Question 21 (OR 1 st question)

Find the particular solution of the following differential equation.

cos y dx + (1 + 2e -x ) sin y dy = 0; y(0) = π/4

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Question 21 (OR 1st question) Find the particular solution of the following differential equation. cos y dx + (1 + 2𝑒^(βˆ’π‘₯)) sin y dy = 0; y(0) = πœ‹/4 cos y dx + (1 + 2𝑒^(βˆ’π‘₯)) sin y dy = 0 cos y dx = – (1 + 2𝑒^(βˆ’π‘₯)) sin y dy 𝑑π‘₯/((1 + γ€–2𝑒〗^(βˆ’π‘₯))) = – sin⁑𝑦/cos⁑𝑦 dy 𝑑π‘₯/((1 + γ€–2𝑒〗^(βˆ’π‘₯))) = – tan y dy 𝑑π‘₯/((1 + 2/𝑒^π‘₯ ) ) = – tan y dy 𝑑π‘₯/(((𝑒^π‘₯ + 2)/𝑒^π‘₯ ) ) = – tan y dy (𝑒^π‘₯ 𝑑π‘₯)/((𝑒^π‘₯ + 2) ) = – tan y dy Integrating both sides ∫1β–’(𝑒^π‘₯ 𝑑π‘₯)/((𝑒^π‘₯ + 2) ) = β€“βˆ«1β–’γ€–tan⁑𝑦 𝑑𝑦〗 ∫1β–’(𝑒^π‘₯ 𝑑π‘₯)/((𝑒^π‘₯ + 2) ) = – log⁑〖 |sec⁑𝑦 |γ€— +𝐢 Let ex + 2 = t ex dx = dt ∫1▒𝑑𝑑/𝑑 = – log⁑〖 |sec⁑𝑦 |γ€— +𝐢 log⁑〖 |𝑑|γ€—= – log⁑〖 |sec⁑𝑦 |γ€— +𝐢 Putting back value of t log⁑〖 |𝑒^π‘₯+2|γ€— = – log⁑〖 |sec⁑𝑦 |γ€— +𝐢 log⁑〖 |𝑒^π‘₯+2|γ€—+ log⁑〖 |sec⁑𝑦 |γ€—=𝐢 As log a + log b = log ab log⁑〖 |(𝑒^π‘₯+2)Γ—sec⁑𝑦 |γ€—=𝐢 log⁑〖 |(𝑒^π‘₯+2)Γ—sec⁑𝑦 |γ€—=log⁑𝐾 Cancelling log (𝑒^π‘₯+2)Γ—sec⁑𝑦 = K (𝑒^π‘₯+2)Γ—1/cos⁑𝑦 = K (𝑒^π‘₯+2) = K cos y log⁑〖 |𝑒^π‘₯+2|γ€— = – log⁑〖 |sec⁑𝑦 |γ€— +𝐢 log⁑〖 |𝑒^π‘₯+2|γ€—+ log⁑〖 |sec⁑𝑦 |γ€—=𝐢 As log a + log b = log ab log⁑〖 |(𝑒^π‘₯+2)Γ—sec⁑𝑦 |γ€—=𝐢 log⁑〖 |(𝑒^π‘₯+2)Γ—sec⁑𝑦 |γ€—=log⁑𝐾 Cancelling log (𝑒^π‘₯+2)Γ—sec⁑𝑦 = K (𝑒^π‘₯+2)Γ—1/cos⁑𝑦 = K (𝑒^π‘₯+2) = K cos y We need to find the particular solution for y(0) = πœ‹/4 Putting x = 0, y = πœ‹/4 in (1) (𝑒^π‘₯+2) = K cos y (𝑒^0+2) = K cos πœ‹/4 (1+2) = K cos 45Β° 3 = K Γ— 1/√2 3√2 = K K = 3√2 Putting value of K in (1) (𝑒^π‘₯+2) = K cos y (𝑒^π‘₯+2) = 3√2 cos y Hence, the particular solution is 𝒆^𝒙+𝟐 = 3√𝟐 cos y

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.