Ex 9.5, 5 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.5, 5 For each of the differential equation given in Exercises 1 to 12, find the general solution : cos^2⁡〖𝑥 𝑑𝑦/𝑑𝑥+𝑦=𝑡𝑎𝑛𝑥(0≤𝑥<𝜋/2)〗 Step 1: Put in form 𝑑𝑦/𝑑𝑥 + Py = Q cos2x.𝑑𝑦/𝑑𝑥 + y = tan x Dividing by cos2x, 𝑑𝑦/𝑑𝑥 + y.1/𝑐𝑜𝑠2𝑥 = tan⁡𝑥/𝑐𝑜𝑠2𝑥 𝒅𝒚/𝒅𝒙 + (sec2x)y = sec2x. tan x Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑𝑥 + Py = Q P = sec2 x and Q = sec2 x. tan x Step 3 : Find integrating factor, I.F I.F = e^∫1▒𝑝𝑑𝑥 I.F = e^(∫1▒〖𝑠𝑒𝑐2𝑥.𝑑𝑥〗 " " ) I.F. = etan x Step 4 : Solution of the equation y × I.F. = ∫1▒〖𝑄×𝐼.𝐹.𝑑𝑥〗+𝑐 Putting values, y.etan x = ∫1▒〖𝒔𝒆𝒄𝟐𝒙.𝒕𝒂𝒏⁡〖𝒙.〗 〗 "etan x.dx + C" Let I = ∫1▒〖𝒔𝒆𝒄𝟐𝒙.𝒕𝒂𝒏⁡〖𝒙.〗 〗 "etan x.dx" Putting t = tan x" " ∴ 𝑠𝑒𝑐2𝑥.dx = dt Putting values of t & dt in equation ∴ I = ∫1▒tan⁡〖𝑥."etan x" .(𝑠𝑒𝑐2𝑥.𝑑𝑥)〗 I =∫1▒〖𝒕.𝒆^𝒕.𝒅𝒕〗 I = t ∫1▒〖𝑒^𝑡 𝑑𝑡〗 − ∫1▒[𝑑𝑡/𝑑𝑡 ∫1▒〖𝑒^𝑡 𝑑𝑡〗] 𝑑𝑡 I = t."et" − ∫1▒〖"et" 𝑑𝑡〗 . Using by parts with ∫1▒〖𝑓(𝑡) 𝑔(𝑡) 𝑑𝑡=𝑓(𝑡) ∫1▒〖𝑔(𝑡) 𝑑𝑥 −∫1▒〖[𝑓^′ (𝑡) ∫1▒〖𝑔(𝑡) 𝑑𝑥] 𝑑𝑥〗〗〗〗 Take f (t) = t & g(t) = 𝑒^𝑡 I = 𝒕"et" − "et" Putting t = tan x I = tan x. etan x – etan x I = etan x ( tan x − 1) Substituting value of I in (2), y etan x = etan x (tan x − 1) + C Dividing by etan x, y = tan x − 1 + C. e–tan x