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y′(x)−axn(y(x)2+1)=0 ✓ Mathematica : cpu = 0.13142 (sec), leaf count = 21
{{y(x)→tan(axn+1n+1+c1)}}
✓ Maple : cpu = 0.06 (sec), leaf count = 23
{y(x)=tan((xn+1+(n+1)_C1)an+1)}
y′−axn(y2+1)=0y′=axn+axny2(1)=P(x)+Q(x)y+R(x)y2
This is Ricatti first order non-linear ODE. P(x)=axn,Q(x)=0,R(x)=axn. But this is separable also. Hencey′(y2+1)=axndy(y2+1)=axndx
Integratingarctan(y(x))=axn+1n+1+C Ory(x)=tan(axn+1n+1+C)
Verification
restart; eq:=diff(y(x),x)-a*x^n*(y(x)^2+1) = 0; sol:=tan(a*x^(n+1)/(n+1)+_C1); odetest(y(x)=sol,eq); 0
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