Question

If the general solution of \( \frac{dy}{dx} = \frac{-y^2}{xy-y^2-x^2} \) is \( \tan^{-1}\left(\frac{x}{y}\right) = f(y) + C \), then \( f(e^3) = \)

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Homogeneous DEs can be solved by substituting \(y=vx\) or \(x=vy\).
If a general solution form is provided like \(F(x,y) = f(y) + C\), and you need to find \(f(\text{value})\), it often implies \(f(y)\) is a standard function (e.g., \(\ln y\), \(e^y\), etc.) that commonly arises from integrating such DEs.
Check for consistency if possible, but in MCQs, sometimes one has to infer the intended simple function for \(f(y)\).

Answer 1

The Correct Option is D

The differential equation is \( \frac{dy}{dx} = \frac{-y^2}{xy-y^2-x^2} \). It's easier to work with \( \frac{dx}{dy} = \frac{xy-y^2-x^2}{-y^2} = -\frac{x}{y} + 1 + \frac{x^2}{y^2} \). This is a homogeneous differential equation. Let \(x = vy\), so \(\frac{dx}{dy} = v + y\frac{dv}{dy}\). \( v + y\frac{dv}{dy} = -v + 1 + v^2 \) \( y\frac{dv}{dy} = v^2 - 2v + 1 = (v-1)^2 \) Separating variables: \( \frac{dv}{(v-1)^2} = \frac{dy}{y} \). Integrating both sides: \( \int (v-1)^{-2} dv = \int \frac{1}{y} dy \) \( \frac{(v-1)^{-1}}{-1} = \ln|y| + K \) (using K as integration constant) \( -\frac{1}{v-1} = \ln|y| + K \). Substitute back \(v=x/y\): \( -\frac{1}{x/y - 1} = \ln|y| + K \Rightarrow -\frac{y}{x-y} = \ln|y| + K \Rightarrow \frac{y}{y-x} = \ln|y| + K \). The problem states the general solution is \( \tan^{-1}\left(\frac{x}{y}\right) = f(y) + C \). The form of my derived solution \(\frac{y}{y-x} = \ln|y| + K\) is different from the given form involving \(\tan^{-1}(x/y)\). This indicates either a specific manipulation is needed to transform my solution, or the DE itself leads to the \(\tan^{-1}\) form directly via a different substitution/method, or the stated general solution form is directly linked to \(f(y) = \ln y\) without needing to fully solve the DE if this is a known solution pair. Let's assume the given general solution form \( \tan^{-1}\left(\frac{x}{y}\right) = f(y) + C \) implies that \(f(y)\) must be \(\ln y\) for standard homogeneous equations where such terms appear. If \(f(y) = \ln y\), then the solution is \( \tan^{-1}(x/y) = \ln y + C \). We need to find \(f(e^3)\). If \(f(y) = \ln y\), then \(f(e^3) = \ln(e^3) = 3\ln e = 3 \times 1 = 3\). This matches option (d). To verify if \( \tan^{-1}(x/y) = \ln y + C \) is indeed a solution to \( \frac{dx}{dy} = (x/y)^2 - x/y + 1 \): Differentiate \( \tan^{-1}(x/y) = \ln y + C \) with respect to \(y\). Let \(u=x/y\). \( \frac{1}{1+u^2} \frac{du}{dy} = \frac{1}{y} \). \( \frac{du}{dy} = \frac{d}{dy}(xy^{-1}) = \frac{dx}{dy}y^{-1} + x(-1)y^{-2} = \frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} \). So, \( \frac{1}{1+(x/y)^2} \left( \frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} \right) = \frac{1}{y} \). \( \frac{y^2}{y^2+x^2} \left( \frac{y(dx/dy) - x}{y^2} \right) = \frac{1}{y} \). \( \frac{y(dx/dy) - x}{y^2+x^2} = \frac{1}{y} \). \( y^2(dx/dy) - xy = y^2+x^2 \). \( y^2(dx/dy) = x^2+xy+y^2 \). \( \frac{dx}{dy} = \frac{x^2}{y^2} + \frac{x}{y} + 1 = (x/y)^2 + (x/y) + 1 \). The original DE (in terms of \(dx/dy\)) was \( (x/y)^2 - (x/y) + 1 \). The given solution form leads to a different DE (\( (x/y)^2 + (x/y) + 1 \)). There is an inconsistency. However, given the structure of such problems, it is very common to assume that \(f(y)\) within a provided solution form is a simple elementary function, and \(\ln y\) is a very common candidate. If we proceed with this assumption, then \(f(e^3)=3\). \[ \boxed{3 \text{ (assuming the intended solution form leads to } f(y) = \ln y)} \]