Solve the differential equation $\large\frac{dy}{dx}$$=cos(x+y)+sin(x+y) and find the particular solution if y(0)=0 1 Answer Toolbox: • sinx=\large\frac{2tan\large\frac{x}{2}}{1+tan^2\frac{x}{2}} • cosx=\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}} • 1+tan^2\large\frac{x}{2}=sec^2\large\frac{x}{2} \large\frac{dy}{dx}$$=sin(x+y)+cos(x+y)$
Let $x+y=z$
$\Rightarrow\:\large\frac{dy}{dx}=\frac{dz}{dx}$$-1 Substituting the the values in the equation \large\frac{dz}{dx}$$-1=sinz+cosz$
$\Rightarrow\:\large\frac{dz}{dx}$$=sinz+cosz+1 \Rightarrow\:\large\frac{dz}{1+sinz+cosz}$$=dx$
$\Rightarrow\:\int\large\frac{dz}{1+sinz+cosz}$$=\int dx 1+sinz+cosz=1+\large\frac{2tan\large\frac{z}{2}}{1+tan^2\frac{z}{2}}+\large\frac{1-tan^2\large\frac{z}{2}}{1+tan^2\large\frac{z}{2}} \Rightarrow\:1+sinz+cosz=\large\frac{1+tan^2\frac{z}{2}+2tan\frac{z}{2}+1-tan^2\frac{z}{2}}{1+tan^2\large\frac{z}{2}} \Rightarrow\:1+sinz+cosz=\large\frac{2+2tan\frac{z}{2}}{1+tan^2\large\frac{z}{2}}=\frac{2(1+tan\large\frac{z}{2})}{sec^2\frac{z}{2}} \Rightarrow\:\int\large\frac{dz}{1+sinz+cosz}$$=\int dx$
$\Rightarrow\:\int \large\frac{sec^2\large\frac{z}{2}}{2(1+tan\frac{z}{2})}$$dz=\int dx Put tan\large\frac{z}{2}$$=t$
Differentiating both the sides,
$\large\frac{1}{2}$$sec^2 \large\frac{z}{2}$$dz=dt$
$\Rightarrow\:\int \large\frac{dt}{1+t}$$=x+c \Rightarrow\:log|1+t|=x+c substituting the value of t log|1+tan\large\frac{z}{2}$$|=x+c$
Again substituting the value of $z$,
$log|1+tan\large\frac{x+y}{2}$$|=x+c Given: y(0)=0 \Rightarrow\:x=y=0 Substituting the values of x\:\;and \:\:y c=0 \therefore\: The solution is log|1+tan\large\frac{x+y}{2}$$|=x$