Ex 9.4, 17 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.4, 17 Find the equation of a curve passing through the point 0 , −2﷯ , given that at any point 𝑥 , 𝑦﷯ on the curve , the product of the slope of its tangent and 𝑦 coordinate of the point is equal to the 𝑥 coordinate of the point . Slope of tangent to the curve = 𝑑𝑦﷮𝑑𝑥﷯ Given at any point (x, y) y 𝑑𝑦﷮𝑑𝑥﷯ = x y dy = x dx Integrating both sides ﷮﷮𝑦 𝑑𝑦= ﷮﷮𝑥 𝑑𝑥 ﷯ ﷯ 𝑦﷮2﷯﷮2﷯ = 𝑥﷮2﷯﷮2﷯ + C The curve passes through point (0, −2) Put x = 0 & y = −2 −2﷯﷮2﷯﷮2﷯ = 0﷮2﷯﷮2﷯ + C 4﷮2﷯ = C C = 2 Put value in equation 𝑦﷮2﷯﷮2﷯ = 𝑥﷮2﷯﷮2﷯ + 2 𝑦﷮2﷯﷮2﷯ = 𝑥﷮2﷯ + 4﷮2﷯ 𝑦﷮2﷯ = 𝑥﷮2﷯ + 4 𝑦﷮2﷯ − 𝑥﷮2﷯= 4 Hence, equation of the curve is 𝒚﷮𝟐﷯ − 𝒙﷮𝟐﷯= 4