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)
Show Answer
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
$$ \begin{aligned} & y \cdot \log x=\int \log x \cdot 2 d x \\ & \Rightarrow \quad y \cdot \log x=2 \int \log x d x \\ & \Rightarrow \quad y \cdot \log x=2[x \log x-x]+c \end{aligned} $$
[
]
