A curve has equation
y2=ye3x+2x y^2 = y e^{3x} + 2x y2=ye3x+2xShow that
dydx=3ye3x+22y−e3x \frac{dy}{dx} = \frac{3y e^{3x} + 2}{2y - e^{3x}} dxdy=2y−e3x3ye3x+2The curve crosses the yyy-axis at the origin (0,0)(0, 0)(0,0) and at a second point PPP. The tangent to the curve at the origin and the tangent to the curve at PPP meet at the point RRR. Find the coordinates of RRR.
