What is the solution of differential equation (cosxcosy-cotx) dx-sinxsinydy=0?

Answer 1

THis is equation of the form

Mdx+Ndy=0

M=cos(x)cos(y)-cot(x)

N=-sin(x)sin(y)

Partial derivative of M w.r.t. y, M_y=-cos(x) sin(y)

Partial derivative of N w.r.t. x, N_x=-cos(x) sin(y)

Hence, M_y=N_x

Such an equation is called exact differential equation

It has solution of the form

F(x,y)=C=constant

dF=F_xdx+F_ydy=0=Mdx+Ndy

F_x=cos(x)cos(y)-cot(x)

Integrating treating y as a constant gives

F=sin(x)cos(y)-ln(sin(x))+g(y)

F_y=-sin(x)sin(y)+g'(y)=N=-sin(x)sin(y)

g'=0 hence, g=d =constant

F=sin(x)cos(y)-ln(sin(x))=C

is the solution