Determine whether the given differential equation is exact. If it is exact, solve it. (tan(x)-sin(x)sin*y))dx+cos(x)cos(y)dy=0 g

Respuesta :

The differential equation

[tex]M(x,y) \, dx + N(x,y) \, dy = 0[/tex]

is considered exact if [tex]M_y = N_x[/tex] (where subscripts denote partial derivatives). If it is exact, then its general solution is an implicit function [tex]f(x,y)=C[/tex] such that [tex]f_x=M[/tex] and [tex]f_y=N[/tex].

We have

[tex]M = \tan(x) - \sin(x) \sin(y) \implies M_y = -\sin(x) \cos(y)[/tex]

[tex]N = \cos(x) \cos(y) \implies N_x = -\sin(x) \cos(y)[/tex]

and [tex]M_y=N_x[/tex], so the equation is indeed exact.

Now, the solution [tex]f[/tex] satisfies

[tex]f_x = \tan(x) - \sin(x) \sin(y)[/tex]

Integrating with respect to [tex]x[/tex], we get

[tex]\displaystyle \int f_x \, dx = \int (\tan(x) - \sin(x) \sin(y)) \, dx[/tex]

[tex]\implies f(x,y) = -\ln|\cos(x)| + \cos(x) \sin(y) + g(y)[/tex]

and differentiating with respect to [tex]y[/tex], we get

[tex]f_y = \cos(x) \cos(y) = \cos(x) \cos(y) + \dfrac{dg}{dy}[/tex]

[tex]\implies \dfrac{dg}{dy} = 0 \implies g(y) = C[/tex]

Then the general solution to the exact equation is

[tex]f(x,y) = \boxed{-\ln|\cos(x)| + \cos(x) \sin(y) = C}[/tex]

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