Solve a dydx yx xex b dydx sinx yexy xexy1 with step by st

Solve

a.) (dy/dx) = (y/x) +xe^x b.) (dy/dx) = (sinx -ye^xy)/( xe^xy+1)

with step by step solution please

Solution

a)Given: dy/dx=(y/x)+xe^x

                           if it is f(x)=dy/dx=(y/x)+xe^x

                                f\'\'(x)=yd/dx(1/x)+xd/dx(e^x)+e^x(x)

                            =(-y/x²)+x.e^x+e^x.

b)dy/dx=(sinx -ye^xy)/( xe^xy+1)

      if it is f(x)=dy/dx=(sinx -ye^xy)/( xe^xy+1)

then f\'(x)=[(x.e^xy+1)d/dx(sin(x)-y.e^xy)-d/dx(x.e^xy+1)(sin(x)-y.e^xy)]/(x.e^xy+1)²

                       =[(x.e^xy+1)d/dxsin(x)-yd/dxe^xy-d/dxx.e^xy(sin(x)-y.e^xy)]/(x.e^xy+1)²

                      =[(x.e^xy+1)(cos(x))-ye^xyd/dx(xy)-(xd/dxe^xy+e^xyd/dxx)(sin(x)-y.e^xy)]/(x.e^xy+1)²

                      =[(x.e^xy+1)(cos(x))-y.ye^xy-(xe^xyd/dx(xy)+e^xy)(sin(x)-y.e^xy)]/(x.e^xy+1

                      =[(x.e^xy+1)(cos(x))-y²e^xy-(xye^xy+e^xy)(sin(x)-y.e^xy)]/(x.e^xy+1)²

                     =[(x.e^xy+1)]/(x.e^xy+1) - [y²e^xy-(xye^xy+e^xy)(sin(x)-y.e^xy)]/(x.e^xy+1)²