Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 9 Class 12 Differential Equations
Ex 9.1, 11 (MCQ) Important
Ex 9.1, 12 (MCQ)
Question 4 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Example 8
Ex 9.3, 14
Example 12 Important
Example 13 Important
Ex 9.4, 8
Ex 9.4, 15
Example 17 Important
Ex 9.5, 7 Important You are here
Ex 9.5, 13
Ex 9.5, 14 Important
Question 6 Deleted for CBSE Board 2024 Exams
Example 21 Important
Example 22 Important
Misc 4
Misc 9
Misc 10 Important
Misc 11
Chapter 9 Class 12 Differential Equations
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.