Respuesta :

Answer:

Step-by-step explanation:

[tex]\frac{dx}{dy} -sin (x+y)=0\\put~x+y=t\\diff .~w.r.t. ~y\\\frac{dx}{dy} +1=\frac{dt}{dy} \\\frac{dx}{dy} =\frac{dt}{dy} -1\\\frac{dt}{dy}-1-sin ~t=0\\\frac{dt}{dy} =1+sin~t\\separate~the ~variables~and~integrating\\\ \int\ {\frac{1}{1-sin~t} } \, dt =\int\ {1} \, dy \\\int\ {\frac{1+ sin~t}{1-sin^2 ~t} } \, dt=y+c\\\int\ {(sec^2t-sec~t~tan~t)} \, dt=y+c\\ tan t-sec t=y+c\\tan(x+y)-sec(x+y)=y+c[/tex]