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

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Example 13 Find the equation of a curve passing through the point (โˆ’2 ,3), given that the slope of the tangent to the curve at any point (๐‘ฅ , ๐‘ฆ) is 2๐‘ฅ/๐‘ฆ^2 Slope of tangent = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆด ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ/๐‘ฆ2 ๐‘ฆ2 dy = 2x dx Integrating both sides โˆซ1โ–’๐‘ฆ2 ๐‘‘๐‘ฆ= โˆซ1โ–’ใ€–2๐‘ฅ ๐‘‘๐‘ฅใ€— ๐‘ฆ^3/3 = 2.๐‘ฅ^2/2 + C ๐‘ฆ^3/3 = ๐‘ฅ^2 + C ๐‘ฆ^3 = ใ€–3๐‘ฅใ€—^2+3๐ถ ๐‘ฆ^3 = ใ€–3๐‘ฅใ€—^2+๐ถ1 where ๐ถ1 = 3C Given that equation passes through (โˆ’2, 3) Putting x = โˆ’2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(โˆ’2)2 + C1 27 = 3 ร— 4 + C1 27 โˆ’ 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " ใ€–"15)" ใ€—^(๐Ÿ/๐Ÿ‘) " "is the particular solution of the equation.

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