Question

i am having a difficult time solving this differential equation: (dy/dx) = sin (x + y) + cos(x +y)

Answer 1

first use substitution u=x+y => du/dx=1+dy/dx dy/dx=du/dx-1 and so our DE becomes:
du/dx-1=sinu+cosu
dy/dx=sinu+cosu+1
\[(1/sinu+cosu+1)du=dx\]
\[\int\limits (1/sinu+cosu+1)du=\int\limits dx\]
\[\int\limits (sinu+cosu-1/(\sin^2u+2sinucosu+\cos^2u-1))du=x+c\]
\[\int\limits\limits (sinu+cosu-1/(2sinucosu+1-1))du=x+c\]
\[\int\limits\limits (sinu+cosu-1/(2sinucosu+1-1))du=x+c\]
\[\int\limits\limits (sinu+cosu-1/(2sinucosu))du=x+c\]

Answer 2

\[\int\limits\limits (sinu/(2sinucosu)+cosu/(2sinucosu)-1/(2sinucosu))du=x+c\]
\[\int\limits\limits\limits (/(2cosu)+1/(2sinu)-1/(\sin2u))du=x+c\]
\[\int\limits\limits\limits (secu/2+cscu/2-\csc2u)du=x+c\]
\[1/2(\ln|secu+tanu|-\ln|cscu+cotu|+\ln|\csc2u+\cot2u|)+k=x+c\]
\[1/2(\ln|\sec(x+y)+\tan(x+y)|-\ln|\csc(x+y)+\cot(x+y)|+\]
\[\ln|\csc2(x+y)+\cot2(x+y)|)+k=x+c\]
just simplify using logarithm laws... can you do it from here or should I continue?

Answer 3

that would be great if u can continue