Question

The solution of differential equation $\frac{dy}{dx}\frac{y}{x} = -\frac{\phi(y/x)}{\phi'(y/x)}$ is

• A. $x\phi(y/x) = k$
• B. $\phi(y/x) = kx$
• C. $y\phi(y/x) = k$
• D. $\phi(y/x) = ky$

Answer 1

Answer: (A)

A

Answer 2

The given differential equation is $\frac{dy}{dx}\frac{y}{x} = -\frac{\phi(y/x)}{\phi'(y/x)}$.

Let's test option (A): $x\phi(y/x) = k$. To verify this, we differentiate both sides with respect to $x$. Let $v = y/x$. So, $x\phi(v) = k$. Using the product rule and chain rule: $\frac{d}{dx}(x\phi(v)) = 1 \cdot \phi(v) + x \cdot \phi'(v) \frac{dv}{dx} = 0$. Substitute $\frac{dv}{dx} = \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y}{x^2}$: $\phi(v) + x\phi'(v)\left(\frac{x\frac{dy}{dx} - y}{x^2}\right) = 0$. $\phi(v) + \frac{\phi'(v)}{x}(x\frac{dy}{dx} - y) = 0$. Multiply by $x$: $x\phi(v) + \phi'(v)(x\frac{dy}{dx} - y) = 0$. $x\phi(v) + x\phi'(v)\frac{dy}{dx} - y\phi'(v) = 0$. Rearrange to solve for $\frac{dy}{dx}$: $x\phi'(v)\frac{dy}{dx} = y\phi'(v) - x\phi(v)$. $\frac{dy}{dx} = \frac{y\phi'(v) - x\phi(v)}{x\phi'(v)} = \frac{y}{x} - \frac{\phi(v)}{\phi'(v)}$. Substituting back $v = y/x$: $\frac{dy}{dx} = \frac{y}{x} - \frac{\phi(y/x)}{\phi'(y/x)}$.

Now, we compare this derived $\frac{dy}{dx}$ with the one from the given differential equation: Derived: $\frac{dy}{dx} = \frac{y}{x} - \frac{\phi(y/x)}{\phi'(y/x)}$. Given: $\frac{dy}{dx} = -\frac{x}{y}\frac{\phi(y/x)}{\phi'(y/x)}$.

For these two expressions for $\frac{dy}{dx}$ to be equal, we would need: $\frac{y}{x} - \frac{\phi(y/x)}{\phi'(y/x)} = -\frac{x}{y}\frac{\phi(y/x)}{\phi'(y/x)}$. $\frac{y}{x} = \frac{\phi(y/x)}{\phi'(y/x)} - \frac{x}{y}\frac{\phi(y/x)}{\phi'(y/x)}$. $\frac{y}{x} = \frac{\phi(y/x)}{\phi'(y/x)}\left(1 - \frac{x}{y}\right)$. $\frac{y}{x} = \frac{\phi(y/x)}{\phi'(y/x)}\left(\frac{y-x}{y}\right)$. $\frac{y^2}{x(y-x)} = \frac{\phi(y/x)}{\phi'(y/x)}$. Let $v = y/x$. $\frac{v^2x^2}{x(vx-x)} = \frac{\phi(v)}{\phi'(v)}$. $\frac{v^2x^2}{x^2(v-1)} = \frac{\phi(v)}{\phi'(v)}$. $\frac{v^2}{v-1} = \frac{\phi(v)}{\phi'(v)}$. This implies $\frac{\phi'(v)}{\phi(v)} = \frac{v-1}{v^2} = \frac{1}{v} - \frac{1}{v^2}$. Integrating this: $\int \frac{\phi'(v)}{\phi(v)} dv = \int \left(\frac{1}{v} - \frac{1}{v^2}\right) dv$. $\ln|\phi(v)| = \ln|v| + \frac{1}{v} + C_1$. $\phi(v) = C_2 v e^{1/v}$. So, option (A) is the solution only if $\phi(y/x)$ takes the specific form $C_2 (y/x) e^{x/y}$. Since $\phi(y/x)$ is a generic function, this is not generally true.

However, in the context of multiple-choice questions, especially in exams, it's common for there to be a slight variation in the problem statement, or the question expects a standard form. The form $\frac{dy}{dx} = \frac{y}{x} - \frac{\phi(y/x)}{\phi'(y/x)}$ is a standard homogeneous differential equation whose solution is $x\phi(y/x) = k$. It is highly probable that the given question intended to be this standard form.