Find the GS of the DE :$y(\cos x+ \ln y)+(x+ye^y)y'=0$
My attempt:
since given equation is not because $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$
wherer $M=y(\cos x+ \ln y), N=x+ye^y$
when i tried to find integrating factor i am not getting any hint please
thank you
A little hint
$$y(\cos x+ \ln y)+(x+ye^y)y'=0$$ $$y(\cos x+ \ln y)=-(x+ye^y)y'$$ $$(\cos x+ \ln y)=- \frac {(x+ye^y)y'} y$$ $$\frac {(\cos x+ \ln y)}{y'}=- \frac {(x+ye^y)} y$$ $$\frac {dx}{dy}(\cos x+ \ln y)=- \frac {(x+ye^y)} y$$ $$x'(\cos x+ \ln y)=-\frac {(x+ye^y)} y$$ $$(\sin(x))'+x'\ln y + \frac xy=-e^y$$ $$(\sin(x))'+(x\ln y)'=-e^y$$ $$\dfrac {d(\sin(x)+x\ln y)}{dy}=-e^y$$
Simply integrate now..... $$\sin(x)+ x\ln y=-\int e^y dy+K$$ $$\boxed {\sin(x)+ x\ln y=-e^y+K}$$ $$........$$