Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations (Important Question)

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Ex 9.5, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : 𝑥𝑙𝑜𝑔𝑥 𝑑𝑦/𝑑𝑥+𝑦=2/𝑥 𝑙𝑜𝑔𝑥 Step 1: Put in form 𝑑𝑦/𝑑𝑥 + Py = Q xlog x 𝑑𝑦/𝑑𝑥 + y = 2/𝑥 log x Dividing by x log x, 𝑑𝑦/𝑑𝑥+𝑦" × " 1/(𝑥 log⁡𝑥 ) = 2/𝑥 𝑙𝑜𝑔 𝑥" × " 1/(𝑥 log⁡𝑥 ) 𝒅𝒚/𝒅𝒙 + (𝟏/(𝒙 𝒍𝒐𝒈⁡𝒙 ))𝒚=𝟐/𝒙^𝟐 Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑𝑥 + Py = Q P = 𝟏/(𝒙 𝒍𝒐𝒈⁡𝒙 ) & Q = 𝟐/𝒙𝟐 Step 3: Find Integration factor, I.F IF = e^∫1▒〖𝑝 𝑑𝑥〗 IF = 𝐞^∫1▒〖𝟏/(𝒙 𝐥𝐨𝐠⁡𝒙 ) 𝒅𝒙〗 Let t = log x dt = 1/𝑥 dx dx = x dt So, IF = e^∫1▒〖1/(𝑥 𝑡) × 𝑥𝑑𝑡〗 IF = e^∫1▒〖1/𝑡 𝑑𝑡〗 IF = e^log⁡〖|𝑡|〗 IF = |𝒕| Putting back t = log x IF = |log x| IF = log x Step 4: Solution of the equation y × I.F = ∫1▒〖𝑄×𝐼.𝐹. 𝑑𝑥+𝐶〗 Putting values, y × log x = ∫1▒𝟐/𝒙𝟐 . log x. dx + C Let I = 2 ∫1▒𝒍𝒐𝒈⁡〖𝒙 𝒙^(−𝟐) 𝒅𝒙〗 Solving I I = 2 ∫1▒𝒍𝒐𝒈⁡〖𝒙 𝒙^(−𝟐) 𝒅𝒙〗 I = 2["log x. " ∫1▒〖𝒙^(−𝟐) 𝒅𝒙−∫1▒ 𝟏/𝒙 [∫1▒〖 𝒙^(−𝟐) 𝒅𝒙〗] 〗 𝒅𝒙" " ] I = 2 ["log x . " 𝑥^(−1)/((−1)) " − " ∫1▒〖 1/𝑥〗 " . " ((𝑥^(−1)))/((−1)) ".dx " ] = 2["− log x. " 1/𝑥 " + " ∫1▒〖1/𝑥^2 .𝑑𝑥〗] = 2[(−1)/𝑥 " .log x − " 1/𝑥] = (−𝟐)/𝒙 (1 + log x) Putting value of I in (2) y log x = I + C y. log x = (−𝟐)/𝒙 (1 + log x) + C Which is the general solution of the given equation.