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Question Number 159982 by ArielVyny last updated on 23/Nov/21

(1 + b f (x)) f^{″} (x) = \frac{p}{λ a}

s o l v e t h i s e q u a t i o n : f i n d f (x)

Answered by mr W last updated on 23/Nov/21

l e t y^{'} = u

y^{″} = \frac{d u}{d x} = \frac{d u}{d y} \times \frac{d y}{d x} = u \frac{d u}{d y}

(1 + b y) u \frac{d u}{d y} = \frac{p}{λ a}

u d u = \frac{p d y}{λ a (1 + b y)}

\int u d u = \frac{p}{λ a} \int \frac{d y}{1 + b y}

\frac{u^{2}}{2} = \frac{p}{λ a b} \ln C (1 + b y)

u = \pm \sqrt{\frac{2 p}{λ a b}} \sqrt{\ln C (1 + b y)}

\frac{d y}{d x} = \pm \sqrt{\frac{2 p}{λ a b}} \sqrt{\ln C (1 + b y)}

\frac{d y}{\sqrt{\ln C (1 + b y)}} = \pm \sqrt{\frac{2 p}{λ a b}} d x

\int \frac{d y}{\sqrt{\ln C (1 + b y)}} = \pm \sqrt{\frac{2 p}{λ a b}} x + D

\int \frac{d (1 + b y)}{\sqrt{\ln C (1 + b y)}} = \pm \sqrt{\frac{2 p b}{λ a}} x + D

\frac{\sqrt{π} e r f i (\sqrt{\ln C (1 + b y)})}{C} = \pm \sqrt{\frac{2 p b}{λ a}} x + D

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