JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y = y(x)\) be the solution curve of the differential equation \((1 + \sin x)\dfrac{dy}{dx} + (y+1)\cos x = 0\), \(y(0) = 0\). If the curve \(y = y(x)\) passes through the point \(\left(\alpha, \dfrac{-1}{2}\right)\), then a value of \(\alpha\) is :
- A \(\dfrac{\pi}{6}\)
- B \(\dfrac{\pi}{4}\)
- C \(\dfrac{\pi}{3}\)
- D \(\dfrac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
The given differential equation can be written as: \((1 + \sin x)dy + (y+1)\cos x dx = 0\) \(d((y+1)(1 + \sin x)) = 0\) Integrating both sides: \((y+1)(1 + \sin x) = C\) Given \(y(0) = 0\), substituting \(x = 0\) and \(y = 0\): \((0+1)(1 + \sin 0) = C \Rightarrow C = 1\) The…
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