Example 2 - 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 You are here
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
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 2 Verify that the function π¦=π^(β3π₯) is a solution of the differential equation (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦=0 π¦=π^(β3π₯) π π/π π=π(π^(β3π₯) )/ππ₯ ππ¦/ππ₯=γβ3 πγ^(β3π₯) (π ^π π)/(π π^π )=π/ππ₯ (ππ¦/ππ₯) =π(γβ3 πγ^(β3π₯) )/ππ₯ =β3 π(π^(β3π₯) )/ππ₯ =β3 Γ (γβ3 πγ^(β3π₯) ) = γ9 πγ^(β3π₯) Now, we have to verify (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦=0 Solving L.H.S (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦ Putting values = γ9 πγ^(β3π₯)+(β3π^(β3π₯) )β6(π^(β3π₯) ) =γ9 πγ^(β3π₯)β3π^(β3π₯)β6π^(β3π₯) =γ9 πγ^(β3π₯)β9π^(β3π₯) =0 = R.H.S β΄ Hence Verified
