Question

The equation of the curve which passes through the point \( (1,0) \) and has tangent with slope \[ 1+\frac{y}{x}+\left(\frac{y}{x}\right)^2 \] is

• A. \( \tan^{-1}\!\left(\frac{x}{y}\right)=\log|x| \)
• B. \( \tan^{-1}\!\left(\frac{x}{y}\right)=\log|y| \)
• C. \( \tan^{-1}\!\left(\frac{y}{x}\right)=\log|y| \)
• D. \( \tan^{-1}\!\left(\frac{y}{x}\right)=\log|x| \)

Hint:

When the slope depends on \( \frac{y}{x} \), always try the substitution \( y=vx \).

Answer 1

The Correct Option is D

Step 1: Write the differential equation.
\[ \frac{dy}{dx} = 1+\frac{y}{x}+\left(\frac{y}{x}\right)^2 \]

Step 2: Use substitution \( y=vx \).
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \] Substituting, \[ v + x\frac{dv}{dx} = 1+v+v^2 \] \[ x\frac{dv}{dx} = 1+v^2 \]

Step 3: Separate and integrate.
\[ \frac{dv}{1+v^2} = \frac{dx}{x} \] \[ \tan^{-1}v = \log|x| + C \]

Step 4: Replace \( v=\frac{y}{x} \).
\[ \tan^{-1}\!\left(\frac{y}{x}\right)=\log|x| + C \]

Step 5: Use the point \( (1,0) \).
\[ 0 = \log 1 + C \Rightarrow C=0 \]

Step 6: Conclusion.
\[ \boxed{\tan^{-1}\!\left(\frac{y}{x}\right)=\log|x|} \]