First we solve the quadratic equation2XSO(ªXdydxto obtain the two possible valuesdy y± √√√x² + y²dxXThis can be rewritten asYX=dydxso it is homogeneous. Putting y = xv givesdvdx1²XY= ± 1+Xdx- 2y.dydx+ v=v± √1+v²,= ±= ++ Swhich givesx = 0Thus ln x + lnc = ±sinheln(cx)-2(²1) ²,dv1+v²±sinh¯¹ v or v = ±sinh (ln(cx)). Hencee2- ln(cx)2cy = ± (c²x²-1).how come ??

Answered: Image /qna-images/answer/ecbd56f1-14cc-4f48-9e7e-8ee17cec01fc.jpg

First we solve the quadratic equation 2 dy (d) dx to obtain the two possible values dy y± √√√x² + y² dx X This can be rewritten as X SO X = y X dy dx so it is homogeneous. Putting y = xv gives +v=v± √1+v², dv dx Y = + 1+ X dx X = dy dx - 2y- = ± ± which gives 2cy= =S₁ Thus In x + lnc = ±sinh v or v =±sinh (ln(cr)). Hence -1 eln(cx) how come ?? - e x = 0 dv 1+ v² +v². 2 2 (²) ², = ± (c²x² - 1). - ln(cx)