Consider the differential equation dy dx = f ( y x )

MATHEMATICS · Multi-part · A Level · Mathematics · Differentiation · Differentiation

Problem

Consider the differential equation dydx=f(yx),x>0\frac{dy}{dx} = f\left(\frac{y}{x}\right), \quad x > 0 The curve y=f(x)y = f(x) for x>0x > 0 has a gradient function given by dydx=y2+3xy+2x2x2.\frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}. The curve passes through the point (1,−1)(1, -1). (1) Use the substitution y=vxy = vx to show that ∫dvf(v)−v=ln⁡x+C\int \frac{dv}{f(v)-v} = \ln x + C where CC is an arbitrary constant. (2) By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation y=x(tan⁡(ln⁡x)−1)y = x(\tan(\ln x) - 1). (3) The curve has a point of inflexion at (x1,y1)(x_1, y_1) where e−π2<x1<eπ2e^{-\frac{\pi}{2}} < x_1 < e^{\frac{\pi}{2}}. Determine the coordinates of this point of inflexion. (4) Use the differential equation dydx=y2+3xy+2x2x2\frac{dy}{dx} = \frac{y^2+3xy+2x^2}{x^2} to show that the points of zero gradient on the curve lie on two straight lines of the form y=mxy = mx where the values of mm are to be determined.

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