Question

The solution of the differential equation

$x\frac{dy}{dx} = y(\log y - \log x + 1)$ is

• A. $y = xe^{cx}$
• B. $y + xe^{cx} = 0$
• C. $y + e^{x} = 0$
• D. None

Answer 1

Answer: (A)

$y = xe^{cx}$

Answer 2

Given equation may be expressed as

$\frac{dy}{dx} = \frac{y}{x}\left\lbrack \log\left( \frac{y}{x} \right) + 1 \right\rbrack$ …….(i)

Let $\frac{y}{x} = v$ ⇒ y = vx ⇒ $\frac{dy}{dx} = v + x\frac{dv}{dx}$

∴ From (i), $v + x\frac{dv}{dx} = v(\log v + 1)$ ⇒ $x\frac{dv}{dx} = v\log v$

⇒ $\frac{dv}{v\log v} = \frac{dx}{x}$ ⇒ $\int_{}^{}{\frac{1}{\log v}d(\log v) = \int_{}^{}\frac{dx}{x}}$

∴ log (log v) = log x + log c ⇒ log (log v) = log (cx) ⇒ log v = cx ⇒ $v = e^{cx}$ ⇒ $\frac{y}{x} = e^{cx}$, ∴ $y = xe^{cx}$