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# Give the correct answer and solve the given equation dx + (frac{x}{y} ​− sin y)dy = 0 # Give the correct answer and solve the given equation dx + (frac{x}{y} ​− sin y)dy = 0

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Integrals asked 2021-03-05
Give the correct answer and solve the given equation $$dx + (\frac{x}{y} ​− \sin y)dy = 0$$

## Answers (1) 2021-03-06
Multiply this equation by y:
$$y dx + (x ​− y \sin y)dy = 0$$
Now we will try to find a function F: $$RR^{2} \rightarrow R$$ such that
$$\frac{\partial F}{\partial x}=y$$
and $$\frac{\partial F}{\partial x} = x - y \sin y$$ (1)
Start with $$\frac{\partial F}{\partial x} = y$$
and integrate with respect to x:
$$F(x, y) = \int y dx = xy + C(y) (2)$$
where C(y) is a constant with respect to x (an integrating constant!). From (2)
$$\frac{\partial F}{\partial x} = x + C'(y)$$
Using (1) now, we get that
$$x + C'(y) = x - y \sin y \Rightarrow C'(y) = -y \sin y$$
Thus,
$$C(y) = \int -y \sin y dy$$
$$= - \int y \sin y dy$$
$$= {(u = y \Rightarrow du = dy),(dv = \sun\ y\ dy \Rightarrow v = -\cos y)}$$
$$= -(y(-\cos\ y) - \int - \cos\ y\ dy)$$
$$= y \cos y - \int \cos y dy$$
$$= y \cos y - \sim y + D,$$
where D is a constant.
Therefore,
$$F(x, y) = xy + C(y) = xy + y \cos y - \sin y + D$$
Now the solution of the initial differential equation is given by $$F(x, y) = 0$$,
so $$xy + y \cos y - \sin y = - D = C,$$
where C is some constant.

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