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Question Number 180550 by a.lgnaoui last updated on 13/Nov/22
Resoudre   af^′ (x)+(b/(f(x)))+c=0    (a,b,c)∈R^3
$${Resoudre}\: \\ $$$${af}^{'} \left({x}\right)+\frac{{b}}{{f}\left({x}\right)}+{c}=\mathrm{0}\:\:\:\:\left({a},{b},{c}\right)\in\mathbb{R}^{\mathrm{3}} \\ $$
Answered by mr W last updated on 13/Nov/22
a(dy/dx)=−((b+cy)/y)  ((aydy)/(b+cy))=−dx  (1−(b/(cy+b)))dy=−(c/a)dx  ∫(1−(b/(cy+b)))dy=−(c/a)∫dx  y−(b/c)ln ∣cy+b∣=−((cx)/a)+k  ⇒y+((cx)/a)−(b/c)ln ∣cy+b∣=k
$${a}\frac{{dy}}{{dx}}=−\frac{{b}+{cy}}{{y}} \\ $$$$\frac{{aydy}}{{b}+{cy}}=−{dx} \\ $$$$\left(\mathrm{1}−\frac{{b}}{{cy}+{b}}\right){dy}=−\frac{{c}}{{a}}{dx} \\ $$$$\int\left(\mathrm{1}−\frac{{b}}{{cy}+{b}}\right){dy}=−\frac{{c}}{{a}}\int{dx} \\ $$$${y}−\frac{{b}}{{c}}\mathrm{ln}\:\mid{cy}+{b}\mid=−\frac{{cx}}{{a}}+{k} \\ $$$$\Rightarrow{y}+\frac{{cx}}{{a}}−\frac{{b}}{{c}}\mathrm{ln}\:\mid{cy}+{b}\mid={k} \\ $$
Commented by a.lgnaoui last updated on 14/Nov/22
thanks
$${thanks}\: \\ $$

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