Variable separable equations

MATVariable Separable

A variable-separable equation can be rearranged so that all $y$ -terms (with $dy$ ) sit on one side and all $x$ -terms (with $dx$ ) on the other, after which both sides are integrated independently. This is the most basic and most frequently tested solution technique, and recognising when an equation separates (often after factoring) is the key skill.

Initial conditions then pin down the arbitrary constant.

Separable form

\dfrac{dy}{dx}=f(x)\,g(y)\ \Rightarrow\ \int\dfrac{dy}{g(y)}=\int f(x)\,dx+C

valid where

g(y)\ne 0

;

C

is a single arbitrary constant.

Standard inverse-tan result

\dfrac{dy}{dx}=\dfrac{1+y^{2}}{1+x^{2}}\ \Rightarrow\ \tan^{-1}y=\tan^{-1}x+C

obtained from

\displaystyle\int\dfrac{dy}{1+y^{2}}=\int\dfrac{dx}{1+x^{2}}

Factoring before separating

\dfrac{dy}{dx}=e^{x-y}+x^{2}e^{-y}=e^{-y}\big(e^{x}+x^{2}\big)\ \Rightarrow\ \int e^{y}\,dy=\int\big(e^{x}+x^{2}\big)dx

factor out

e^{-y}

so the variables separate; gives

e^{y}=e^{x}+\dfrac{x^{3}}{3}+C

Trigonometric separable example

\sec^{2}x\,\tan y\,dx+\sec^{2}y\,\tan x\,dy=0\ \Rightarrow\ \int\dfrac{\sec^{2}y}{\tan y}\,dy=-\int\dfrac{\sec^{2}x}{\tan x}\,dx

divide by

\tan x\,\tan y

; integrates to

\ln|\tan x\,\tan y|=C

(i.e.

\tan x\,\tan y=C

Try to write $\dfrac{dy}{dx}=f(x)g(y)$ ; sometimes you must factor (e.g. $e^{x-y}+x^2 e^{-y}=e^{-y}(e^x+x^2)$ ) before the variables split.
After separating, integrate each side with respect to its own variable and add a single constant $C$ .
Use $\displaystyle\int\dfrac{dx}{1+x^{2}}=\tan^{-1}x$ and $\displaystyle\int\dfrac{f'(x)}{f(x)}\,dx=\ln|f(x)|$ — these recur constantly here.
$\dfrac{dy}{dx}=2xy$ separates to $\dfrac{dy}{y}=2x\,dx$ , giving $\ln|y|=x^{2}+C$ , i.e. $y=Ae^{x^{2}}$ .
For an IVP, substitute the given $(x,y)$ after integrating to determine $C$ (e.g. $y=1$ at $x=0$ ).
Keep absolute values inside logarithms during integration; convert $\ln|y|=\ldots$ to $y=Ae^{(\cdot)}$ only at the end.
Check that you are not dividing by a factor that could be zero; constant solutions like $y=0$ may be lost in the process.
The answer may be left in implicit form (e.g. $\tan^{-1}y=\tan^{-1}x+C$ ) — solving explicitly for $y$ is optional unless asked.

Where the marks go

Failing to factor an expression that IS separable (e.g. not spotting $e^{x-y}=e^{x}e^{-y}$ ), and concluding wrongly that the equation is not separable.
Forgetting the $+C$ , or adding a separate constant on each side instead of one combined constant.
Dropping the modulus in $\ln|y|$ and mishandling signs when exponentiating to recover $y$ .
Substituting the initial condition before completing the integration, or losing the singular solution obtained when the dividing factor equals zero.

How the board asks it

Numericalthe separable form $\frac{dy}{dx}=f(x)g(y)$
Solve the differential equation $\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}$ .
Numericalsubstituting the initial condition to determine $C$
Find the particular solution of $\frac{dy}{dx}=2xy$ given that $y=1$ when $x=0$ .
Numericalfactoring before separating the variables
Solve the differential equation $\frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$ .
Numericalseparating a trigonometric differential form
Solve $\sec^{2}x\,\tan y\,dx+\sec^{2}y\,\tan x\,dy=0$ .
Applicationforming and solving a separable equation from a slope condition
The slope of the tangent to a curve at any point $(x,y)$ is $\frac{dy}{dx}=\frac{x}{y}$ . Find the equation of the curve if it passes through the point $(0,2)$ .

Practise this topic

PreviousOrder & degree, and forming a differential equation NextHomogeneous differential equations

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.