Question

Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tan(x) − sin(x) sin(y)) dx + cos(x) cos(y) dy = 0

Answer 1

Answer:

$f(x,y)=ln secx+cosx siny+C$

Step-by-step explanation:

We are given that DE

$(tanx-sinx siny)dx+cosxcosydy=0$

We have to determine given DE is exact or not.

Compare it with Mdx+Ndy=0

$M=tanx-sinx siny$

$N=cosxcosy$

$\frac{\partial M}{\partial y}=$$M_y=-sinxcosy$

$\frac{\partial N}{\partial x}=N_x=-sinxcosy$

Therefore, $M_y=N_x$

If DE is exact then $M_y=N_x$

Hence,it is exact.

$M=\frac{\partial f}{\partial x}=tanx-sinxsiny$

Integrate w.r.t x on both sides

$f(x,y)=\int(tanx-sinxsiny)dx$

$f(x,y)=lnsecx+cosxsiny+\phi(y)$...(1)

By using the formula

$\int tanxdx=lnsecx+C,\int sinxdx=-cosx+C$

Differentiate partially  equation (1) w.r.t y

$\frac{\partial f}{\partial y}=cosxcosy+\phi'(y)$

By using the formula:

$\frac{d(sinx)}{dx}=cosx$

$N=\frac{\partial f}{\partial y}=cosxcosy=cosxcosy+\phi'(y)$

$\phi'(y)=cosxcosy-cosxcosy=0$

$\phi'(y)=0$

Integrate  w.r.t y

$\phi(y)=C$

Substitute the value

$f(x,y)=ln secx+cosx siny+C$