Question

Consider the differential equation, $x \frac{dy}{dx} = y(\log_e y - \log_e x + 1)$, then which of the following are true?
(A) It is a linear differential equation
(B) It is a homogenous differential equation
(C) Its general solution is $\log_e(\frac{y}{x}) = Cx$, where C is constant of integration
(D) Its general solution is $\log_e(\frac{x}{y}) = Cy$, 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:

• A. (A), (D) and (E) only
• B. (A) and (D) only
• C. (B) and (C) only
• D. (B), (C) and (E) only

Hint:

To quickly identify a homogeneous differential equation, check if all terms have the same degree, or if the equation can be written entirely in terms of $y/x$. The substitution $y=vx$ is the standard method for solving them.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to analyze the given differential equation based on its type (linear, homogeneous), find its general solution, and then determine a particular solution given an initial condition.
Step 3: Detailed Explanation:
The given equation is $x \frac{dy}{dx} = y(\log_e y - \log_e x + 1)$.
Using properties of logarithms, we can rewrite it as:
\[ x \frac{dy}{dx} = y\left(\log_e\left(\frac{y}{x}\right) + 1\right) \] \[ \frac{dy}{dx} = \frac{y}{x}\left(\log_e\left(\frac{y}{x}\right) + 1\right) \] (A) Is it linear? A linear DE is of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Due to the term $\log_e y$, this equation is not linear. So, (A) is false.
(B) Is it homogeneous? A DE is homogeneous if it can be written in the form $\frac{dy}{dx} = F(\frac{y}{x})$. Our equation is in this form. So, (B) is true.
(C) & (D) General Solution: Since it is homogeneous, we use the substitution $y = vx$. This implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation:
\[ v + x \frac{dv}{dx} = v(\log_e v + 1) = v\log_e v + v \] \[ x \frac{dv}{dx} = v\log_e v \] Separating the variables:
\[ \frac{dv}{v\log_e v} = \frac{dx}{x} \] Integrating both sides: $\int \frac{dv}{v\log_e v} = \int \frac{dx}{x}$.
For the left integral, let $u = \log_e v$, so $du = \frac{1}{v} dv$.
\[ \int \frac{du}{u} = \ln(u) = \ln(\log_e v) \] For the right integral, $\int \frac{dx}{x} = \ln(x) + \ln(C) = \ln(Cx)$.
Equating the results:
\[ \ln(\log_e v) = \ln(Cx) \implies \log_e v = Cx \] Substitute back $v = \frac{y}{x}$:
\[ \log_e\left(\frac{y}{x}\right) = Cx \] This matches statement (C). So, (C) is true and (D) is false.
(E) Particular Solution: Given the initial condition $y(1) = 1$.
Using the general solution from (C): $\log_e(\frac{y}{x}) = Cx$.
Substitute $x=1$ and $y=1$:
\[ \log_e\left(\frac{1}{1}\right) = C(1) \implies \log_e(1) = C \implies C=0 \] The particular solution is $\log_e(\frac{y}{x}) = 0$.
Taking the exponential of both sides:
\[ \frac{y}{x} = e^0 = 1 \implies y=x \] So, (E) is true.
Step 4: Final Answer:
The true statements are (B), (C), and (E). This corresponds to option (4).