Question

$\left(x+y+1\right)\frac{dy}{dx}=1$

Answer 1

We have,
$$\begin{gathered} \left(x+y+1\right)\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\left(x+y+1\right)} \end{gathered}$$

$$\begin{gathered} \mathrm{Let}x+y+1=v \\ \therefore 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1 \\ \therefore \frac{dv}{dx}-1=\frac{1}{v} \\ \Rightarrow \frac{dv}{dx}=\frac{1}{v}+1 \\ \Rightarrow \frac{v}{v+1}dv=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{v}{v+1}dv=\int dx \\ \Rightarrow \int \frac{v+1-1}{v+1}dv=\int dx \\ \Rightarrow \int \left(1-\frac{1}{v+1}\right)dv=\int dx \\ \Rightarrow v-\log \left|v+1\right|=x+K \\ \Rightarrow x+y+1-\log \left|x+y+1+1\right|=x+K \\ \Rightarrow y-\log \left|x+y+2\right|=K-1 \\ \Rightarrow y-\log \left|x+y+2\right|=C_1\left(C_1=K-1\right) \\ \Rightarrow y-C_1=\log \left|x+y+2\right| \\ \Rightarrow e^{y-C_1}=x+y+2 \\ \Rightarrow \frac{e^y}{e^{C_1}}=x+y+2 \\ \Rightarrow e^{-C_1}{e}^y=x+y+2 \\ \Rightarrow C{e}^y=x+y+2\left(C=e^{-C_1}\right) \\ \Rightarrow x=C{e}^y-y-2 \end{gathered}$$