Slide20.JPG

Slide21.JPG
Slide22.JPG
Slide23.JPG
Slide24.JPG

Go Ad-free

Transcript

Ex 9.5, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : π‘₯π‘™π‘œπ‘”π‘₯ 𝑑𝑦/𝑑π‘₯+𝑦=2/π‘₯ π‘™π‘œπ‘”π‘₯ Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q xlog x 𝑑𝑦/𝑑π‘₯ + y = 2/π‘₯ log x Dividing by x log x, 𝑑𝑦/𝑑π‘₯+𝑦" Γ— " 1/(π‘₯ log⁑π‘₯ ) = 2/π‘₯ π‘™π‘œπ‘” π‘₯" Γ— " 1/(π‘₯ log⁑π‘₯ ) π’…π’š/𝒅𝒙 + (𝟏/(𝒙 π’π’π’ˆβ‘π’™ ))π’š=𝟐/𝒙^𝟐 Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 𝟏/(𝒙 π’π’π’ˆβ‘π’™ ) & Q = 𝟐/π’™πŸ Step 3: Find Integration factor, I.F IF = e^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝐞^∫1β–’γ€–πŸ/(𝒙 π₯𝐨𝐠⁑𝒙 ) 𝒅𝒙〗 Let t = log x dt = 1/π‘₯ dx dx = x dt So, IF = e^∫1β–’γ€–1/(π‘₯ 𝑑) Γ— π‘₯𝑑𝑑〗 IF = e^∫1β–’γ€–1/𝑑 𝑑𝑑〗 IF = e^log⁑〖|𝑑|γ€— IF = |𝒕| Putting back t = log x IF = |log x| IF = log x Step 4: Solution of the equation y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹. 𝑑π‘₯+𝐢〗 Putting values, y Γ— log x = ∫1β–’πŸ/π’™πŸ . log x. dx + C Let I = 2 ∫1β–’π’π’π’ˆβ‘γ€–π’™ 𝒙^(βˆ’πŸ) 𝒅𝒙〗 Solving I I = 2 ∫1β–’π’π’π’ˆβ‘γ€–π’™ 𝒙^(βˆ’πŸ) 𝒅𝒙〗 I = 2["log x. " ∫1▒〖𝒙^(βˆ’πŸ) π’…π’™βˆ’βˆ«1β–’ 𝟏/𝒙 [∫1β–’γ€– 𝒙^(βˆ’πŸ) 𝒅𝒙〗] γ€— 𝒅𝒙" " ] I = 2 ["log x . " π‘₯^(βˆ’1)/((βˆ’1)) " βˆ’ " ∫1β–’γ€– 1/π‘₯γ€— " . " ((π‘₯^(βˆ’1)))/((βˆ’1)) ".dx " ] = 2["βˆ’ log x. " 1/π‘₯ " + " ∫1β–’γ€–1/π‘₯^2 .𝑑π‘₯γ€—] = 2[(βˆ’1)/π‘₯ " .log x βˆ’ " 1/π‘₯] = (βˆ’πŸ)/𝒙 (1 + log x) Putting value of I in (2) y log x = I + C y. log x = (βˆ’πŸ)/𝒙 (1 + log x) + C Which is the general solution of the given equation.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo