16 Jun 2025

First-Order Homogeneous and Non-Homogeneous Differential Equations

Homogeneous first-order equations and the integrating-factor solution of linear homogeneous and non-homogeneous equations.

bsc semester-ii mathematical-physics differential-equations first-order-equations

A first-order ordinary differential equation contains $y(x)$ and $dy/dx$, but no higher derivative. Two meanings of homogeneous occur in first-order work, so they must be kept distinct.

Homogeneous in the ratio (y/x)

An equation

\[\frac{dy}{dx}=F\!\left(\frac yx\right)\]

is homogeneous because a common scaling $(x,y)\mapsto(\lambda x,\lambda y)$ leaves $y/x$ unchanged. Put

\[v=\frac yx, \qquad y=xv(x).\]

The product rule, which must not be omitted, gives

\[\frac{dy}{dx}=v+x\frac{dv}{dx}.\]

Hence

\[v+x\frac{dv}{dx}=F(v), \qquad \frac{dv}{F(v)-v}=\frac{dx}{x}.\]

After integration, substitute $v=y/x$.

For example,

\[\frac{dy}{dx}=1+\frac yx\]

becomes $v+xv’=1+v$, so $xv’=1$. Therefore

\[v=\ln\lvert x\rvert+C, \qquad \boxed{y=x\bigl(\ln\lvert x\rvert+C\bigr)}.\]

Differentiating the answer gives $y’=\ln\lvert x\rvert+C+1=1+y/x$, so the original equation is recovered.

Linear homogeneous and non-homogeneous equations

A first-order linear equation is

\[\boxed{y'+P(x)y=Q(x)}.\]

It is linear homogeneous when $Q=0$ and non-homogeneous when $Q\ne0$. To combine the two terms on the left into one derivative, multiply by a function $\mu(x)$:

\[\mu y'+\mu Py=\frac{d}{dx}(\mu y) =\mu y'+\mu' y.\]

The coefficients of $y$ agree when

\[\mu'=P\mu, \qquad \frac{d\mu}{\mu}=P(x)dx.\]

Thus an integrating factor is

\[\boxed{\mu(x)=e^{\int P(x)dx}}.\]

The multiplied equation is $(\mu y)’=\mu Q$. Integrating between a reference point $x_0$ and $x$ gives

\[\mu(x)y(x)-\mu(x_0)y(x_0) =\int_{x_0}^{x}\mu(s)Q(s)\,ds.\]

Equivalently,

\[\boxed{ y(x)=\frac{1}{\mu(x)} \left[C+\int^x\mu(s)Q(s)\,ds\right]}.\]

For $Q=0$, this reduces to $y_h=C/\mu$. For $Q\ne0$, the answer is $y=y_h+y_p$.

Example

Solve

\[y'+2y=e^{-x}.\]

Here $P=2$, so $\mu=e^{2x}$. Then

\[e^{2x}y'+2e^{2x}y=(e^{2x}y)'=e^x.\]

Therefore

\[e^{2x}y=e^x+C, \qquad \boxed{y=e^{-x}+Ce^{-2x}}.\]

Every curve satisfies \(y'+2y=e^{-x}\). The \(Ce^{-2x}\) part is the homogeneous freedom, while \(e^{-x}\) is one particular solution.

Substitution gives

\[y'+2y=(-e^{-x}-2Ce^{-2x}) +2(e^{-x}+Ce^{-2x})=e^{-x}.\]

The substitutions in this post are checked in the Unit II Maxima worksheet; every reported residual is zero.