Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

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### Transcript

Ex 9.4, 9 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=sin^(β1)β‘π₯ ππ¦/ππ₯=sin^(β1)β‘π₯ ππ¦ = sin^(β1)β‘π₯ dx Integrating both sides β«1βγππ¦ γ= β«1βγsin^(β1)β‘γπ₯.1 ππ₯γ γ y = sinβ1 x β«1βγ1 ππ₯ ββ«1β[1/β(1 β π₯^2 ) β«1βγ1.ππ₯ γ] γ dx Integrating by parts, using formula β«1βγπ (π₯)π(π₯)ππ₯ γ= π(π₯) β«1βγπ(π₯)ππ₯ ββ«1βγ[πβ²(π₯)β«1βπ(π₯)ππ₯] ππ₯ γ γ Take f(x) = sinβ1 x and g(x) = 1 ("β΄" (πγ(sinγ^(β1) π₯))/ππ₯=1/β(1 β π₯^2 )) y = x γπ ππγ^(β1) π₯ β β«1βπ₯/β(1 β π₯^2 ) dx Let t = 1 β x2 dt = β2xdx x dx = (βππ‘)/2 Hence, our equation becomes y = x sinβ1 x β β«1β(βππ‘)/(2βπ‘) y = x sinβ1 x + β«1βππ‘/(2βπ‘) y = x sinβ1 x + 1/2 β«1βγπ‘^((β1)/2) ππ‘γ y = x sinβ1 x + 1/2 π‘^((β1)/2 + 1)/((β1)/2 + 1) + C y = x sinβ1 x + 1/2 (π‘^(1/2) )/((1/2) )+πΆ y = x sinβ1 x + βπ‘ + C Putting back value of t y = x sinβ1 x + β(πβπ^π ) + C