Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 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γπ¬π’π§γ^(βπ)β‘γπ.π ππγ γ 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 y = x γπππγ^(βπ) π β β«1βπ/β(π β π^π ) 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 + π/π β«1βγπ^((βπ)/π) ππγ y = x sinβ1 x + π/π π^((βπ)/π + π)/((βπ)/π + π) + 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 y = sinβ1 x β«1βγπ ππ ββ«1β[π/β(π β π^π ) β«1βγπ.ππ γ] γ dx