Q.Solve $(e^y + 1)\cos x \, dx + e^y \sin x \, dy = 0$

$\int dy = \int \cos x\,dx$ gives the general solution $y=\sin x + c$.

The equation $\dfrac{dy}{dx}=\cos x$ is directly variable-separable:

$dy = \cos x\,dx.$

Integrating both sides:

$\int dy = \int \cos x\,dx \;\Rightarrow\; y = \sin x + c,$

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$y=\int\cos x\,dx=\sin x+c$.

Separating and integrating $\dfrac{dy}{dx}=\cos x$:

$y=\int\cos x\,dx=\sin x+c,$

where $c$ is an arbitrary constant (a first-order equation's general solution must contain one arbitrary constant). Options with fixed …

Use $x = vy$.

When a differential equation has the homogeneous form $\dfrac{dx}{dy} = f\!\left(\dfrac{x}{y}\right)$, the right side depends only on the ratio $\dfrac{x}{y}$. The standard technique is to set that ratio equal to a new variable:

$x = vy \quad\Rightarrow\quad \frac{x}{y} = v, \qquad \frac{dx}{dy} = v + y\frac{dv}{dy}.$

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