Example 21 - Chapter 9 Class 12 Differential Equations (Term 2)
Last updated at Dec. 11, 2019 by Teachoo
Examples
Example 1 (i)
Example 1 (ii) Important
Example 1 (iii) Important
Example 2
Example 3 Important
Example 4 Deleted for CBSE Board 2023 Exams
Example 5 Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Deleted for CBSE Board 2023 Exams
Example 9
Example 10
Example 11
Example 12 Important
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19
Example 20 Important
Example 21 You are here
Example 22 Important
Example 23 Important
Example 24
Example 25 Deleted for CBSE Board 2023 Exams
Example 26
Example 27 Important
Example 28 Important
Chapter 9 Class 12 Differential Equations
Serial order wise
Transcript
Example 21 Find the general solution of the differential equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 Given equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 π¦ ππ₯=(π₯+2π¦^2 )ππ¦ ππ¦/ππ₯ = π¦/(π₯ + 2π¦^2 ) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We find ππ₯/ππ¦ ππ₯/ππ¦ = (π₯ + 2π¦^2)/π¦ ππ₯/ππ¦ = (π₯ )/π¦ + (2π¦^2)/π¦ ππ₯/ππ¦ β (π₯ )/π¦ = 2y Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = (β1)/π¦ & Q1 = 2y Now, IF = π^β«1βγπ_1 ππ¦γ IF = e^β«1βγ(β1)/π¦ ππ¦" " γ IF = e^(βlogβ‘π¦ ) IF = e^logβ‘(1/π¦) IF = 1/π¦ IF = e^logβ‘(1/π¦) IF = 1/π¦ Solution is x(IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆ γ x Γ 1/π¦=β«1βγ2π¦Γ1/π¦γ ππ¦+πΆ π₯/π¦ = β«1βγ2ππ¦+πΆγ π₯/π¦ = 2π¦+πΆ x = y (2y + C) x = 2y2 + Cy (As a log x = log xπ) (As π^πππβ‘π₯ = x)
