The solution of the differential equation $\frac{d y}{d x}=\frac{x+y}{x}$ satisfying the condition $y(1)=1$ is
The solution of the differential equation $\frac{d y}{d x}=\frac{x+y}{x}$ satisfying the condition $y(1)=1$ is
-
$y=\ln x+x$
-
$y=x \ln x+x^2$
-
$y=x e^{(x-1)}$
-
$y=x \ln x+x$
Solution
$
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
$
Asked in: JEE Main 2008
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