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The solution of the differential equation $\frac{d y}{d x}=\frac{x+y}{x}$ satisfying the condition $y(1)=1$ is
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Verified Answer
The correct answer is:
$y=x \ln x+x$
$y=x \ln x+x$
$$
y=v x
$$
$$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& v+x \frac{d v}{d x}=1+v \\
& \Rightarrow d v=\frac{d x}{x} \\
& \therefore v=\log x+c \\
& \Rightarrow \frac{y}{x}=\log x+c
\end{aligned}
$$
Since, $y(1)=1$, we have
$$
y=x \log x+x
$$
y=v x
$$
$$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& v+x \frac{d v}{d x}=1+v \\
& \Rightarrow d v=\frac{d x}{x} \\
& \therefore v=\log x+c \\
& \Rightarrow \frac{y}{x}=\log x+c
\end{aligned}
$$
Since, $y(1)=1$, we have
$$
y=x \log x+x
$$
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