# Example 13 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Nov. 14, 2019 by Teachoo

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Example 13 You are here

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Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Nov. 14, 2019 by Teachoo

Example 13 Find the equation of a curve passing through the point (β2 ,3), given that the slope of the tangent to the curve at any point (π₯ , π¦) is 2π₯/π¦^2 Slope of tangent = ππ¦/ππ₯ β΄ ππ¦/ππ₯ = 2π₯/π¦2 π¦2 dy = 2x dx Integrating both sides β«1βπ¦2 ππ¦= β«1βγ2π₯ ππ₯γ π¦^3/3 = 2.π₯^2/2 + C π¦^3/3 = π₯^2 + C π¦^3 = γ3π₯γ^2+3πΆ π¦^3 = γ3π₯γ^2+πΆ1 where πΆ1 = 3C Given that equation passes through (β2, 3) Putting x = β2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(β2)2 + C1 27 = 3 Γ 4 + C1 27 β 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " γ"15)" γ^(π/π) " "is the particular solution of the equation.