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I am trying to find an integrating factor and solve the following differential equation:

$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$

These are my steps:

$$(2x \sin(x + y) + \cos(x + y)) + \cos(x + y)dy/dx = 0$$

I check if the equation is exact:

\begin{equation} \partial U_{xy} = \partial U_{xy} \end{equation}

\begin{equation} 2x\cos \left(x+y\right)-\sin \left(x+y\right) \neq -\sin \left(x+y\right), \end{equation}

Its not so I need to find an integrating factor such that

$$\frac{d}{dy} \left( μ(x)2x\cos \left(x+y\right)-\sin \left(x+y\right) \right)= \frac{d}{dx} \left( μ(x)-\sin \left(x+y\right) \right)$$

And at this point I simply get stuck. Any help or advice would be appreciated.

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$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$ $$2x \sin(x + y)dx +( \cos(x + y))(dx +dy) = 0$$ Substitute $v=x+y$ $$2x \sin(v)dx + \cos(v)dv = 0$$ It's not exact. Multiply by $\mu=e^{x^2}$ as integrating factor $$2xe^{x^2} \sin(v)dx + e^{x^2}\cos(v)dv = 0$$ The diffrential is exact.. $$\boxed{e^{x^2} \sin(x+y)=K}$$

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Divide by $\cos(x+y)$. Then we have $$ \big(2x+\cot(x+y)\big)\,dx+\cot(x+y)\,dy=0 \tag{1} $$ Note that $$ \big(\log(\sin x)\big)'=\cot x. $$ So $(1)$ is equivalent to $$ \frac{d}{dx}\Big(x^2+\log\big(\sin \big(x+y(x)\big)\big)\Big)=0 $$

Note. You get the answer by looking for an integrating factor of the form $\mu=\mu(x+y)$.

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