Example 8 - Chapter 9 Class 12 Differential Equations (Important Question)

Last updated at Dec. 16, 2024 by

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Example 8 Find the equation of a curve passing through the point (−2 ,3), given that the slope of the tangent to the curve at any point (𝑥 , 𝑦) is 2𝑥/𝑦^2 Slope of tangent = 𝑑𝑦/𝑑𝑥 ∴ 𝒅𝒚/𝒅𝒙 = 𝟐𝒙/𝒚𝟐 𝑦2 dy = 2x dx Integrating both sides ∫1▒𝒚𝟐 𝒅𝒚= ∫1▒〖𝟐𝒙 𝒅𝒙〗 𝑦^3/3 = 2.𝑥^2/2 + C 𝑦^3/3 = 𝑥^2 + C 𝑦^3 = 〖3𝑥〗^2+3𝐶 𝒚^𝟑 = 〖𝟑𝒙〗^𝟐+𝑪𝟏 where 𝐶1 = 3C Given that equation passes through (−2, 3) Putting x = −2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(−2)2 + C1 27 = 3 × 4 + C1 27 − 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " 〖"15)" 〗^(𝟏/𝟑) " "is the particular solution of the equation.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo