# Misc 2 (iv) - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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 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 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 2 For each of the exercise given below , verify that the given function (𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡) is a solution of the corresponding differential equation . (iv) 𝑥^2=2𝑦^2 log𝑦 : (𝑥^2+𝑦^2 ) 𝑑𝑦/𝑑𝑥−𝑥𝑦=0 𝑥^2=2𝑦^2 log𝑦 Differentiating Both sides w.r.t. x (𝑥^2 )^′=(2𝑦^2 log𝑦 )′ 2𝑥=(2𝑦^2 )^′ log𝑦 +2𝑦^2 (log𝑦 )^′ 2𝑥=2×2𝑦𝑦^′ log𝑦 + 2𝑦^2× 1/𝑦 𝑦′ 2𝑥=4𝑦𝑦^′ log𝑦 + 2𝑦𝑦′ 𝑥=2𝑦𝑦^′ log𝑦 + 𝑦𝑦′ 𝒙=〖𝒚𝒚〗^′ (𝟐 𝒍𝒐𝒈𝒚+𝟏) Now, from our equation 𝑥^2=2𝑦^2 log𝑦 𝑥^2/(2𝑦^2 )=log𝑦 𝒍𝒐𝒈𝒚=𝒙^𝟐/(𝟐𝒚^𝟐 ) Putting value of 𝑙𝑜𝑔𝑦 in (1) 𝑥=〖𝑦𝑦〗^′ (2 𝒍𝒐𝒈𝒚+1) 𝑥=〖𝑦𝑦〗^′ (2×𝒙^𝟐/(𝟐𝒚^𝟐 ) " " +1) 𝑥=〖𝑦𝑦〗^′ (𝑥^2/𝑦^2 " " +1) 𝒙=〖𝒚𝒚〗^′ ((𝒙^𝟐 + 𝒚^𝟐)/𝒚^𝟐 ) 𝑥=𝑦^′ ((𝑥^2 + 𝑦^2)/𝑦) 𝑥𝑦=𝑦^′ (𝑥^2 + 𝑦^2 ) 𝒚^′ (𝒙^𝟐 + 𝒚^𝟐 )−𝒙𝒚=𝟎 Thus, Given Function is a solution of the Differential Equation