dydx+yg'(x)=g(x)g'(x)I.F= e∫g'(x) dx=eg(x)Solution isy.(I.F)=∫g(x)g'(x)(I.F) dx⇒ y eg(x)= ∫g(x)eg(x)g'(x)dxLet g(x)=v⇒ g'(x)dx=dv= ∫v evdv=v.∫evdv−∫((v)'∫evdv)dv=v.ev−∫evdv= v.ev−ev+cyeg(x)=(g(x)−1)eg(x)y+1−g(x)c=0.e− g(x)⇒ −g(x)=log (1+y−g(x)c)⇒ g(x)+log(1+y−g(x))=c

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General solution of differnetial equation $\frac{dy}{dx} + {yg}^{΄} (x) = g (x) g^{΄} (x)$ where $g (x)$ is a function $x)$

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$g (x) - \log (1 + y - g (x)) = c$

$g (x) + \log (1 + y + g (x)) = c$

$g (x) + \log (y + g (x)) = c$

$g (x) + \log (1 + y - g (x)) = c$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} + y g' (x) = g (x) g' (x) \\ I . F = e^{\int g' (x) d x} = e^{g (x)} \\ S o l u t i o n i s \\ y . (I . F) = \int g (x) g' (x) (I . F) d x \\ \Rightarrow y e^{g (x)} = \int g (x) e^{g (x)} g' (x) d x \\ L e t g (x) = v \\ \Rightarrow g' (x) d x = d v \\ = \int v e^{v} d v \end{array}$

$\begin{array}{l} = v . \int e^{v} d v - \int ((v)' \int e^{v} d v) d v \\ = v . e^{v} - \int e^{v} d v \\ = v . e^{v} - e^{v} + c \\ y e^{g (x)} = (g (x) - 1) e^{g (x)} \\ \frac{y + 1 - g (x)}{c} = 0. e^{- g (x)} \\ \Rightarrow - g (x) = \log (\frac{1 + y - g (x)}{c}) \\ \Rightarrow g (x) + \log (1 + y - g (x)) = c \end{array}$