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Miscellaneous

Misc 1 (i)

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

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Misc 15 Important

Misc 16 (MCQ)

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Misc 18 (MCQ)

Chapter 9 Class 12 Differential Equations

Serial order wise

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