Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 11 Find a particular solution satisfying the given condition : (π₯^3+π₯^2+π₯+1) ππ¦/ππ₯=2π₯^2+π₯; π¦=1 when π₯=0(π₯^3+π₯^2+π₯+1) ππ¦/ππ₯=2π₯^2+π₯ ππ = (ππ^π + π)/(π^π + π^π + π + π) ππ Integrating both sides β«1βγππ¦=β«1β(2π₯^2 + π₯)/(π₯3 + π₯2 + π₯ + 1)γ dx y = β«1β(ππ^π + π)/( (π + π)(π^π + π)) dx Rough x = β1 is a solution of x3 + x2 + x + 1 as (-1)2 + (-1)2 + (β1) + 1 = 0 Hence (x + 1) is one of its factors. So, we can write x3 + x2 + x + 1 = (x + 1) (x2 + 1) Integrating by partial fractions, using formula (ππ^π +π)/((π + π)(π^π+π)) = π¨/(π + π)+(π©π + πͺ)/(π^π + π) (2π₯^2 +π₯)/((π₯ + 1)(π₯^2+1)) = (π΄(π₯^2+1) + (π΅π₯ + π)(π₯ + 1))/((π₯ + 1)(π₯^2 + 1)) 2π₯^2 + x = A (π₯^2+ 1) + (Bx + C) (x + 1) Putting x = β1 2(β1)2 β 1 = A ((β1)2 + 1) + (B(β1) + C)(β1 + 1) 2 β 1 = A(2) + (βB + C)(0) 1 = 2A A = π/π Putting x = 0 0 = A (0 + 1) + (B(0) + C)(0 + 1) 0 = A + C(1) A = βC Since A = π/π β΄ C = (βπ)/π Putting x = 1 2(1) + 1 = A (12 + 1) + (B(1) + C) (1 + 1) 3 = 2A + 2B + 2C Putting A = π/π, C = (βπ)/π 3 = 2 Γ 1/2 + 2B + 2 (β1/2 ) 3 = 2B B = π/π Hence, (ππ^π + π)/((π + π)(π^π+π)) = 1/(2(π₯ + 1)) + (3/2 π₯ β 1/2)/(π₯^2 + 1) = π/(π(π + π)) + (ππ βπ)/(γπ(πγ^(π )+ π)) Now, our equation becomes y = β«1βγ(2π₯^2 + π₯)/((π₯ + 1)(π₯^2 + 1)) ππ₯γ y = β«1βγπ/(π(π + π))+ππ/(π(π^(π )+π)) β π/π(π^π + π) ππγ y = β«1βγ1/(2(π₯ + 1)) ππ₯γ+β«1βγ3π₯/(2(π₯^(2 )+1)) ππ₯γββ«1βγ 1/2(π₯^2 + 1) ππ₯γ y = π/π log (x + 1) +β«1βγππ/(π(π^(π )+π)) ππγβ π/π tanβ1 x Integrating β«1βγππ/(π(π^π + π)) ππγ Put t = x2 + 1 dt = 2x dx β΄ dx = ππ‘/2π₯ So, β«1βγππ/(π(π^π + π)) ππγ = 3/2 β«1βπ₯/π‘Γππ‘/2π₯ = 3/4 β«1βππ‘/π‘ = π/π log |π|+π Putting back value of t β«1βγ3π₯/(2(π₯^2+1)) ππ₯γ = 3/4 log (x2 + 1) + C Now, From (1) y = 1/2 log (x + 1) +β«1βγ3π₯/(2(π₯^(2 )+1)) ππ₯γβ 1/2 tanβ1 x y = 1/2 log (x + 1) +3/4 " log (x2 + 1)"β 1/2 tanβ1 x + C Putting x = 0 and y = 1 1 = 1/2 log (0 + 1) + 3/4 log (0 + 1) β 1/2 tanβ1 0 + C 1 = 1/2 log (1) + 3/4 log (1) β 1/2 tanβ1 0 + C 1 = 0 + 0 β 0 + C 1 = C Putting value of C in (1) y = π/π log (x + 1) + π/π log (x2 + 1) β π/π tanβ1 x + 1 y = π/4 log (x + 1) + 3/4 log (x2 + 1) β 1/2 tanβ1 x + 1 y = 1/4 log (x + 1)2 + 1/4 log (x2 + 1)3 β 1/2 tanβ1 x + 1 y = 1/4 [logβ‘γ (π₯+1)^2 γ+logβ‘γ(π₯^2+1)^3 γ ] "β " 1/2 " tanβ1 x + 1 " As log π + log b = log πb y = π/π πππβ‘γγ [(π+π)γ^π (π^π+π)^π] γ "β " π/π " tanβ1 x + 1 "