y-y-0-5-1-y-c-1-2-c-2-0-5-y-

Question Number 138564 by DomaPeti last updated on 14/Apr/21

$${y}\centerdot{y}'=\mathrm{0}.\mathrm{5}\centerdot\left(\mathrm{1}+{y}\centerdot{c}_{\mathrm{1}} \right)^{\mathrm{2}} \centerdot{c}_{\mathrm{2}} +\mathrm{0}.\mathrm{5} \\ $$$$ \\ $$$${y}=? \\ $$

Answered by mr W last updated on 15/Apr/21

2yy′=c_2 (1+c_1 y)^2 +1=c_2 c_1 ^2 y^2 +2c_1 c_2 y+c_2 +1 =ay^2 +by+c ∫((2ydy)/(ay^2 +by+c))=∫dx x=∫((2ydy)/(ay^2 +by+c)) ...

$$\mathrm{2}{yy}'={c}_{\mathrm{2}} \left(\mathrm{1}+{c}_{\mathrm{1}} {y}\right)^{\mathrm{2}} +\mathrm{1}={c}_{\mathrm{2}} {c}_{\mathrm{1}} ^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{1}} {c}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ay}^{\mathrm{2}} +{by}+{c} \\ $$$$\int\frac{\mathrm{2}{ydy}}{{ay}^{\mathrm{2}} +{by}+{c}}=\int{dx} \\ $$$${x}=\int\frac{\mathrm{2}{ydy}}{{ay}^{\mathrm{2}} +{by}+{c}} \\ $$$$… \\ $$

Commented by mr W last updated on 15/Apr/21