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

Find particular solution cos y dx + (1 + 2e^-x) sin y dy = 0 ; y(0) =

Question 21 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 21 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 21 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 21 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 5

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.