A curve has equation
5y2+2x=5ye−3x 5y^2 + 2x = 5ye^{-3x} 5y2+2x=5ye−3xFind dydx\frac{dy}{dx}dxdy in terms of xxx and yyy.
The curve crosses the yyy-axis at the origin and at the point PPP.
Find the equation of the normal to the curve at PPP, writing your answer in the form y=mx+cy = mx + cy=mx+c where mmm and ccc are constants to be found.