ConceptComplete

Linear First-Order Equations

Linear first-order differential equations form a crucial class of ODEs that can always be solved explicitly using the integrating factor method. These equations appear frequently in applications including electrical circuits, mixing problems, and cooling/heating processes.

DefinitionLinear First-Order ODE

A linear first-order differential equation has the standard form:

dydx+p(x)y=q(x)\frac{dy}{dx} + p(x)y = q(x)

where p(x)p(x) and q(x)q(x) are given functions of xx. The equation is called homogeneous if q(x)=0q(x) = 0 and nonhomogeneous if q(x)0q(x) \neq 0.

The key insight is that by multiplying the equation by a carefully chosen function called an integrating factor, we can transform the left side into an exact derivative.

Integrating Factor Method

The integrating factor is defined as:

μ(x)=ep(x)dx\mu(x) = e^{\int p(x)dx}

Multiplying the standard form by μ(x)\mu(x):

μ(x)dydx+μ(x)p(x)y=μ(x)q(x)\mu(x)\frac{dy}{dx} + \mu(x)p(x)y = \mu(x)q(x)

The left side becomes ddx[μ(x)y]\frac{d}{dx}[\mu(x)y], giving:

ddx[μ(x)y]=μ(x)q(x)\frac{d}{dx}[\mu(x)y] = \mu(x)q(x)

Integrating both sides:

μ(x)y=μ(x)q(x)dx+C\mu(x)y = \int \mu(x)q(x)dx + C

ExampleSimple Linear Equation

Solve dydx+2y=e3x\frac{dy}{dx} + 2y = e^{3x}.

Solution: Here p(x)=2p(x) = 2 and q(x)=e3xq(x) = e^{3x}.

Integrating factor: μ(x)=e2dx=e2x\mu(x) = e^{\int 2dx} = e^{2x}

Multiply through:

e2xdydx+2e2xy=e5xe^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{5x}

ddx[e2xy]=e5x\frac{d}{dx}[e^{2x}y] = e^{5x}

Integrate:

e2xy=15e5x+Ce^{2x}y = \frac{1}{5}e^{5x} + C

y=15e3x+Ce2xy = \frac{1}{5}e^{3x} + Ce^{-2x}

The solution consists of a particular solution 15e3x\frac{1}{5}e^{3x} plus the general solution Ce2xCe^{-2x} of the homogeneous equation.

Remark

The general solution of a linear first-order ODE always has the form:

y=yh+ypy = y_h + y_p

where yhy_h is the general solution of the homogeneous equation and ypy_p is any particular solution of the nonhomogeneous equation. This is the principle of superposition for linear equations.

ExampleNewton's Law of Cooling

Newton's law of cooling states that the rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature:

dTdt=k(TTa)\frac{dT}{dt} = -k(T - T_a)

Rewriting in standard form:

dTdt+kT=kTa\frac{dT}{dt} + kT = kT_a

Integrating factor: μ(t)=ekt\mu(t) = e^{kt}

Solution:

T(t)=Ta+(T0Ta)ektT(t) = T_a + (T_0 - T_a)e^{-kt}

where T0=T(0)T_0 = T(0) is the initial temperature. The temperature exponentially approaches the ambient temperature TaT_a.

The integrating factor method is algorithmic and always succeeds for linear first-order equations, making it one of the most reliable techniques in ODE theory.