JEE Mains · Maths · STD 12 - 9. differential equations
let \(y = y\left( x \right)\) be the solution of the differential equation \(\sin x\frac{{dy}}{{dx}} + ycos\;x = 4x\;\), \(x \in \left( {0,\pi } \right)\) . If \(y\left( {\frac{\pi }{2}} \right) = 0\) then \(y\left( {\frac{\pi }{6}} \right) = .\;.\;..\;\) .
- A \( - \frac{8}{{9\sqrt 3 }}{\pi ^2}\)
- B \( - \frac{8}{9}{\pi ^2}\;\;\;\;\;\;\)
- C \( - \frac{4}{9}{\pi ^2}\)
- D \(\frac{4}{{9\sqrt 3 }}{\pi ^2}\)
Answer & Solution
Correct Answer
(B) \( - \frac{8}{9}{\pi ^2}\;\;\;\;\;\;\)
Step-by-step Solution
Detailed explanation
(2) Consider the given differential equation the \(\sin x d y+y \cos x d x=4 x d x\) \(\Rightarrow d(y \sin x)=4 x d x\) Integrate both sides \(\Rightarrow \quad y \cdot \sin x=2 x^{2}+C\) ......\((1)\)…
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