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Example 13 - Find curve (-2 ,3), slope of tangent is 2x/y2 - Examples

Example 13 - Chapter 9 Class 12 Differential Equations - Part 2

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Transcript

Example 13 Find the equation of a curve passing through the point (βˆ’2 ,3), given that the slope of the tangent to the curve at any point (π‘₯ , 𝑦) is 2π‘₯/𝑦^2 Slope of tangent = 𝑑𝑦/𝑑π‘₯ ∴ 𝑑𝑦/𝑑π‘₯ = 2π‘₯/𝑦2 𝑦2 dy = 2x dx Integrating both sides ∫1▒𝑦2 𝑑𝑦= ∫1β–’γ€–2π‘₯ 𝑑π‘₯γ€— 𝑦^3/3 = 2.π‘₯^2/2 + C 𝑦^3/3 = π‘₯^2 + C 𝑦^3 = γ€–3π‘₯γ€—^2+3𝐢 𝑦^3 = γ€–3π‘₯γ€—^2+𝐢1 where 𝐢1 = 3C Given that equation passes through (βˆ’2, 3) Putting x = βˆ’2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(βˆ’2)2 + C1 27 = 3 Γ— 4 + C1 27 βˆ’ 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " γ€–"15)" γ€—^(𝟏/πŸ‘) " "is the particular solution of the equation.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.