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JulijaS [17]
3 years ago
6

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
Mathematics
1 answer:
never [62]3 years ago
7 0

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

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