Differential Equations 2 Question 16
16.
Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)$. Then, $y(e)$ is equal to
(a) e
(b) 0
(c) 2
(d) $2 e$
(2015 Main)
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Answer:
Correct Answer: 16. (c)
Solution:
- Given differential equation is
$ \begin{aligned} & (x \log x) \frac{d y}{d x}+y=2 x \log x \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=2 \end{aligned} $
This is a linear differential equation.
$ \therefore \quad IF=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x $
Now, the solution of given differential equation is given by
$ y \cdot \log x=\int \log x \cdot 2 d x $
$ \Rightarrow \quad y \cdot \log x=2 \int \log x d x $
$ \Rightarrow \quad y \cdot \log x=2[x \log x-x]+c$
At $\quad x=1 \Rightarrow c=2 $
$\Rightarrow \quad y.log x=2 [x logx-x ] +2$
At $\quad x= e, y= 2 (e-e) +2$
$y=2$
