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The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$ is given by
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Verified Answer
The correct answer is:
$y=x+\log (x+y)+c$
$$
\frac{d y}{d x}=\frac{x+y+1}{x+y-1}
$$
Put $x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x}$
$$
\begin{aligned}
& \therefore \frac{\mathrm{du}}{\mathrm{dx}}-1=\frac{\mathrm{u}+1}{\mathrm{u}-1} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{du}+1}{\mathrm{dx}-1}+1=\frac{2 \mathrm{u}}{\mathrm{u}-1} \\
& \therefore\left(\frac{\mathrm{u}-1}{\mathrm{u}}\right) \mathrm{du}=2 \mathrm{dx}
\end{aligned}
$$
Integrating both sides, we get
$$
\begin{aligned}
& \int d u-\int \frac{d u}{u}=\int 2 d x \\
& \therefore u-\log u=2 x+c \\
& \therefore x+y-\log (x+y)=2 x+c \Rightarrow y=x+\log (x+y)+c
\end{aligned}
$$
\frac{d y}{d x}=\frac{x+y+1}{x+y-1}
$$
Put $x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x}$
$$
\begin{aligned}
& \therefore \frac{\mathrm{du}}{\mathrm{dx}}-1=\frac{\mathrm{u}+1}{\mathrm{u}-1} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{du}+1}{\mathrm{dx}-1}+1=\frac{2 \mathrm{u}}{\mathrm{u}-1} \\
& \therefore\left(\frac{\mathrm{u}-1}{\mathrm{u}}\right) \mathrm{du}=2 \mathrm{dx}
\end{aligned}
$$
Integrating both sides, we get
$$
\begin{aligned}
& \int d u-\int \frac{d u}{u}=\int 2 d x \\
& \therefore u-\log u=2 x+c \\
& \therefore x+y-\log (x+y)=2 x+c \Rightarrow y=x+\log (x+y)+c
\end{aligned}
$$
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