# Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

Ex 9.5, 1
Important

Ex 9.5, 2

Ex 9.5, 3 Important

Ex 9.5, 4

Ex 9.5, 5 Important

Ex 9.5, 6

Ex 9.5, 7 Important You are here

Ex 9.5, 8 Important

Ex 9.5, 9

Ex 9.5, 10

Ex 9.5, 11

Ex 9.5, 12 Important

Ex 9.5, 13

Ex 9.5, 14 Important

Ex 9.5, 15

Ex 9.5, 16 Important

Ex 9.5, 17 Important

Ex 9.5, 18 (MCQ)

Ex 9.5, 19 (MCQ) Important

Last updated at April 16, 2024 by Teachoo

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 log𝑥 ) = 2/𝑥 𝑙𝑜𝑔 𝑥" × " 1/(𝑥 log𝑥 ) 𝒅𝒚/𝒅𝒙 + (𝟏/(𝒙 𝒍𝒐𝒈𝒙 ))𝒚=𝟐/𝒙^𝟐 Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑𝑥 + Py = Q P = 𝟏/(𝒙 𝒍𝒐𝒈𝒙 ) & Q = 𝟐/𝒙𝟐 Step 3: Find Integration factor, I.F IF = e^∫1▒〖𝑝 𝑑𝑥〗 IF = 𝐞^∫1▒〖𝟏/(𝒙 𝐥𝐨𝐠𝒙 ) 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.