Example 12 - Chapter 9 Class 12 Differential Equations (Term 2)
Last updated at Aug. 20, 2021 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 You are here
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 12 Find the equation of the curve passing through the point (1 , 1) whose differential equation is π₯ ππ¦= (2π₯^2+1)ππ₯(π₯β 0) π₯ ππ¦ = (2x2 + 1)dx dy = "(2x2 + 1)" /π₯ dx dy = ("2x2" /π₯+1/π₯) dx dy = (2π₯+1/π₯) dx Integrating both sides. β«1βππ¦ = β«1β(2π₯+1/π₯) ππ₯ β«1βππ¦ = β«1βγ2π₯ ππ₯+γ β«1βγ1/π₯ ππ₯γ y = 2 π₯2/2 + log |π₯| + C y = π₯2 + log |π₯| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |π| + C 1 = 1 + 0 + C 1 β 1 = C β΄ C = 0 Put C = 0 in (1) i.e y = x2 + log |π₯| + C y = x2 + log |π₯| + 0 y = x2 + log |π₯| Hence, the equation of curve is y = x2 + log |π|
