Differential Equations Question 6
Question 6 - 29 January - Shift 2
Let $y=y(x)$ be the solution of the differential equation $x \log _e x \frac{d y}{d x}+y=x^{2} \log _e x,(x>1)$. If $y(2)=2$, then $y(e)$ is equal to
(1) $\frac{4+e^{2}}{4}$
(2) $\frac{1+e^{2}}{4}$
(3) $\frac{2+e^{2}}{2}$
(4) $\frac{1+e^{2}}{2}$
Show Answer
Answer: (1)
Solution:
Formula: Linear differential equations of first order
$x \log _e x \frac{d y}{d x}+y=x^{2} \log _e x,(x>1)$.
$\Rightarrow \frac{dy}{dx}+\frac{y}{x \ln x}=x$
Linear differential equation
I.F. $=e^{\int \frac{1}{x \ln x} d x}=|\ln x|$
$\therefore$ Solution of differential equation
$y|\ln x|=\int x|\ln x| d x$
$=|\ln x| \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$
$\Rightarrow y|\ln x|=|\ln x|(\frac{x^{2}}{2})-\frac{x^{2}}{4}+c$
For constant
$y(2)=2 \Rightarrow c=1$
So, $y(x)=\frac{x^{2}}{2}-\frac{x^{2}}{4|\ln x|}+\frac{1}{|\ln x|}$
Hence, $y(e)=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1=1+\frac{e^{2}}{4}$
