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f(x)y(x)2+g(x)y(x)+y′(x)=0 ✓ Mathematica : cpu = 0.090378 (sec), leaf count = 54
{{y(x)→exp(∫1x−g(K[1])dK[1])−∫1x−exp(∫1K[2]−g(K[1])dK[1])f(K[2])dK[2]+c1}} ✓ Maple : cpu = 0.024 (sec), leaf count = 28
{y(x)=e∫−g(x)dx∫e∫−g(x)dxf(x)dx+_C1}
y2f+gy+y′=0y′=−gy−y2f(1)=P(x)+Q(x)y+R(x)y2
This is Bernoulli first order non-linear ODE. P(x)=0,Q(x)=−g,R(x)=f. First step is to divide by y2(2)y′y2=−g1y−f
Let u=1y, then u′=−y′y2 and (2) becomes−u′=−gu−fu′−gu=f
Integrating factor is e−∫gdx henced(e−∫gdxu)=fe−∫gdxe−∫gdxu=∫fe−∫gdxdx+Cu=e∫gdx(∫fe−∫gdxdx+C)
Hence y=1e∫gdx(∫fe−∫gdx+C)=e−∫gdx∫fe−∫gdxdx+C
Let β=e−∫gdx theny=β∫fβdx+C
Verification
restart; eq:=diff(y(x),x)+f(x)*y(x)^2+g(x)*y(x) = 0; beta:=exp(-Int(g(x),x)): my_sol:=beta/(Int(f(x)*beta,x)+_C1); odetest(y(x)=my_sol,eq); 0
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