Q.The velocity $v$, of a parachute falling vertically satisfies the equation $v\dfrac{dv}{dx}=g\left(1-\dfrac{v^2}{k^2}\right)$, where $g$ and $k$ are constants. If $v$ and $x$ are both initially zero, find $v$ in terms of $x$.

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.
• A constant $a$ on one side (e.g. $\sin\dfrac{dy}{dx}=a$) simply gives $\dfrac{dy}{dx}=\sin^{-1}a$, a constant slope, integrating to a straight line.

Substitution method. When the right side is a function of a single linear combination $ax+by+c$ rather than of $x$ and $y$ separately — i.e. $\dfrac{dy}{dx}=f(ax+by+c)$ — put $z=ax+by+c$. Then $\dfrac{dz}{dx}=a+b\dfrac{dy}{dx}$, so $\dfrac{dy}{dx}=\dfrac{1}{b}\left(\dfrac{dz}{dx}-a\right)$ (for $b\ne0$), and substituting turns the equation into one relating $z$ and $x$ only — which is separable. Solve for $z$ by the separable method, then replace $z$ by $ax+by+c$ to get the answer back in $x,y$. (If $a=0$ or $b=0$, the original equation is already separable and no substitution is needed.)