Ex 9.5

Chapter 9 Class 12 Differential Equations
Serial order wise

Transcript

Ex 9.5, 1 For each of the differential equation given in Exercises 1 to 12, find the general solution : ππ¦/ππ₯+2π¦=π πππ₯ Step 1: Put in form ππ¦/ππ₯ + Py = Q ππ¦/ππ₯+2π¦=sinβ‘π₯ Step 2: Find P and Q Comparing (1) with ππ¦/ππ₯ + Py = Q β΄ P = 2 and Q = sin x Step 3: Find integrating factor, IF IF = e^β«1βπππ₯ IF = π^β«1β2ππ₯ IF = π^ππ Step 4 : Solution of the equation y Γ I.F = β«1βγπΓπΌ.πΉ.ππ₯+π γ Putting values, π Γ π^ππ = β«1βγπ¬π’π§β‘π π^ππ ππγ+π πΏππ‘ πΌ= β«1βγsinβ‘π₯ π^2π₯ ππ₯γ Solving I πΌ= β«1βγsinβ‘γπ₯.π^2π₯ γ.ππ₯ γ Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = sin x & g (x) = π^2π₯ πΌ = sin x β«1βγπ^ππ.ππββ«1βγ[π/ππ πππβ‘π β«1βγπ^ππ ππ γ] γ γ πΌ = sin x π^2π₯/2 β β«1βcosβ‘π₯ π^2π₯/2 dx Again using by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = cos x & g(x) = π^2π₯ πΌ = π/π πππβ‘γπ π^ππ γβπ/π [πππβ‘π β«1βπ^ππ ππ ββ«1ββ(π@ππ) πππβ‘π β«1βπ^ππ ππ ]dx πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [cosβ‘π₯ β«1βπ^2π₯/2 ββ«1βγ(βsin x)γ β«1βπ^2π₯/2 ππ₯ ] πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [cosβ‘π₯ π^2π₯/2+1/2 β«1βγπ¬π’π§ π± π^ππ ππγ] πΌ = 1/2 sinβ‘γπ₯ π^2π₯ γβ1/2 [(cosβ‘π₯ π^2π₯)/2+1/2 π°] + C I = 1/2 sin x π^2π₯ β1/4 cos x π^2π₯ β π/π I + C I + π/π I = 1/4 [2 sinβ‘γπ₯ π^2π₯ βcosβ‘γπ₯ π^2π₯ γ γ ] + C 5πΌ/4 = π^2π₯/4 [2 sinβ‘γπ₯βcosβ‘π₯ γ ] + C π° = π^ππ/π [π πππβ‘γπβπππβ‘π γ ] + C Now, Putting value of I in (2) y π^2π₯ = π^2π₯/5 [2 sinβ‘γπ₯ βcosβ‘π₯ γ ] + C Dividing by e2x y = π/π [π π¬π’π§β‘γπ βππ¨π¬β‘π γ ]+πͺπ^(βππ)