First-Order Linear Differential Equations, calculus lesson

A tank that drains in proportion to how full it is while being refilled at a varying rate, a body that cools toward a changing room temperature, a current pushed by a time dependent voltage. All of these lead to the same kind of equation, where the unknown appears only to the first power. For this class there is a complete recipe that always works.

Key terms

Linear equation in standard form

Definition A first order linear differential equation is one that can be written in the standard form

y' + P(x)\,y = Q(x),

where $P$ and $Q$ are functions of $x$ alone. Throughout, $P$ and $Q$ are assumed continuous on an open interval $I$ . An equation $a(x)y' + b(x)y = c(x)$ is put in standard form by dividing by $a(x)$ , which is allowed only on intervals where $a(x) \neq 0$ .

In plain words The unknown $y$ and its derivative appear on their own, never squared, never inside a function like $\sin y$ and never multiplied together.

Example The equation $xy' + 2y = 4x^2$ is linear. Dividing by $x$ gives $y' + \frac{2}{x}y = 4x$ , with $P(x) = \frac{2}{x}$ and $Q(x) = 4x$ , valid on $(0, \infty)$ or on $(-\infty, 0)$ but not across $x = 0$ . The equation $y' + y^2 = x$ is not linear.

Homogeneous and nonhomogeneous equations

Definition The equation $y' + P(x)y = Q(x)$ is homogeneous if $Q(x) = 0$ for all $x$ in $I$ and nonhomogeneous otherwise. The homogeneous equation $y' + P(x)y = 0$ has general solution

y_h = Ce^{-\int P(x)\,dx}.

If $y_p$ is any one solution of the nonhomogeneous equation, then every solution is $y = y_p + y_h$ .

In plain words First solve the problem with no forcing. Then any single solution of the full problem, plus all the solutions of the unforced one, gives you everything.

Example For $y' + 3y = 6$ , the constant $y_p = 2$ works, since $0 + 6 = 6$ . The homogeneous equation $y' + 3y = 0$ gives $y_h = Ce^{-3x}$ . So the general solution is $y = 2 + Ce^{-3x}$ . The difference of any two solutions solves $y' + 3y = 0$ , which is why nothing else can occur.

Reverse product rule

Definition By the product rule, for differentiable functions $\mu$ and $y$ ,

\left(\mu y\right)' = \mu y' + \mu' y.

If the left side of an equation already has the form $\mu y' + \mu' y$ , it can be integrated at once.

In plain words Sometimes the left side of an equation is secretly the derivative of a product. Spotting that turns the problem into a single integration.

Example For $x^2 y' + 2xy = \cos x$ , the left side is $\left(x^2 y\right)'$ . So $x^2 y = \sin x + C$ , and on $(0, \infty)$ or on $(-\infty, 0)$

y = \frac{\sin x + C}{x^2}.

Integrating factor

Definition For $y' + P(x)y = Q(x)$ with $P$ continuous on $I$ , the function

\mu(x) = e^{\int P(x)\,dx}

(any fixed antiderivative will do) is an integrating factor. It is positive and satisfies $\mu' = P\mu$ , so multiplying the equation by $\mu$ gives $\left(\mu y\right)' = \mu Q$ . Hence

y = \frac{1}{\mu(x)}\left(\int \mu(x)Q(x)\,dx + C\right).

In plain words Multiply by a cleverly chosen function so that the left side becomes the derivative of $\mu y$ , then integrate both sides.

Example Solve $y' - 2xy = x$ . Here $P(x) = -2x$ , so $\mu = e^{-x^2}$ and

\left(e^{-x^2}y\right)' = xe^{-x^2} \quad\Longrightarrow\quad e^{-x^2}y = -\frac{1}{2}e^{-x^2} + C \quad\Longrightarrow\quad y = -\frac{1}{2} + Ce^{x^2}.

As a check, $y' = 2Cxe^{x^2}$ and

y' - 2xy = 2Cxe^{x^2} + x - 2Cxe^{x^2} = x.

