Question
Sat November 19, 2011

# 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

Â½ln|secz+tanz| + ?2udu/(2u2-1) - ?du/u = x

Â½ln|secz+tanz| + Â½?4udu/(2u2-1) - lnu = x

Â½ln|secz+tanz| + Â½ln(2u2-1) - lnu = x

Â½ln|secz+tanz| + Â½ln(2cos2(z/2)-1) - lncos(z/2) = x

Â½ln|secz+tanz| + Â½lncosz - Â½lncos2(z/2) = x

Â½ln(seczcosz+tanzcosz) - Â½lncos2(z/2) = x

Â½ln(1+sinz) - Â½lncos2(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
Related Questions
Wed February 08, 2017

# The amount of pollution content added in air in a city due to x-diesel vehicles is given by P(x) = 0.005(x)^3 + 0.02(x)^2 + 30(x). Find the marginal increase in pollution content when 3 diesel vehicles are added.

Tue January 24, 2017