Q.Show that the solution of the differential equation $yx^3 dx + e^{-x} dy = 0$ is $(x^3 - 3x^2 + 6x - 6)e^x + \log y = c$.

A first-order differential equation is separable if it can be rearranged so that every $y$ (and $dy$) is on one side and every $x$ (and $dx$) is on the other — that is, written as $h(y)\,y'=g(x)$, or equivalently as $f_1(x)g_1(y)\,dx+f_2(x)g_2(y)\,dy=0$.

Method.

1. Rearrange the given equation into the separated form

2. Integrate both sides independently:

Only one arbitrary constant $C$ is needed — the two constants that would arise from integrating each side separately combine into a single overall constant.

3. If an initial condition is given (e.g. $y=y_0$ at $x=x_0$), substitute it into the integrated equation to evaluate $C$ and obtain the particular solution.

"Solving" a differential equation is therefore also called "integrating" it, since the whole process reduces to two ordinary integrations once the variables are separated.

Recognising a separable equation in disguise. Many equations that do not look separable at first become separable after a short algebraic step:

• Product-to-sum trig identities — e.g. $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ — turn a mixed trigonometric equation into one where $x$ and $y$ split apart.
• Recognising an exact differential — noticing that $\dfrac{y\,dx-x\,dy}{y^2}=d\!\left(\dfrac{x}{y}\right)$ lets an equation be separated in the single combined variable $\dfrac{x}{y}$ directly, without a full substitution. …