Variation of the parameter

Definition Let $y_h$ be a nowhere zero solution of the homogeneous equation on $I$ , for example

y_h(x) = e^{-\int P(x)\,dx}.

Look for a solution of the form $y = u(x)\,y_h(x)$ . Substituting and using $y_h' + Py_h = 0$ leaves

u'(x)\,y_h(x) = Q(x), \qquad\text{so}\qquad u(x) = \int \frac{Q(x)}{y_h(x)}\,dx + C.

For the choice above $\frac{1}{y_h} = \mu$ , so this gives the same formula as the integrating factor.

In plain words Take the homogeneous solution $Cy_h$ and let the constant $C$ become a function. The equation then tells you exactly which function.

Example Solve $y' + y = e^x$ . Here $y_h = e^{-x}$ , so $u'e^{-x} = e^x$ , which means $u' = e^{2x}$ and $u = \frac{1}{2}e^{2x} + C$ . Then

y = \left(\frac{1}{2}e^{2x} + C\right)e^{-x} = \frac{1}{2}e^{x} + Ce^{-x}.

Initial conditions and the existence theorem

Definition If $P$ and $Q$ are continuous on an open interval $I$ , $x_0$ is in $I$ and $y_0$ is any real number, then the problem

y' + P(x)y = Q(x), \qquad y(x_0) = y_0

has exactly one solution, and it is defined on all of $I$ . That solution is

y(x) = \frac{1}{\mu(x)}\left(y_0 + \int_{x_0}^{x}\mu(t)Q(t)\,dt\right), \qquad \mu(x) = e^{\int_{x_0}^{x} P(t)\,dt}.

In plain words Linear equations never blow up on their own. The solution lasts as long as the coefficients stay continuous, and the starting value fixes the constant.

Example Solve $xy' + 2y = 4x^2$ , $y(1) = 2$ . On $I = (0, \infty)$ the standard form is $y' + \frac{2}{x}y = 4x$ , and $\mu = e^{2\ln x} = x^2$ . Then $\left(x^2 y\right)' = 4x^3$ , so $x^2 y = x^4 + C$ and $y = x^2 + \frac{C}{x^2}$ . From $y(1) = 1 + C = 2$ we get $C = 1$ , so

y = x^2 + \frac{1}{x^2}, \qquad x > 0.

The interval is $(0, \infty)$ because it is the largest interval containing $x_0 = 1$ on which $P(x) = \frac{2}{x}$ is continuous.

Check yourself

Solve $y' + 2y = 4$ with $y(0) = 5$ .

Show the answer

The integrating factor $\mu = e^{2x}$ gives $\left(e^{2x}y\right)' = 4e^{2x}$ , so $e^{2x}y = 2e^{2x} + C$ and $y = 2 + Ce^{-2x}$ . Then $y(0) = 2 + C = 5$ , so $y = 2 + 3e^{-2x}$ .

Solve $y' + \frac{1}{x}y = 3x$ for $x > 0$ .

Show the answer

The integrating factor is $\mu = x$ , so $(xy)' = 3x^2$ and $xy = x^3 + C$ . The general solution is $y = x^2 + \frac{C}{x}$ on $(0, \infty)$ .

On what interval is the solution of $(x - 3)y' + y = \sin x$ , $y(1) = 0$ guaranteed to exist?

Show the answer

In standard form $P(x) = \frac{1}{x - 3}$ and $Q(x) = \frac{\sin x}{x - 3}$ , both continuous except at $x = 3$ . The interval containing $x_0 = 1$ is $(-\infty, 3)$ .

First-Order Linear Differential Equations

Key terms

Linear equation in standard form

Homogeneous and nonhomogeneous equations

Reverse product rule

Integrating factor

Variation of the parameter

Initial conditions and the existence theorem

Check yourself

How this lesson connects

Builds on

Used later in