Ex 9.5, 6 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 9.5, 6 For each of the differential equation given in Exercises 1 to 12, find the general solution : π₯ ππ¦/ππ₯+2π¦=π₯^2 ππππ₯ Step 1 : Convert into ππ¦/ππ₯ + py = Q π₯ ππ¦/ππ₯+2π¦=π₯^2 ππππ₯ Dividing both sides by x π π/π π + ππ/π = x log x Step 2 : Find P and Q Differential equation is of the form ππ¦/ππ₯+ππ¦=π where P = π/π and Q = x log x Step 3 : Finding integrating factor IF = π^β«1βγπ ππ₯γepdx IF = π^β«1βγ2/π₯ ππ₯γ IF = π^(2β«1βγ1/π₯ ππ₯γ) IF = π^(2 logβ‘π₯ ) IF = π^logβ‘γπ₯^2 γ IF = π^π Step 4 : Solution of the equation Solution is y (IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yx2 = β«1βγπ₯ logβ‘γπ₯Γπ₯^2 ππ₯+πγ γ yx2 = β«1βγπππβ‘π π^π γ + π Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = sin x & g (x) = π^2π₯ yx2 = log x β«1βγπ₯^3 ππ₯ββ«1β[π/ππ₯ logβ‘γπ₯ γ β«1βγπ₯^3 ππ₯γ] γdx yx2 = log x (π₯^4/4)ββ«1β1/π₯ (π₯^4/4)ππ₯+π yx2 = (π₯^4 logβ‘π₯)/4 β β«1βπ₯^3/4 ππ₯+π yx2 = (π₯^4 logβ‘π₯)/4 β π₯^4/(4 Γ 4)+π yx2 = (π^π πππβ‘π)/π β π^π/ππ+π y = (π₯^4 logβ‘π₯)/(4π₯^2 ) β π₯^4/(16π₯^2 ) + πΆ/π₯^2 y = (π₯^2 logβ‘γ|π₯|γ)/4 β π₯^2/16 + ππ₯^(β2) y = π^π/ππ (4 log |"x" | β 1) + ππ^(βπ)
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Ex 9.5, 6 You are here
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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