Solve all your doubts with Teachoo Black (new monthly pack available now!)
Example 11 - Chapter 9 Class 12 Differential Equations (Term 2)
Last updated at Dec. 24, 2018 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 You are here
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
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 11 Find the particular solution of the differential equation ππ¦/ππ₯=β4π₯π¦^2 given that π¦=1 , π€βππ π₯=0 Given differential equation , ππ¦/ππ₯=β4π₯π¦^2 ππ¦/π¦^2 = (β4 x) dx Integrating both sides. β«1βππ¦/π¦^2 = β«1βγβ4π₯ ππ₯γ β«1βππ¦/π¦^2 = β4 β«1βγπ₯ ππ₯γ π¦^(β2+1)/(β2+1) = β4.π₯^2/2 + c π¦^(β1)/(β1) = β2x2 + c β 1/π¦ = β2x2 + c y = (β1)/(β2π₯2 + π) y = (β1)/(β(2π₯2 β π)) y = 1/(2π₯2 β π) Given that at x = 0, y = 1 Putting x = 0, y = 1, in (1) 1 = 1/(2(0)^2 ) β c 1 = 1/(βπΆ) c = β1 Put c = β1 in (1) y = 1/(2π₯^2 ) β(β1) y = 1/(2π₯^2 + 1) Hence, the particular solution of the equation is y = π/(ππ^π + π)
