Misc 2 (iii) - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii) You are here
Misc 2 (iv) Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11
Misc 12 Important
Misc 13 (MCQ)
Misc 14 (MCQ) Important
Misc 15 (MCQ)
Question 1
Question 2 Important
Question 3 Important
Last updated at Dec. 16, 2024 by Teachoo
Misc 2 For each of the exercise given below , verify that the given function (ππππππππ‘ ππ ππ₯ππππππ‘) is a solution of the corresponding differential equation . (iii) π¦=π₯ sinβ‘3π₯ : (π^2 π¦)/(ππ₯^2 )+9π¦β6 cosβ‘γ3π₯=0γ π¦=π₯ sinβ‘3π₯ Differentiating w.r.t x π¦^β²=(π₯ π ππβ‘3π₯ )^β² π¦^β²=π₯^β² sinβ‘3π₯+π₯(sinβ‘3π₯)β² π¦^β²=sinβ‘3π₯+π₯Γ3 cosβ‘3π₯ π^β²=πππβ‘ππ+ππ πππβ‘ππ Differentiating again w.r.t. x π¦^β²β²=(sinβ‘3π₯ )^β²+(3π₯ cosβ‘3π₯ )^β² π¦^β²β²=3 cosβ‘3π₯+γ3(π₯)γ^β² cosβ‘3π₯+3π₯ (cosβ‘3π₯ )^β² π¦^β²β²=3 cosβ‘3π₯+3 cosβ‘3π₯+3π₯(β3 sinβ‘3π₯) π^β²β²=π πππβ‘ππβππ πππβ‘ππ Putting π¦=π₯ sinβ‘3π₯ π¦^β²β²=6 cosβ‘3π₯β9π¦ π^β²β²+ππβπ πππβ‘ππ=π Thus, Given Function is a solution of the Differential Equation