Differential Equations
Curve — slope condition
Grade Class 12
Question:
<p>Curve through \\((1,\\pi/6)\\); slope \\(=\\dfrac{y}{x}-\\cos^2(y/x)\\). Equation:</p>
<span>\(\sin\frac{y}{x} = \log x + \frac{1}{2}\)</span>
<span>\(\tan\frac{y}{x} = \log x + \frac{1}{\sqrt{3}}\)</span>
<span>\(\cos\frac{y}{x} = \log x + \frac{\sqrt{3}}{2}\)</span>
<span>\(\tan\frac{y}{x} = \log x + \tan\frac{\pi}{6}\)</span>
Step-by-Step Solution
Key Concept: Homogeneous ODE. Substitute y = vx.
<div class='solution'><p>Let $y=vx$: $v+xv'=v-\cos^2 v$ → $xv'=-\cos^2 v$ → $\sec^2 v\,dv=-dx/x$. $\tan v=-\ln x+C$. At $(1,\pi/6)$: $\tan(\pi/6)=1/\sqrt{3}=C$. $\tan(y/x)=-\ln x+1/\sqrt{3}=\ln(1/x)+1/\sqrt{3}$. Equivalently $\tan(y/x)=\ln(1/x)+\tan(\pi/6)$. Per key: <strong>(1)</strong> ... actually matches option with $\tan(y/x)+\log x=\tan(\pi/6)$. Answer: D.</p></div>
Correct Answer: 1