Question
Sat November 19, 2011 By: Aakarsh Sharma

sin(x+y)+ cos(x+y)=dy/dx solve by variable seprable

Expert Reply
Mon November 21, 2011
sin(x+y)+cos(x+y)=dy/dx

Let us  x+y=z
 
Differentiate this w.r.t. x we get
 
1+ dy/dx = dz/dx or, dy/dx = dz/dx -1
 
Replacing the values of x+y by z and their derivatives in the original eqn we get
 
sinz+cosz= dz/dx -1
 
or dz/dx = 1+sinz+cosz
 
or dz/(1+sinz+cosz) = dx
 
Above is the variable separable form
 
now sinz=2sin(z/2)cos(z/2) and cosz=2cos2(z/2)-1putting these values we get
 
dz/(1+2sin(z/2)cos(z/2)+2cos2(z/2)-1) = dx
 
dz/2cos(z/2)[sin(z/2)+cos(z/2)] =dx
 
Multiply by [cos(z/2)-sin(z/2)] on the Numerator and Denominator of LHS, we get
 
[cos(z/2)-sin(z/2)]dz/2cos(z/2)[cos(z/2)-sin(z/2)][sin(z/2)+cos(z/2)] =dx
 
[cos(z/2)-sin(z/2)]dz/2cos(z/2)[cos2(z/2)-sin2(z/2)] =dx
 
cos(z/2)dz/2cos(z/2)[cos2(z/2)-sin2(z/2)] - sin(z/2)dz/2cos(z/2)[cos2(z/2)-sin2(z/2)] =dx
 
dz/2cosz - sin(z/2)dz/2cos(z/2)[2cos2(z/2)-1] = dx
 
Let u= cos(z/2) hence du=-sin(z/2)dz/2, putting this we get
 
seczdz/2 + du/u(2u2-1) = dx
 
Integrating both sides we get
 
?seczdz/2 + ?du/u(2u2-1) = ?dx
 
½ln|secz+tanz| + ?2udu/(2u2-1) - ?du/u = x
 
½ln|secz+tanz| + ½?4udu/(2u2-1) - lnu = x
 
½ln|secz+tanz| + ½ln(2u2-1) - lnu = x
 
½ln|secz+tanz| + ½ln(2cos2(z/2)-1) - lncos(z/2) = x
 
½ln|secz+tanz| + ½lncosz - ½lncos2(z/2) = x
 
½ln(seczcosz+tanzcosz) - ½lncos2(z/2) = x
 
½ln(1+sinz) - ½lncos2(z/2) = x
 
ln(1+sinz)/cos2(z/2) = 2x
 
put z=x+y
 
ln(1+sin(x+y))/cos2(x+y/2) = 2x
 
This is the required solution
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