JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution of the differential equation \(\frac{d y}{d x}=2(y+2 \sin x-5) x-2 \cos x\) such that \(\mathrm{y}(0)=7\). Then \(\mathrm{y}(\pi)\) is equal to :
- A \(2 \mathrm{e}^{\pi^{2}}+5\)
- B \(\mathrm{e}^{\pi^{2}}+5\)
- C \(3 \mathrm{e}^{\pi^{2}}+5\)
- D \(7 \mathrm{e}^{\pi^{2}}+5\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{e}^{\pi^{2}}+5\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}-2 x y=2(2 \sin x-5) x-2 \cos x\) \(I F=e^{-x^{2}}\) \(\text { so, } y \cdot e^{-x^{2}}=\int e^{-x^{2}}(2 x(2 \sin x-5)-2 \cos x) d x\) \(\Rightarrow y \cdot e^{-x^{2}}=e^{-x^{2}}(5-2 \sin x)+c\) \(\Rightarrow y=5-2 \sin x+c \cdot e^{x^{2}}\) Given at…
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