Q41 / 155 · WBJEE 2016 · Mathematics
If the solution of the differential equation $x \frac{d y}{d x}+y=x e^{x}$ be $x y=e^{x} \phi(x)+C,$ then $\phi(x)$ is equal to
- A$x+1$
- Correct answerB$x-1$
- C$1-x$
- D$x$
Solution
Given, $\quad x \frac{d y}{d x}+y=x e^{x}$
$\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=e^{x}$
On comparing Eq. by $\frac{d y}{d x}+P y=Q$, we get
$\therefore$
$$
\begin{array}{l}
P=\frac{1}{x} \text { and } Q=e^{x} \\
IF=e^{\int \frac{1}{x} d x}
\end{array}
$$
$$
=e^{log^{x}}=x
$$
Hence, solution of differential equation,
$$
\begin{array}{ll}
\quad y \cdot x=\int x e^{x} d x+C \\
\Rightarrow \quad x y=x e^{x}-\int e^{x} d x+C \\
\Rightarrow \quad x y=x e^{x}-e^{x}+C
\end{array}
$$
$\Rightarrow \quad x y=e^{x}(x-1)+C$
$\because xy=e^{x} \phi(x)+c$ [given]
$$
\phi(x)=(x-1)
$$
Hence, the correct answer is option: 'B'.