1.If xdydx+y=xf(xy)f′(xy)x\frac{dy}{dx} + y = x\frac{f(xy)}{f'(xy)}xdxdy+y=xf′(xy)f(xy), then ∣f(xy)∣|f(xy)|∣f(xy)∣ is equal toa.Cex2/2Ce^{x^2/2}Cex2/2b.Cex2Ce^{x^2}Cex2c.Ce2x2Ce^{2x^2}Ce2x2d.Cex2/3Ce^{x^2/3}Cex2/3Login to continueOnly logged in users canattempt or see the solution.
