Question

Question 3 The orthogonal trajectories of the family of curves go? = 2y - 1+ Ce-2y, where is an arbitrary constant, is given by the family of curves + kwhere is an arbitrary constant o 2 + y2 + k where K is an arbitrary constant 4 None of them C zy +k where x is an arbitrary constant 4 24 2 + y = + k, where k is an arbitrary constant 4

Answer 1

First option is correct

solletion 2 Ezy-ltcery where Constant to x c is ar arbitary - x² 2y zy ey ey to Now differentiate above equation with respect (Note we are going to use chain Rele) 2x ey + x² E2 e dy] 2 édy +2y [2edyn] 22 dy to 2x e + 2x² e 24 dy 2 e 24 dy + 44 e dy da da da 2 e dy 44 dy dy uy duty - 2x ey + 2x² e 24 dy dy Ay e dy 2x + 2x² dy dx 2x + 2x² dy dx [2x²-44] dy dx Now put dy da in above equation (2x² ay 1-dx I dy = (2x 4y) dx dy za dy - Jx -22 2x dx (2x²44) = 22²-44 dy x- dy dx 2x de x

dy 2 0 da y = a It is linear differential equation in the form of dy & Py = di w p= 2 2 7 2. general solution of ® is SP dx 19. SP dx dx tk (kis arbitory constant) = g. dx = (x. 2x da tk Now Se dx = 2 1 ar ena (2 dx = ln x² ina2 = y. é da tk yox? 1 x + x² da tk y.x² = x² dx tk =) xy fx. pinaz 24 4 a4 4 where K is constant farbitery It is an equation of far familly Orthogonal trajectories