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, $y(e)=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1=1+\frac{e^{2}}{4}$