Homogeneous differential equations

MATHomogeneous Equations

A first-order equation is homogeneous if it can be written as $\dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right)$ , i.e. both sides are functions of the ratio $\dfrac{y}{x}$ only.

The substitution $y=vx$ converts it into a variable-separable equation in $v$ and $x$ . Recognising homogeneity (every term of the same total degree) and applying the substitution cleanly is the examined skill.

Homogeneous test & substitution

\dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right),\qquad y=vx,\qquad \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}

v=\dfrac{y}{x}

; substituting reduces the equation to one separable in

v

and

x

Separated form after substitution

v+x\dfrac{dv}{dx}=F(v)\ \Rightarrow\ \int\dfrac{dv}{F(v)-v}=\int\dfrac{dx}{x}+C

valid where

F(v)-v\ne 0

; integrate, then replace

v=\dfrac{y}{x}

Worked example

\dfrac{dy}{dx}=\dfrac{x+y}{x}=1+\dfrac{y}{x}\ \Rightarrow\ v+x\dfrac{dv}{dx}=1+v\ \Rightarrow\ x\dfrac{dv}{dx}=1

gives

v=\ln|x|+C

, so

y=x\big(\ln|x|+C\big)

Linear-in-

x

homogeneous case

\dfrac{dx}{dy}=G\!\left(\dfrac{x}{y}\right),\qquad x=vy,\qquad \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}

use

x=vy

(with

v=\dfrac{x}{y}

) when the equation is more naturally a function of

\dfrac{x}{y}

An equation is homogeneous when $\dfrac{dy}{dx}=\dfrac{P(x,y)}{Q(x,y)}$ with $P,Q$ homogeneous of the same degree (every term has the same total degree in $x,y$ ).
Substitute $y=vx$ and $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ ; the $x$ 's cancel, leaving a separable equation in $v$ .
Separate as $\dfrac{dv}{F(v)-v}=\dfrac{dx}{x}$ , integrate, then back-substitute $v=\dfrac{y}{x}$ to return to $x,y$ .
If the equation is more naturally a function of $\dfrac{x}{y}$ , use $x=vy$ instead — choose the substitution that simplifies the algebra.
Example $\dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}$ becomes $v+x\dfrac{dv}{dx}=\dfrac{1+v^{2}}{v}\Rightarrow xv\dfrac{dv}{dx}=1$ , giving $\dfrac{v^{2}}{2}=\ln|x|+C$ .
Always replace $v$ by $\dfrac{y}{x}$ at the end so the final answer is in the original variables.
Constants/expressions like $\dfrac{1}{2}\ln|x|$ can be absorbed into $C$ ; present the cleanest implicit form.
Homogeneous form $\dfrac{dy}{dx}=F(y/x)$ is distinct from linear form — do not apply the integrating-factor method here.

Where the marks go

Misclassifying the equation: applying the homogeneous substitution to a non-homogeneous equation (terms of unequal degree).
Differentiating $y=vx$ incorrectly — forgetting the product rule and writing $\dfrac{dy}{dx}=x\dfrac{dv}{dx}$ instead of $v+x\dfrac{dv}{dx}$ .
Forgetting to substitute $v=\dfrac{y}{x}$ back, leaving the answer in terms of $v$ .
Sign/algebra slips while forming $F(v)-v$ , especially dropping the modulus in $\ln|x|$ and the constant $C$ .

How the board asks it

Numericalthe $y=vx$ substitution
Solve the differential equation $\dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}$ .
Derive / provethe same-degree homogeneity test
Show that the differential equation $\left(x^{2}+xy\right)dy=\left(x^{2}+y^{2}\right)dx$ is homogeneous, and hence solve it.
Numericalback-substitution with a boundary condition
Find the particular solution of $x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx$ , given that $y=0$ when $x=1$ .
Numericalreducing to variable-separable form by a suitable substitution
Solve the differential equation $\left(x-y\right)dy=\left(x+y\right)dx$ .
Conversionframing a slope condition as $\dfrac{dy}{dx}=F(y/x)$
Find the equation of the curve passing through the point $(1,1)$ for which the slope of the tangent at any point $(x,y)$ is $\dfrac{x^{2}+y^{2}}{2xy}$ .

Practise this topic

PreviousVariable separable equations NextLinear differential equations (integrating factor)

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.