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


Ex 9.5, 5 For each of the differential equation given in Exercises 1 to 12, find the general solution : cos^2⁑〖π‘₯ 𝑑𝑦/𝑑π‘₯+𝑦=π‘‘π‘Žπ‘›π‘₯(0≀π‘₯<πœ‹/2)γ€— Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q cos2x.𝑑𝑦/𝑑π‘₯ + y = tan x Dividing by cos2x, 𝑑𝑦/𝑑π‘₯ + y.1/π‘π‘œπ‘ 2π‘₯ = tan⁑π‘₯/π‘π‘œπ‘ 2π‘₯ π’…π’š/𝒅𝒙 + (sec2x)y = sec2x. tan x Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = sec2 x and Q = sec2 x. tan x Step 3 : Find integrating factor, I.F I.F = e^∫1▒𝑝𝑑π‘₯ I.F = e^(∫1▒〖𝑠𝑒𝑐2π‘₯.𝑑π‘₯γ€— " " ) I.F. = etan x Step 4 : Solution of the equation y Γ— I.F. = ∫1▒〖𝑄×𝐼.𝐹.𝑑π‘₯γ€—+𝑐 Putting values, y.etan x = ∫1β–’γ€–π’”π’†π’„πŸπ’™.𝒕𝒂𝒏⁑〖𝒙.γ€— γ€— "etan x.dx + C" Let I = ∫1β–’γ€–π’”π’†π’„πŸπ’™.𝒕𝒂𝒏⁑〖𝒙.γ€— γ€— "etan x.dx" Putting t = tan x" " ∴ 𝑠𝑒𝑐2π‘₯.dx = dt Putting values of t & dt in equation ∴ I = ∫1β–’tan⁑〖π‘₯."etan x" .(𝑠𝑒𝑐2π‘₯.𝑑π‘₯)γ€— I =∫1▒〖𝒕.𝒆^𝒕.𝒅𝒕〗 I = t ∫1▒〖𝑒^𝑑 𝑑𝑑〗 βˆ’ ∫1β–’[𝑑𝑑/𝑑𝑑 ∫1▒〖𝑒^𝑑 𝑑𝑑〗] 𝑑𝑑 I = t."et" βˆ’ ∫1β–’γ€–"et" 𝑑𝑑〗 . Using by parts with ∫1▒〖𝑓(𝑑) 𝑔(𝑑) 𝑑𝑑=𝑓(𝑑) ∫1▒〖𝑔(𝑑) 𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓^β€² (𝑑) ∫1▒〖𝑔(𝑑) 𝑑π‘₯] 𝑑π‘₯γ€—γ€—γ€—γ€— Take f (t) = t & g(t) = 𝑒^𝑑 I = 𝒕"et" βˆ’ "et" Putting t = tan x I = tan x. etan x – etan x I = etan x ( tan x βˆ’ 1) Substituting value of I in (2), y etan x = etan x (tan x βˆ’ 1) + C Dividing by etan x, y = tan x βˆ’ 1 + C. e–tan x

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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.