Question

The particular solution of the differential equation $(1+\log x)\frac{dx}{dy}-x\log x=0$ when $x=e,y=e^2$ is:

• A. $y=\log \left(x\log x\right)+(e^2-1)$
• B. $ey=x\log x+e^3-e$
• C. $xy=e\log x-e+e^3$
• D. $y\log x=ex$

Answer 1

The correct option is A $y=\log \left(x\log x\right)+(e^2-1)$

Given : $(1+\log x)\frac{dx}{dy}-x\log x=0$

$⟹\frac{dy}{dx}=\frac{1+\log x}{x\log x}$

$⟹\frac{1+\log x}{x\log x}dx=dy$

Integrating both sides we get

$\int \frac{1+\log x}{x\log x}dx=\int dy$

Let $\log x=t$

$\therefore \frac{dx}{x}=dt$

$⟹\int \frac{1+t}{t}dt=\int dy$

$⟹y=\log \left(t\right)+t+k$ ...............[where k is a constant]

$⟹y=\log \left(\log x\right)+\log x+k$

Put $x=eandy=e^2$

$⟹e^2=\log \left(log\right(e\left)\right)+log\left(e\right)+k$

$⟹k=e^2-1$

$\therefore y=\log \left(x\log x\right)+(e^2-1)$ ....... $[\because \log a+\log b=\log (ab\left)\right]$