Last updated at Dec. 16, 2024 by Teachoo
Ex 9.3, 9 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=sin^(β1)β‘π₯ππ¦/ππ₯=sin^(β1)β‘π₯ ππ¦ = sin^(β1)β‘π₯ dx Integrating both sides β«1βγπ π γ= β«1βγγπ¬π’π§γ^(βπ)β‘γπ.π π πγ γ Integrating by parts, using formula β«1βγπ (π₯)π(π₯)ππ₯ γ= π(π₯) β«1βγπ(π₯)ππ₯ ββ«1βγ[πβ²(π₯)β«1βπ(π₯)ππ₯] ππ₯ γ γ Take f(x) = sinβ1 x and g(x) = 1 y = x γπππγ^(βπ) π β β«1βπ/β(π β π^π ) dx Let t = 1 β x2 dt = β2xdx x dx = (βππ‘)/2 Hence, our equation becomes y = x sinβ1 x β β«1β(βππ‘)/(2βπ‘) y = x sinβ1 x + β«1βππ‘/(2βπ‘) y = x sinβ1 x + π/π β«1βγπ^((βπ)/π) π πγ y = x sinβ1 x + π/π π^((βπ)/π + π)/((βπ)/π + π) + C y = x sinβ1 x + 1/2 (π‘^(1/2) )/((1/2) )+πΆ y = x sinβ1 x + βπ‘ + C Putting back value of t y = x sinβ1 x + β(πβπ^π ) + C y = sinβ1 x β«1βγπ π π ββ«1β[π/β(π β π^π ) β«1βγπ.π π γ] γ dx
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About the Author
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