Ex 9.4, 3 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving homogeneous differential equation
Ex 9.4, 16 (MCQ)
Ex 9.4, 2
Ex 9.4, 15
Ex 9.4, 4 Important
Example 11
Ex 9.4, 14 Important
Ex 9.4, 13
Ex 9.4, 8
Ex 9.4, 7 Important
Example 21 Important
Ex 9.4, 6 Important
Ex 9.4, 5
Example 13 Important
Ex 9.4, 9
Ex 9.4, 12 Important
Ex 9.4, 1 Important
Ex 9.4, 3 You are here
Ex 9.4, 11
Example 10 Important
Misc 3 Important
Example 12 Important
Ex 9.4, 10 Important
Misc 8 Important
Misc 9
Solving homogeneous differential equation
Last updated at April 16, 2024 by Teachoo
Ex 9.4, 3 In each of the Exercise 1 to 10, show that the given differential equation is homogeneous and solve each of them. (xβy)πyβ(x+y)ππ₯=0 Step 1: Find ππ¦/ππ₯ (x β y) dy β (x + y) dx = 0 (x β y) dy = (x + y) dx π π/π π = (π + π)/(π β π) Step 2: Put ππ¦/ππ₯ = F(x, y) and find out F(πx, πy) F(x, y) = (π₯ + π¦)/(π₯ β π¦) F(πx, πy) = (ππ₯ + ππ¦)/(ππ₯βππ¦) = (π(π₯ + π¦))/(π (π₯ β π¦)) = (π₯ + π¦)/(π₯ β π¦) = F(x, y) β΄ F(πx, πy) = π0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, ππ¦/ππ₯ is a homogenous differential equation. Step 3: Solving ππ¦/ππ₯ by putting y = vx Putting y = vx. Differentiating w.r.t. x ππ¦/ππ₯ = x ππ£/ππ₯+π£ππ₯/ππ₯ π π/π π = π π π/π π + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = (π₯ + π¦)/(π₯ β π¦) x π π/π π + v = (π + ππ)/(π β ππ) x ππ£/ππ₯ + v = (π₯ (1 + π£))/(π₯(1 β π£)) x ππ£/ππ₯ + v = ((1 + π£))/((1 β π£)) x ππ£/ππ₯ = ((1 + π£))/((1 β π£)) β v x ππ£/ππ₯ = (1 + π£ β π£(1 β π£))/(1 β π£) x ππ£/ππ₯ = (1 + π£ β π£ + π£^2)/(1 β π£) x ππ£/ππ₯ = (1 + π£^2)/(1 β π£) (π β π)π π/(π + π^π ) = π π/π Integrating both sides β«1β((1 β π£)/(1 + π£^2 )) ππ£=β«1βππ₯/π₯ β«1βππ£/(1 + π£^2 )ββ«1β(π£ ππ£)/(1 + π£^2 )=β«1βππ₯/π₯ tanβ1 v β β«1βπ/(π + π^π ) = log|π|+π Let I = β«1βπ/(π + π^π ) dv Putting t = 1 + π^π Diff w.r.t. v π/ππ£(1 + v2) = ππ‘/ππ£ 2v = ππ‘/ππ£ dv = π π/ππ Therefore I = β«1βπ£/(1 + π£^2 ) dv = β«1βππ‘/2π‘ = 1/2 πππ|π‘|+π Putting back t = 1 + v2 = 1/2 πππ|1+π£^2 | + c (As β«1β1/(1 + π₯^2 ) dx = tanβ1 x) Putting value of I in (2) tanβ1 v β β«1βπ/(π + π^π ) = log|π₯|+π tanβ1 v "β " π/π log |π+ππ| = log |π₯| + c tanβ1 v "= " 1/2 log |1+π£2| + log |π₯| + c tanβ1 v "= " 1/2 log |1+π£2| + 2/2 log |π₯| + c tanβ1 v "= " 1/2 ["log " |1+π£2|" + " 2" log " |π₯|] + c tanβ1 v "= " π/π "log" [" " |π+ππ|.|π|^π ] + c Putting v = π¦/π₯ tanβ1 π/π "= " π/π "log" [" " (π+(π/π)^π )Γπ^π ]+π tanβ1 π¦/π₯ "= " 1/2 "log" [(π₯^2 + π¦^2)/π₯^2 Γπ₯^2 ]+c tanβ1 π¦/π₯ "= " 1/2 "log" [π₯^2+π¦^2 ]+c tanβ1 π/π "= " π/π "log" [π^π+π^π ]+π is the required solution