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

 

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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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo