Q119 / 180 · KCET 2012 · Mathematics
Solution of $e^{d y / d x}=x$ when $x=1$ and $y=0$ is
- A$y=x(\log x-1)+4$
- B$y=x(\log x-1)+3$
- C$y=x(\log x+1)+1$
- Correct answerD$y=x(\log x-1)+1$
Solution
Given, $e^{d y / d x}=x$
Taking logarithm on both sides, we get
$$
\begin{aligned}
\log e^{d y} / d x &=\log x \\
\frac{d y}{d x} &=\log x \\
d y &=\log x d x
\end{aligned}
$$
On integrating, we get
$$
\begin{aligned}
&\int \begin{aligned}
\int \mathrm{dy} &=\int \log \mathrm{xdx} \\
&=\log \mathrm{x} \int 1 \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x} \int 1 \mathrm{dx}\right] \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \frac{1}{\mathrm{x}} \times \mathrm{x} \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\mathrm{x} \\
\mathrm{y} &=\mathrm{x}(\log \mathrm{x}-1)+\mathrm{C} \\
\text { when } \mathrm{x} &=1 \text { and } \mathrm{y}=0 \\
\Rightarrow \quad 0=1(\log 1-1)+\mathrm{C}
\end{aligned}
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & 0=(0-1)+C \\
\Rightarrow & C=1
\end{array}
$$
$\therefore \mathrm{Eq}$. (i) becomes
$$
y=x(\log x-1)+1
$$
Hence, the correct answer is option: 'D'.