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Ex 9.5
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
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 Aug. 11, 2023 by Teachoo
Ex 9.5, 1 For each of the differential equation given in Exercises 1 to 12, find the general solution : ππ¦/ππ₯+2π¦=π πππ₯ Step 1: Put in form ππ¦/ππ₯ + Py = Q ππ¦/ππ₯+2π¦=sinβ‘π₯ Step 2: Find P and Q Comparing (1) with ππ¦/ππ₯ + Py = Q β΄ P = 2 and Q = sin x Step 3: Find integrating factor, IF IF = e^β«1βπππ₯ IF = π^β«1β2ππ₯ IF = π^ππ Step 4 : Solution of the equation y Γ I.F = β«1βγπΓπΌ.πΉ.ππ₯+π γ Putting values, π Γ π^ππ = β«1βγπ¬π’π§β‘π π^ππ π πγ+π πΏππ‘ πΌ= β«1βγsinβ‘π₯ π^2π₯ ππ₯γ Solving I πΌ= β«1βγsinβ‘γπ₯.π^2π₯ γ.ππ₯ γ 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₯ πΌ = sin x β«1βγπ^ππ.π πββ«1βγ[π /π π πππβ‘π β«1βγπ^ππ π π γ] γ γ πΌ = sin x π^2π₯/2 β β«1βcosβ‘π₯ π^2π₯/2 dx Again using 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) = cos x & g(x) = π^2π₯ πΌ = π/π πππβ‘γπ π^ππ γβπ/π [πππβ‘π β«1βπ^ππ π π ββ«1ββ(π @π π) πππβ‘π β«1βπ^ππ π π ]dx πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [cosβ‘π₯ β«1βπ^2π₯/2 ββ«1βγ(βsin x)γ β«1βπ^2π₯/2 ππ₯ ] πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [cosβ‘π₯ π^2π₯/2+1/2 β«1βγπ¬π’π§ π± π^ππ π πγ] πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [(cosβ‘π₯ π^2π₯)/2+1/2 π°] + C I = 1/2 sin x π^2π₯ β1/4 cos x π^2π₯ β π/π I + C I + π/π I = 1/4 [2 sinβ‘γπ₯ π^2π₯ βcosβ‘γπ₯ π^2π₯ γ γ ] + C 5πΌ/4 = π^2π₯/4 [2 sinβ‘γπ₯βcosβ‘π₯ γ ] + C π° = π^ππ/π [π πππβ‘γπβπππβ‘π γ ] + C Now, Putting value of I in (2) y π^2π₯ = π^2π₯/5 [2 sinβ‘γπ₯ βcosβ‘π₯ γ ] + C Dividing by e2x y = π/π [π π¬π’π§β‘γπ βππ¨π¬β‘π γ ]+πͺπ^(βππ)