Ex 9.5, 11 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving Linear differential equations - Equation given
Ex 9.5, 19 (MCQ) Important
Misc 14 (MCQ) Important
Ex 9.5, 2
Ex 9.5, 10
Ex 9.5, 3 Important
Ex 9.5, 4
Misc 15 (MCQ)
Ex 9.5, 13
Ex 9.5, 8 Important
Misc 10 Important
Misc 11
Ex 9.5, 14 Important
Ex 9.5, 6
Ex 9.5, 5 Important
Ex 9.5, 9
Ex 9.5, 7 Important
Ex 9.5, 15
Example 14
Ex 9.5, 1 Important
Ex 9.5, 12 Important
Ex 9.5, 11 You are here
Example 16
Example 17 Important
Example 22 Important
Solving Linear differential equations - Equation given
Last updated at April 16, 2024 by Teachoo
Ex 9.5, 11 For each of the differential equation find the general solution : π¦ ππ₯+(π₯βπ¦^2 )ππ¦=0 Step 1 : Put in form ππ¦/ππ₯ + Py = Q or ππ₯/ππ¦ + P1 x = Q1, y dx + (x β y2) dy = 0 y dx = β (x β y2)dy ππ¦/ππ₯ = (βπ¦)/(π₯βπ¦^2 ) This is not of the form ππ¦/ππ₯ + Py = Q β΄ We find π π/π π ππ₯/ππ¦ = (π¦^2 β π₯)/π¦ ππ₯/ππ¦ = y β π₯/π¦ π π/π π + π/π = y Step 2 : Find P1 and Q1 Comparing (1) with ππ₯/ππ¦ + P1 x = Q1 Where P1 = π/π & Q1 = y Step 3 : Find Integrating factor, IF = π^β«1βγππ π πγ = π^β«1βππ¦/π¦ = π^logβ‘π¦ = y Step 4 : Solution of the equation Solution is x (IF) = β«1βγ(π1ΓπΌπΉ)ππ¦+πγ xy = β«1βγπΓπ π π+πγ xy = β«1βγπ¦^2 ππ¦+πγ xy = π¦^3/3+πΆ x = π¦^3/3π¦+πΆ/π¦ x = π^π/π+πͺ/π