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
CBSE Class 12 Sample Paper for 2021 Boards
Last updated at Dec. 16, 2024 by Teachoo
Transcript
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