Question:
The slope of the tangent at \((x,y)\) to a curve passing through \(\left(1, \frac{\pi}{4}\right)\) is given by \(\frac{y}{x} - \cos^2\left(\frac{y}{x}\right)\) then the equation of the curve is
The slope of the tangent at \((x,y)\) to a curve passing through \(\left(1, \frac{\pi}{4}\right)\) is given by \(\frac{y}{x} - \cos^2\left(\frac{y}{x}\right)\) then the equation of the curve is
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Use homogeneous substitution \(y = vx\).
Updated On: Apr 23, 2026
- \(y = \tan^{-1}\left(\log \frac{c}{x}\right)\)
- \(y = x\tan^{-1}\left(\log \frac{x}{c}\right)\)
- \(y = x\tan^{-1}\left(\log \frac{c}{x}\right)\)
- None of these
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The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ \frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \]
Step 2: Calculation / Simplification}
Let \(y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(v + x\frac{dv}{dx} = v - \cos^2 v \Rightarrow x\frac{dv}{dx} = -\cos^2 v\)
\(\sec^2 v dv = -\frac{dx}{x} \Rightarrow \tan v = -\log x + \log c = \log(c/x)\)
\(\tan(y/x) = \log(c/x) \Rightarrow y = x\tan^{-1}\left(\log \frac{c}{x}\right)\)
At \((1, \pi/4)\): \(\tan(\pi/4) = 1 = \log c \Rightarrow c = e\)
Step 3: Final Answer
\[ y = x\tan^{-1}\left(\log \frac{c}{x}\right) \]
\[ \frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \]
Step 2: Calculation / Simplification}
Let \(y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(v + x\frac{dv}{dx} = v - \cos^2 v \Rightarrow x\frac{dv}{dx} = -\cos^2 v\)
\(\sec^2 v dv = -\frac{dx}{x} \Rightarrow \tan v = -\log x + \log c = \log(c/x)\)
\(\tan(y/x) = \log(c/x) \Rightarrow y = x\tan^{-1}\left(\log \frac{c}{x}\right)\)
At \((1, \pi/4)\): \(\tan(\pi/4) = 1 = \log c \Rightarrow c = e\)
Step 3: Final Answer
\[ y = x\tan^{-1}\left(\log \frac{c}{x}\right) \]
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