Question

The general solution of the differential equation $\frac{dy}{dx}+y$ g' (x) = g (x) g' (x), where g (x) is a given function of x, is
(a) g (x) + log {1 + y + g (x)} = C
(b) g (x) + log {1 + y − g (x)} = C
(c) g (x) − log {1 + y − g (x)} = C
(d) none of these

Answer 1

(b) g (x) + log {1 + y − g (x)} = C


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+yg\text{'}\left(x\right)=g\left(x\right)g\text{'}\left(x\right).....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=g\text{'}\left(x\right)\mathrm{and}Q=g\left(x\right)g\text{'}\left(x\right). \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int g\text{'}\left(x\right)dx} \\ =e^{g\left(x\right)} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.,\mathrm{we}\mathrm{get} \\ {e}^{g\left(x\right)}\left(\frac{dy}{dx}+yg\text{'}\left(x\right)\right)=e^{g\left(x\right)}g\left(x\right)g\text{'}\left(x\right) \\ \Rightarrow e^{g\left(x\right)}\frac{dy}{dx}+e^{g\left(x\right)}yg\text{'}\left(x\right)=e^{g\left(x\right)}g\left(x\right)g\text{'}\left(x\right) \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^{g\left(x\right)}=\int e^{g\left(x\right)}g\left(x\right)g\text{'}\left(x\right)dx+K \\ \Rightarrow y{e}^{g\left(x\right)}=I+K \\ \mathrm{where}I=\int e^{g\left(x\right)}g\left(x\right)g\text{'}\left(x\right)dx \\ \mathrm{Now}, \\ I=\int e^{g\left(x\right)}g\left(x\right)g\text{'}\left(x\right)dx \\ \mathrm{Putting}g\left(x\right)=t,\mathrm{we}\mathrm{get} \\ g\text{'}\left(x\right)dx=dt \\ \therefore I=\int \underset{\mathrm{I}}{t}\underset{\mathrm{II}}{e^t}dt \\ =t\int e^{t}dt-\int \left[\frac{d}{dx}\left(t\right)\int e^{t}dt\right]dt \\ =t{e}^t-e^t \\ =g\left(x\right)e^{g\left(x\right)}-e^{g\left(x\right)} \\ \therefore y{e}^{g\left(x\right)}=g\left(x\right)e^{g\left(x\right)}-e^{g\left(x\right)}+K \\ \Rightarrow y{e}^{g\left(x\right)}+e^{g\left(x\right)}-g\left(x\right)e^{g\left(x\right)}=K \\ \Rightarrow y+1-g\left(x\right)=K{e}^{-g\left(x\right)} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log \left\{y+1-g\left(x\right)\right\}=-g\left(x\right)+\log K \\ \Rightarrow g\left(x\right)+\log \left\{1+y-g\left(x\right)\right\}=C\left(\mathrm{Where},C=\log K\right) \end{gathered}$$