Question:
The general solution of the differential equation \(secy+(x-e^{siny})\frac{dy}{dx} = 0\) is...
The general solution of the differential equation \(secy+(x-e^{siny})\frac{dy}{dx} = 0\) is...
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Separate variables after noticing that d(e^x cos^2 x) appears.
Updated On: Oct 1, 2026
- \(e^{siny} = x+c\)
- \(xe^{siny} = \frac{e^{2siny}}{2}+c\)
- \(2xcosy = e^x+c\)
- \(siny-e^{siny} = c\)
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The Correct Option is B
Solution and Explanation
Step 1: Separate the variables:
\(e^{-x}(y + 1)\,dy = -(\cos^2x - \sin2x)y\,dx\). Divide by \(y\) and multiply by \(e^x\):
\[ \frac{y + 1}{y}\,dy = -e^x(\cos^2x - \sin2x)\,dx \]
Step 2: Recognise the derivative:
\(\frac{d}{dx}\left(e^x\cos^2x\right) = e^x\cos^2x - e^x\cdot2\sin x\cos x = e^x(\cos^2x - \sin2x)\).
So the right hand side is \(-d(e^x\cos^2x)\).
Step 3: Integrate:
\[ y + \log y = -e^x\cos^2x + C \Rightarrow \log y + y + e^x\cos^2x = C \]
At \(x = 0\), \(y = 1\): \(0 + 1 + 1 = 2 = C\). So \(\log y + y + e^x\cos^2x = 2\), option (B).
Final Answer:
The solution is \(\log y + y + e^x\cos^2x = 2\), option (B). \[ \boxed{\log y + y + e^x\cos^2x = 2} \]
\(e^{-x}(y + 1)\,dy = -(\cos^2x - \sin2x)y\,dx\). Divide by \(y\) and multiply by \(e^x\):
\[ \frac{y + 1}{y}\,dy = -e^x(\cos^2x - \sin2x)\,dx \]
Step 2: Recognise the derivative:
\(\frac{d}{dx}\left(e^x\cos^2x\right) = e^x\cos^2x - e^x\cdot2\sin x\cos x = e^x(\cos^2x - \sin2x)\).
So the right hand side is \(-d(e^x\cos^2x)\).
Step 3: Integrate:
\[ y + \log y = -e^x\cos^2x + C \Rightarrow \log y + y + e^x\cos^2x = C \]
At \(x = 0\), \(y = 1\): \(0 + 1 + 1 = 2 = C\). So \(\log y + y + e^x\cos^2x = 2\), option (B).
Final Answer:
The solution is \(\log y + y + e^x\cos^2x = 2\), option (B). \[ \boxed{\log y + y + e^x\cos^2x = 2} \]
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