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Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii)
Misc 2 (iv) Important You are here
Misc 3 Deleted for CBSE Board 2023 Exams
Misc 4 Important
Misc 5 Important Deleted for CBSE Board 2023 Exams
Misc 6
Misc 7 Important
Misc 8
Misc 9 Important
Misc 10 Important
Misc 11
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Misc 18 (MCQ)
Last updated at Aug. 20, 2021 by Teachoo
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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 . (iv) π₯^2=2π¦^2 logβ‘π¦ : (π₯^2+π¦^2 ) ππ¦/ππ₯βπ₯π¦=0 π₯^2=2π¦^2 logβ‘π¦ Differentiating Both sides w.r.t. x (π₯^2 )^β²=(2π¦^2 logβ‘π¦ )β² 2π₯=(2π¦^2 )^β² logβ‘π¦ +2π¦^2 (logβ‘π¦ )^β² 2π₯=2Γ2π¦π¦^β² logβ‘π¦ + 2π¦^2Γ 1/π¦ π¦β² 2π₯=4π¦π¦^β² logβ‘π¦ + 2π¦π¦β² π₯=2π¦π¦^β² logβ‘π¦ + π¦π¦β² π₯=γπ¦π¦γ^β² (2 logβ‘π¦+1) Now, from our equation π₯^2=2π¦^2 logβ‘π¦ π₯^2/(2π¦^2 )=logβ‘π¦ πππβ‘π¦=π₯^2/(2π¦^2 ) Putting value of πππβ‘π¦ in (1) π₯=γπ¦π¦γ^β² (2 πππβ‘π+1) π₯=γπ¦π¦γ^β² (2Γπ₯^2/(2π¦^2 ) " " +1) π₯=γπ¦π¦γ^β² (π₯^2/π¦^2 " " +1) π₯=γπ¦π¦γ^β² ((π₯^2 + π¦^2)/π¦^2 ) π₯=π¦^β² ((π₯^2 + π¦^2)/π¦) π₯π¦=π¦^β² (π₯^2 + π¦^2 ) π¦^β² (π₯^2 + π¦^2 )βπ₯π¦=0 Thus, Given Function is a solution of the Differential Equation