Consider the differential equation, \(x\displaystyle\frac{dy}{dx} = y(\log_e y - \log_e x + 1)\), then which of the following are true?
(A) It is a linear differential equation
(B) It is a homogeneous differential equation
(C) Its general solution is \(\log_e\left(\displaystyle\frac{y}{x}\right) = Cx\), where C is constant of integration
(D) Its general solution is \(\log_e\left(\displaystyle\frac{x}{y}\right) = Cy\), where C is constant of integration
(E) If \(y(1) = 1\), then its particular solution is \(y = x\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B), (C) and (E) only
Since \(\log_e y - \log_e x = \log_e \frac{y}{x}\), the equation can be written as
\(\displaystyle \frac{dy}{dx} = \frac{y}{x}\left(\log_e \frac{y}{x} + 1\right)\)
The right side is a function of \(\frac{y}{x}\) alone. Put y = vx, so that \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\(\displaystyle v + x\frac{dv}{dx} = v(\log_e v + 1)\)
\(\displaystyle x\frac{dv}{dx} = v\log_e v\)
Separating the variables:
\(\displaystyle \int \frac{dv}{v\log_e v} = \int \frac{dx}{x}\)
With \(t = \log_e v\), \(dt = \frac{dv}{v}\), the left side is \(\int \frac{dt}{t}\):
\(\log_e|\log_e v| = \log_e|x| + \log_e|C|\)
\(\log_e v = Cx\)
\(\displaystyle \log_e \frac{y}{x} = Cx\)
So the equation is homogeneous, and its general solution is \(\log_e \frac{y}{x} = Cx\).
Checking the statements
(A) It is a linear differential equation — Incorrect. The term \(y\log_e y\) is not linear in y.
(B) It is a homogeneous differential equation — Correct. \(\frac{dy}{dx}\) is a function of \(\frac{y}{x}\) alone, so it is a homogeneous differential equation.
(C) Its general solution is \(\log_e\left(\frac{y}{x}\right) = Cx\), where C is constant of integration — Correct. With \(v = \frac{y}{x}\), the solution is \(\log_e v = Cx\).
(D) Its general solution is \(\log_e\left(\frac{x}{y}\right) = Cy\), where C is constant of integration — Incorrect. \(\log_e\frac{x}{y} = -\log_e\frac{y}{x} = -Cx\), a constant times x, not a constant times y.
(E) If \(y(1) = 1\), then its particular solution is \(y = x\) — Correct. Putting x = 1, y = 1 in the general solution:
\(\log_e 1 = C(1) \Rightarrow C = 0\)
\(\displaystyle \log_e \frac{y}{x} = 0 \Rightarrow y = x\)