# Ex 9.5, 6 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

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Ex 9.5, 6 You are here

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Ex 9.5, 18 (MCQ)

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Last updated at April 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)