Solve the following differential equation: ππ¦ ππ₯ = π₯ ^{ 3 } πππ ππ π¦, πππ£ππ π‘ℎππ‘ π¦(0) = 0.
CBSE Class 12 Sample Paper for 2021 Boards
Question 1 (Choice 2) Important
Question 2
Question 3 (Choice 1) Important
Question 3 (Choice 2) Important
Question 4
Question 5 β Choice 1
Question 5 (Choice 2)
Question 6 Important
Question 7 (Choice 1)
Question 7 (Choice 2)
Question 8
Question 9 (Choice 1) Important
Question 9 (Choice 2)
Question 10 Important
Question 11
Question 12 Important
Question 13
Question 14
Question 15 Important
Question 16
Question 17 (Case Based Question) Important
Question 18 (Case Based Question) Important
Question 19 Important
Question 20 (Choice 1)
Question 20 (Choice 2)
Question 21
Question 22 Important
Question 23 (Choice 1)
Question 23 (Choice 2)
Question 24
Question 25 You are here
Question 26 Important
Question 27 Important
Question 28 (Choice 1)
Question 28 (Choice 2) Important
Question 29
Question 30
Question 31 (Choice 1)
Question 31 (Choice 2) Important
Question 32 Important
Question 33
Question 34 (Choice 1)
Question 34 (Choice 2)
Question 35
Question 36 (Choice 1) Important
Question 36 (Choice 2)
Question 37 (Choice 1) Important
Question 37 (Choice 2) Important
Question 38 (Choice 1)
Question 38 (Choice 2) Important
CBSE Class 12 Sample Paper for 2021 Boards
Last updated at April 16, 2024 by Teachoo
Question 25 Solve the following differential equation: ππ¦/ππ₯ = π₯3 πππ ππ π¦, πππ£ππ π‘βππ‘ π¦(0) = 0. Given ππ¦/ππ₯ = π₯3 πππ ππ π¦ ππ¦ Γ 1/(πππ ππ π¦) = π₯3 ππ₯ ππ¦ Γ sin y = π₯3 ππ₯ Integrating both sides β«1βγsinβ‘π¦ ππ¦γ = β«1βγπ₯^3 ππ₯γ βπππ π¦ = π₯^4/4+πΆ Since y(0) = 0 Putting x = 0, y = 0 βπππ 0 = 0/4+πΆ β1 = πΆ πͺ=βπ So, our equation becomes βπππ π¦ = π₯^4/4+πΆ βπππ π¦ = π₯^4/4β1 π^π/π+ππ¨π¬β‘πβπ=π