# Solve :$\large\frac{dy}{dx}$$=\cos (x+y)+\sin (x+y).[Hint :Substitute x+y=z] \begin{array}{1 1}(A)\;\log \bigg|1-\tan \large\frac{(x+y)}{2}\bigg|=x+c \\ (B)\;\log \bigg|1+\tan \large\frac{(x+y)}{2}\bigg|=x+c \\(C)\;\log \bigg|1+\tan \large\frac{(x+y)}{2}\bigg|=\log |x+c| \\ (D)\;\log \bigg|1-\tan \large\frac{(x+y)}{2}\bigg|=\log|x+c|\end{array} ## 1 Answer Toolbox: • A linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$ has a general solution:$ye^{\int pdx}=\int Q.e^{\int pdx}.dx+c$
• $\sin x=2 \sin \large\frac{x}{2}$$\cos \large\frac{x}{2} • 1+\cos x=2 \cos ^2 \large\frac{x}{2} Step 1: Given \large\frac{dy}{dx}$$=\cos (x+y)+\sin (x+y)$
Now let us substituted $x+y=z$ on differentiating w.r.t x on both sides we get,
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$
=>$\large\frac{dy}{dx}=\frac{dz}{dx}$$-1 Now substituting for (x+y) and \large\frac{dy}{dx} we get, \large\frac{dz}{dx}$$-1=\cos z+\sin z$
$\large\frac{dz}{dx}$$=\cos z+\sin z 1+\cos z=2 \cos ^2 \large\frac{z}{2} and \sin z= 2\sin \large\frac{z}{2}$$\cos \large\frac{z}{2}$
$\large\frac{dz}{dx}$$=2 \cos ^2 \large\frac{z}{2}$$+2 \sin \large\frac{z}{2} $$\cos \large\frac{z}{2} Step 2: Now seperating the variables we get, \large\frac{dz}{2 \cos ^2 \Large\frac{z}{2}\large+2 \sin \Large\frac{z}{2} \cos \frac{z}{2}}$$=dx$
Divide both the numerator and the denominator by $\cos ^2 \large\frac{z}{2}$
$\large \frac{\Large\frac{dz}{\cos ^2z/2}}{2+2 \tan z/2}$$=dx => \large \frac {\sec ^2 z/2}{2(1+\tan z/2)}$$dz= dx$
Integrating on both sides we get,
$\large \frac{1}{2}\int \frac {\sec ^2 z/2}{(1+\tan z/2)}$$dz=\int dx Consider \int \large \frac {\sec ^2 z/2}{(1+\tan z/2)}$$dz$
Put $1+\tan z/2=t$ on differentiating w.r.t z we get, $\sec ^2\large\frac{ z}{2} $$dz=dt \int \large\frac{dt}{t}$$=\int dx$
$=\log |t|=x+c$
Substituting for t we get,
$\log |1+\tan z/2|=x+c$
Substituting for z we get,