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Example 11 - Chapter 9 Class 12 Differential Equations (Term 2)
Last updated at March 16, 2023 by Teachoo
Examples
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Example 1 (ii) Important
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Example 4 Deleted for CBSE Board 2023 Exams
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Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
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Example 11 You are here
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Example 25 Deleted for CBSE Board 2023 Exams
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Chapter 9 Class 12 Differential Equations
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Transcript
Example 11 Find the particular solution of the differential equation ππ¦/ππ₯=β4π₯π¦^2 given that π¦=1 , π€βππ π₯=0 Given differential equation , ππ¦/ππ₯=β4π₯π¦^2 ππ¦/π¦^2 = (β4 x) dx Integrating both sides. β«1βππ¦/π¦^2 = β«1βγβ4π₯ ππ₯γ β«1βππ¦/π¦^2 = β4 β«1βγπ₯ ππ₯γ π¦^(β2+1)/(β2+1) = β4.π₯^2/2 + c π¦^(β1)/(β1) = β2x2 + c β 1/π¦ = β2x2 + c y = (β1)/(β2π₯2 + π) y = (β1)/(β(2π₯2 β π)) y = 1/(2π₯2 β π) Given that at x = 0, y = 1 Putting x = 0, y = 1, in (1) 1 = 1/(2(0)^2 ) β c 1 = 1/(βπΆ) c = β1 Put c = β1 in (1) y = 1/(2π₯^2 ) β(β1) y = 1/(2π₯^2 + 1) Hence, the particular solution of the equation is y = π/(ππ^π + π)
