JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation: \(\dfrac{dy}{dx}+\left(\dfrac{6x^2+(3x^2+2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})}\right)y=2+e^{-2x}\), \(x \in (-1,2)\), satisfying \(y(0)=\dfrac{3}{2}\). If \(y(1)=\alpha(2+e^{-2})\), then \(\alpha\) is equal to:
- A \(\dfrac{13}{8}\)
- B \(\dfrac{6}{13}\)
- C \(\dfrac{12}{13}\)
- D \(\dfrac{13}{12}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{13}{12}\)
Step-by-step Solution
Detailed explanation
The given differential equation is a linear differential equation of the form \(\dfrac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \dfrac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3+2)(2+e^{-2x})}\) and \(Q(x) = 2 + e^{-2x}\). We can rewrite \(P(x)\) as:…
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