Solve the following differential equation: 𝑑𝑦 𝑑π‘₯ = π‘₯ 3 π‘π‘œπ‘ π‘’π‘ 𝑦, 𝑔𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑦(0) = 0.

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Solve differential equation: dy/dx = x^3 cosec y, given that y(0) = 0

Question 25 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2

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Transcript

Question 25 Solve the following differential equation: 𝑑𝑦/𝑑π‘₯ = π‘₯3 π‘π‘œπ‘ π‘’π‘ 𝑦, 𝑔𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑦(0) = 0. Given 𝑑𝑦/𝑑π‘₯ = π‘₯3 π‘π‘œπ‘ π‘’π‘ 𝑦 𝑑𝑦 Γ— 1/(π‘π‘œπ‘ π‘’π‘ 𝑦) = π‘₯3 𝑑π‘₯ 𝑑𝑦 Γ— sin y = π‘₯3 𝑑π‘₯ Integrating both sides ∫1β–’γ€–sin⁑𝑦 𝑑𝑦〗 = ∫1β–’γ€–π‘₯^3 𝑑π‘₯γ€— βˆ’π‘π‘œπ‘  𝑦 = π‘₯^4/4+𝐢 Since y(0) = 0 Putting x = 0, y = 0 βˆ’π‘π‘œπ‘  0 = 0/4+𝐢 βˆ’1 = 𝐢 π‘ͺ=βˆ’πŸ So, our equation becomes βˆ’π‘π‘œπ‘  𝑦 = π‘₯^4/4+𝐢 βˆ’π‘π‘œπ‘  𝑦 = π‘₯^4/4βˆ’1 𝒙^πŸ’/πŸ’+πœπ¨π¬β‘π’šβˆ’πŸ=𝟎

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