Example 23

Transcript

Example 23 Find the equation of a curve passing the point 0 , 1﷯. if the slope of the tangent to the curve at any point 𝑥, 𝑦﷯ is equal to the sum of the 𝑥 coordinate 𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎﷯ and the product of the coordinate and 𝑦 coordinate 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒﷯ of that point . Slope of the tangent to the curve at (x, y) = 𝑑𝑦﷮𝑑𝑥﷯ Given equation: 𝑑𝑦﷮𝑑𝑥﷯ = x + xy 𝑑𝑦﷮𝑑𝑥﷯ − xy = x Differential equation is of the form 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q where P = –x & Q = x IF = e﷮ ﷮﷮𝑝𝑑𝑥﷯﷯ = e﷮− ﷮﷮𝑥 𝑑𝑥﷯﷯ = e﷮ −𝑥﷮2﷯﷮2﷯﷯ Solution is y (IF) = ﷮﷮ 𝑄×𝐼𝐹﷯𝑑𝑥+𝐶﷯ y 𝑒﷮ − 𝑥﷮2﷯﷮2﷯﷯ = ﷮﷮𝑥 𝑒﷮ −𝑥﷮2﷯﷮2﷯﷯ 𝑑𝑥+𝐶﷯ Putting − 𝑥﷮2﷯﷮2﷯ = t − 2𝑥﷮2﷯ 𝑑𝑥=𝑑𝑡 x dx = − dt Thus, our equations become y e﷮ − 𝑥﷮2﷯﷮2﷯﷯ = ﷮﷮− 𝑒﷮𝑡﷯𝑑𝑡+𝑐﷯ y e﷮ − 𝑥﷮2﷯﷮2﷯﷯ = − 𝑒﷮𝑡﷯+𝑐 Putting t = − 𝑥﷮2﷯﷮2﷯ y e﷮ − 𝑥﷮2﷯﷮2﷯﷯ = − e﷮ − 𝑥﷮2﷯﷮2﷯﷯ + C 𝑦﷮ 𝑒﷮ 𝑥﷮2﷯﷮2﷯﷯﷯ = − 1﷮ 𝑒﷮ 𝑥﷮2﷯﷮2﷯﷯﷯ + C y = −1 + C e﷮ 𝑥﷮2﷯﷮2﷯﷯ Since curve passes through (0, 1) Putting x = 0, y = 1 in (1) y = −1 + C 𝑒﷮ 𝑥﷮2﷯﷮2﷯﷯ 1 = −1 + C e﷮ 0﷮2﷯﷮2﷯﷯ 1 = −1 + C. 1 + 1 = C C = 2 Putting value of C in (1) y = −1 + C e﷮ 𝑥﷮2﷯﷮2﷯﷯ y = −1 + 2 e﷮ 𝑥﷮2﷯﷮2﷯﷯ ∴ Equation of the curve is y = −1 + 𝟐 𝒆﷮ 𝒙﷮𝟐﷯﷮𝟐﷯﷯