English: So, you apply the boundary condition, y(a) = y(b), so you get So, what you get the equation is So, this is equation number-one, if you apply other boundary condition, other periodic boundary condition y^(′)(a) = y^(′)(b), what you get is μ(c₁e^(μa) − c₂e^(−μa)), that is what if you differentiate and put x = a, which is same μ(c₁e^(μb) − c₂e^(−μb)) So μ, μ you can cancel because μ is positive, so nonzero and you get c₁(e^(μa)−e^(μb)) = c₂(e^(−μa)−e^(−μb)) and this when you bring it to this side, it becomes plus and you have finally so this is equation number 2.
Telugu: కాబట్టి మీరు c₁(e^(μa)−e^(μb)) = c₂(e^(−μa)−e^(−μb))ని పొందుతారు మరియు మీరు దానిని మరొక వైపుకు తీసుకువెళ్లినప్పుడు, అది ప్లస్ అవుతుంది మరియు మీకు చివరకు c₁(e^(μa)−e^(μb)) − c₂(e^(−μa)−e^(−μb)) = 0  − (2) లభిస్తుంది.