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Solve the following differential equation: ππ¦ ππ₯ = π₯
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3
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πππ ππ π¦, πππ£ππ π‘βππ‘ π¦(0) = 0.

Last updated at Oct. 26, 2020 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

CBSE Class 12 Sample Paper for 2021 Boards

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Class 12

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.