Question

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

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

For homogeneous differential equations, use the substitution \[ y=vx. \] Then \[ \frac{dy}{dx}=v+x\frac{dv}{dx}. \] This converts the equation into a separable differential equation.

Answer 1

The Correct Option is D

Step 1: Identify the differential equation as homogeneous.
Given, \[ \frac{dy}{dx}=\frac{y^2}{xy-y^2-x^2} \] The right-hand side is a homogeneous function of degree \(0\).
So, put \[ y=vx \] Then, \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \]

Step 2: Substitute \(y=vx\).
Now, \[ \frac{dy}{dx} = \frac{v^2x^2}{x(vx)-v^2x^2-x^2} \] \[ = \frac{v^2x^2}{vx^2-v^2x^2-x^2} \] \[ = \frac{v^2}{v-v^2-1} \] Therefore, \[ v+x\frac{dv}{dx} = \frac{v^2}{v-v^2-1} \]

Step 3: Separate the variables.
\[ x\frac{dv}{dx} = \frac{v^2}{v-v^2-1}-v \] \[ x\frac{dv}{dx} = \frac{v^2-v(v-v^2-1)}{v-v^2-1} \] \[ x\frac{dv}{dx} = \frac{v^3+v}{v-v^2-1} \] \[ x\frac{dv}{dx} = \frac{v(v^2+1)}{v-v^2-1} \] Thus, \[ \frac{v-v^2-1}{v(v^2+1)}\,dv=\frac{dx}{x} \]

Step 4: Integrate both sides.
Now, \[ \frac{v-v^2-1}{v(v^2+1)} = \frac{1}{v^2+1}-\frac{1}{v} \] So, \[ \int \left(\frac{1}{v^2+1}-\frac{1}{v}\right)dv = \int \frac{dx}{x} \] \[ \tan^{-1}v-\log v=\log x+C \]

Step 5: Substitute \(v=\frac{y}{x}\).
Since \[ v=\frac{y}{x}, \] we get \[ \tan^{-1}\left(\frac{y}{x}\right)-\log\left(\frac{y}{x}\right)=\log x+C \] \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\left(\frac{y}{x}\right)+\log x+C \] \[ \tan^{-1}\left(\frac{y}{x}\right) = \log y+C \] Thus, \[ f(y)=\log y \]

Step 6: Find \(f(e^3)\).
\[ f(e^3)=\log(e^3) \] \[ f(e^3)=3 \]

Step 7: Final conclusion.
Therefore, \[ \boxed{3} \]