A first-order differential equation is variable-separable if it can be rearranged so that everything involving $y$ (together with $dy$) sits on one side and everything involving $x$ (together with $dx$) sits on the other:
Once separated, integrate each side independently. The two constants of integration merge into a single arbitrary constant $C$, which is why a first-order DE has exactly one constant in its general solution.
Method for separable equations
- Bring the equation to the form $\dfrac{dy}{h(y)}=g(x)\,dx$.
- Integrate both sides; write a single constant $C$.
- Simplify (often the logarithm constant is written as $\ln C$ so that the answer comes out in product form $y=Cf(x)$).
- If an initial condition is given, substitute it to find $C$ and obtain the particular solution.
A constant on the right may be written as $\ln C$ or $C$ — choose whichever makes the algebra cleanest. For instance integrating $\dfrac{dy}{y}=k\,dx$ gives $\ln|y|=kx+C_1$, and writing $C_1=\ln C$ tidies this to $y=Ce^{kx}$.
Applications: growth, decay, cooling and interest
Many real processes obey "rate of change is proportional to the present amount", which is the separable equation $\dfrac{dy}{dt}=ky$ with solution $y=y_0e^{kt}$ (take $k>0$ for growth, $k<0$ for decay). Common models:
| Process | Model | Solution |
|---|---|---|
| Population / bacteria growth | $\dfrac{dP}{dt}=kP$ | $P=P_0e^{kt}$ |
| Radioactive decay | $\dfrac{dN}{dt}=-kN$ | $N=N_0e^{-kt}$ |
| Continuous interest | $\dfrac{dA}{dt}=rA$ | $A=A_0e^{rt}$ |
| Newton's cooling | $\dfrac{dT}{dt}=-k(T-T_s)$ | $T=T_s+(T_0-T_s)e^{-kt}$ |
In Newton's law of cooling the temperature difference $T-T_s$ above the surroundings $T_s$ decays exponentially; the substitution $u=T-T_s$ turns the equation into the standard $\dfrac{du}{dt}=-ku$.
Homogeneous differential equations
A first-order DE $\dfrac{dy}{dx}=F(x,y)$ is homogeneous when the right-hand side depends on $x$ and $y$ only through the ratio $\dfrac{y}{x}$ — equivalently, when $F(\lambda x,\lambda y)=F(x,y)$ for every $\lambda$. Such equations are usually not separable as they stand, but a substitution makes them so.
A function $f(x,y)$ is homogeneous of degree $n$ if $f(\lambda x,\lambda y)=\lambda^{n}f(x,y)$. An equation written as $\dfrac{dy}{dx}=\dfrac{P(x,y)}{Q(x,y)}$ is homogeneous when $P$ and $Q$ are homogeneous of the same degree, because then the $\lambda$'s cancel and the ratio depends only on $y/x$.
Substitution. Put $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. The equation becomes separable in $v$ and $x$:
Integrate, then replace $v$ by $\dfrac{y}{x}$ to return to the original variables. If instead the ratio appears naturally as $\dfrac{x}{y}$ (i.e. the equation is easier in terms of $x/y$), use $x=vy$ with $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$.
Recognising homogeneity quickly: divide numerator and denominator of the right-hand side by the highest power of $x$. If every remaining term is a function of $\dfrac{y}{x}$ alone, the equation is homogeneous. For example $\dfrac{x^2+y^2}{xy}=\dfrac{1+(y/x)^2}{y/x}$ is purely in $y/x$, so it is homogeneous.
Deeper Insight — separation is the engine, substitution just feeds it. Every technique in this topic ultimately reduces to "separate and integrate". The growth/decay/cooling models are nothing but $\dfrac{dy}{dt}=ky$ in disguise — once you recognise the proportionality, the exponential answer is automatic. Homogeneous equations look harder only because the variables are tangled; the substitution $y=vx$ exists for a single reason — it untangles them so that the same separable machinery applies. So do not memorise four separate "types"; learn to spot when a clever substitution turns a stubborn equation back into the separable form you already know.