (t - 1) = 0$Rearranging the terms:$\frac{{dx}}{{dt}} (e^t + {e^{{x^2}}}) + 2 x {e^{{x^2}}} (t - 1) = 0$Let $z = {e^{{x^2}}}$,then $\frac{{dz}}{{dt}} = {e^{{x^2}}} \cdot 2x \frac{{dx}}{{dt}}$.Substituting $z$ into the equation:$\frac{{dz}}{{dt}} + z + {e^t} + z(t - 1) \frac{{dx}}{{dt}} \cdot \frac{1}{x} \dots$ (This approach is complex,let's use the substitution $z = {e^{{x^2}}}$ directly).Given $\left( {{e^{{x^2}}} + {e^{{y^2}}}} \right) y \frac{{dy}}{{dx}} + {e^{{x^2}}} x (y^2 - 1) = 0$.Let $y^2 = t$,then $2y \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$.$\frac{1}{2} (e^{{x^2}} + e^t) \frac{{dt}}{{dx}} + x e^{{x^2}} (t - 1) = 0$.Let $u = e^{{x^2}}$,then $\frac{{du}}{{dx}} = 2x e^{{x^2}} \implies du = 2x e^{{x^2}} dx$.This is a linear differential equation in $z = e^{{x^2}}$ with respect to $t$:$\frac{{dz}}{{dt}} + \frac{z}{t-1} = -\frac{e^t}{t-1}$.Integrating factor $I.F. = e^{\int \frac{1}{t-1} dt} = t-1$.$z(t-1) = \int -e^t dt = -e^t + C$.Substituting back $z = e^{{x^2}}$ and $t = y^2$:$e^{{x^2}}(y^2 - 1) + e^{{y^2}} = C$.

The solution of the differential equation $\left( {{e^{{x^2}}} + {e^{{y^2}}}} \right) y \frac{{dy}}{{dx}} + {e^{{x^2}}}(x{y^2} - x) = 0$ is

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