Differential Equations Ques 49
- Let $y=y(x)$ be the solution of the differential equation, $x \frac{d y}{d x}+y=x \log _e x,(x>1)$. If $2 y(2)=\log _e 4-1$, then $y(e)$ is equal to
(2019 Main, 12 Jan I)
(a) $-\frac{e}{2}$
(b) $-\frac{e^{2}}{2}$
(c) $\frac{e}{4}$
(d) $\frac{e^{2}}{4}$
Show Answer
Answer:
Correct Answer: 49.(c)
Solution:
Formula:
ELEMENTARY TYPES OF FIRST ORDER \& FIRST DEGREE DIFFERENTIAL EQUATIONS :
- Given differential equation is
$$ \begin{aligned} & & x \frac{d y}{d x}+y=x \log _e x,(x>1) \\ \Rightarrow \quad & \frac{d y}{d x}+\frac{1}{x} y & =\log _e x \end{aligned} $$
Which is a linear differential equation.
So, if $=e^{\int \frac{1}{x} d x}=e^{\log _e x}=x$
Now, solution of differential Eq. (i), is
$$ \begin{array}{cc} & y \times x=\int\left(\log _e x\right) x d x+C \\ \Rightarrow & y x=\frac{x^{2}}{2} \log _e x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C \\ \Rightarrow \quad & \quad \text { [using integrat } \\ & y x=\frac{x^{2}}{2} \log _e x-\frac{x^{2}}{4}+C \end{array} $$
$$ \text { [using integration by parts] } $$
Given that, $\quad 2 y(2)=\log _e 4-1$
On substituting, $x=2$, in Eq. (ii),
we get
$$ 2 y(2)=\frac{4}{2} \log _e 2-\frac{4}{4}+C, $$
[where, $y(2)$ represents value of $y$ at $x=2$ ]
$$ \Rightarrow \quad 2 y(2)=\log _e 4-1+C $$
$\left[\because m \log a=\log a^{m}\right]$ From Eqs. (iii) and (iv), we get
$$ C=0 $$
So, required solution is
$$ \begin{aligned} y x & =\frac{x^{2}}{2} \log _e x-\frac{x^{2}}{4} \\ x & =e, e y(e)=\frac{e^{2}}{2} \log _e e-\frac{e^{2}}{4} \end{aligned} $$
Now, at
[where, $y(e)$ represents value of $y$ at $x=e$ ]
$$ \Rightarrow \quad y(e)=\frac{e}{4} \quad\left[\because \log _e e=1\right] $$
BETA
