Linear First-Order Equations and Integrating Factors, differential equations lesson

Let $t$ be the independent variable and write $y' = dy/dt$ . The equation $y' + 2y = 4t$ describes a quantity that decays in proportion to its size while a steadily growing input pushes it up. Solved for the derivative it reads $y' = 4t - 2y$ , and that rate does not split into a factor of $t$ times a factor of $y$ , so separation does not apply. Multiplying by a well chosen function makes the left side the derivative of a product, and then a single integration solves the equation. This lesson builds that multiplier, the integrating factor, and shows on which interval the resulting solution is guaranteed.

Key terms

Linear first-order equation

Definition A first-order equation is linear if it can be written as

a(t)\,y' + b(t)\,y = g(t),

where the coefficients $a$ and $b$ and the input $g$ depend on $t$ alone and $a$ is not identically zero. The unknown $y$ and its derivative appear only to the first power, each multiplied by a function of $t$ . The equation is homogeneous when $g(t) = 0$ for all $t$ and nonhomogeneous otherwise. This meaning of homogeneous refers to the input, and it differs from the scaling meaning of a homogeneous first-order equation in a later lesson.

In plain words The left side adds $y'$ and $y$ with weights that depend only on $t$ , and the right side is an input that depends only on $t$ . Nothing squares $y$ , multiplies $y$ by $y'$ or puts $y$ inside another function.

Example The equation $y' + 2y = 4t$ is linear, with $a(t) = 1$ , $b(t) = 2$ and $g(t) = 4t$ . It is not separable. For two fixed times $t_1 \ne t_2$ the ratio $(4t_1 - 2y)/(4t_2 - 2y)$ changes with $y$ , while a product of a function of $t$ and a function of $y$ would give a ratio that does not depend on $y$ . The equation $t^2y' + (\sin t)\,y = e^t$ is linear as well. The equations $y' + y^2 = t$ and $yy' = t$ are not.

Normalized linear form

Definition On an open interval $J$ where $a(t) \ne 0$ , dividing by $a(t)$ gives the normalized linear form

y' + p(t)\,y = q(t),

with $p = b/a$ and $q = g/a$ . On $J$ a function solves the normalized equation exactly when it solves the original one, because division by a nonzero number can be undone. A point where $a(t) = 0$ is a singular point of the equation. Normalization is not available there, so each interval $J$ must avoid the singular points, and a solution that passes through a singular point has to be checked in the original equation.

In plain words Make the coefficient of $y'$ equal to $1$ , and record the interval on which that division was allowed.

Example The equation $(t^2 - 1)\,y' + t\,y = 2$ has $a(t) = t^2 - 1$ , which vanishes at $t = \pm 1$ . On each of the intervals $(-\infty, -1)$ , $(-1, 1)$ and $(1, \infty)$ its normalized form is

y' + \frac{t}{t^2 - 1}\,y = \frac{2}{t^2 - 1}.

A problem with initial time $t_0 = 0$ is posed on $(-1, 1)$ , and one with $t_0 = 3$ on $(1, \infty)$ .

Derivative of a product

Definition For differentiable functions $\mu$ and $y$ the product rule gives

(\mu y)' = \mu\,y' + \mu'\,y.

Multiplying the normalized equation by a differentiable function $\mu$ gives $\mu y' + p\mu y = \mu q$ . Its left side equals $(\mu y)'$ for every differentiable $y$ exactly when

\mu' = p\,\mu.

Then the equation reads $(\mu y)' = \mu q$ , and the unknown appears only inside one derivative.

In plain words The product rule has the pattern $\mu y' + \mu' y$ . Multiplying $y' + py$ by $\mu$ produces that pattern exactly when $\mu' = p\mu$ , and then the whole left side collapses into one derivative.

Example Multiply $y' + 2y = 4t$ by $e^{2t}$ . Since $(e^{2t})' = 2e^{2t}$ , the left side is

e^{2t}y' + 2e^{2t}y = \left(e^{2t}y\right)',

so the equation becomes $(e^{2t}y)' = 4te^{2t}$ . Multiplying by $e^{t}$ instead would give $e^{t}y' + 2e^{t}y$ , which is not the derivative $(e^{t}y)' = e^{t}y' + e^{t}y$ .

Integrating factor

Definition Let $p$ be continuous on an open interval $J$ and let $P$ be an antiderivative of $p$ on $J$ . The function

\mu(t) = e^{P(t)}

is an integrating factor for $y' + p(t)y = q(t)$ . It is positive and satisfies $\mu' = p\mu$ , so multiplying by $\mu$ turns the equation into $(\mu y)' = \mu q$ , which has the same solutions because $\mu$ is never zero. If $q$ is also continuous on $J$ , a function on an interval inside $J$ is a solution exactly when

\mu y = \int \mu q\,dt + C

for some constant $C$ , where the integral stands for one fixed antiderivative of $\mu q$ . Dividing by $\mu$ gives $y$ . A nonzero constant multiple $k\mu$ is also an integrating factor, and $k$ cancels from the family of solutions.

In plain words The integrating factor is the exponential of an antiderivative of $p$ . Multiply by it, recognize the left side as $(\mu y)'$ , integrate and divide by $\mu$ .

Example Solve $y' + 2y = 4t$ with $y(0) = 3$ . Here $p(t) = 2$ and $\mu = e^{2t}$ , so $(e^{2t}y)' = 4te^{2t}$ . Integration by parts, differentiating $4t$ and integrating $e^{2t}$ , gives

\begin{aligned} &\int 4te^{2t}\,dt \\[0.4em] &= 2te^{2t} - \int 2e^{2t}\,dt \\[0.4em] &= e^{2t}(2t - 1) + C. \end{aligned}

Dividing by $e^{2t}$ gives the general solution $y = 2t - 1 + Ce^{-2t}$ , and $y(0) = -1 + C = 3$ gives $C = 4$ , so

y = 2t - 1 + 4e^{-2t}.

As a check, $y' = 2 - 8e^{-2t}$ and $2y = 4t - 2 + 8e^{-2t}$ , which add up to $4t$ . Two solutions differ by $(C_1 - C_2)e^{-2t}$ , which decays as $t$ increases, yet every solution approaches the line $y = 2t - 1$ and grows without bound. The decay concerns the difference between solutions, not the solutions themselves.

Continuous coefficients

Definition Suppose $p$ and $q$ are continuous on an open interval $J$ , the time $t_0$ lies in $J$ and $y_0$ is any real number. Then the initial-value problem

\begin{gathered} y' + p(t)\,y = q(t), \\[0.4em] y(t_0) = y_0 \end{gathered}

has exactly one solution on $J$ , and that solution is defined on all of $J$ . A linear equation cannot blow up inside an interval where its normalized coefficients are continuous. For the original form $ay' + by = g$ with $a$ , $b$ and $g$ continuous, $J$ can be taken as the largest open interval containing $t_0$ on which $a \ne 0$ . At a singular point the theorem says nothing, and an initial value there may have no solution, exactly one, or infinitely many.

In plain words As long as the coefficients of the normalized equation stay continuous, each starting value gives one solution, and that solution lasts across the whole interval.

Example Take $xy' + y = 1$ , with $x$ as the independent variable and $y' = dy/dx$ . The leading coefficient vanishes at $x = 0$ . The left side is already a product derivative, $(xy)' = xy' + y$ , so on any interval every solution satisfies $xy = x + C$ . On an interval inside $x > 0$ or inside $x < 0$ this gives

y = 1 + \frac{C}{x}.

On an interval containing $0$ , setting $x = 0$ in $xy = x + C$ forces $C = 0$ , so the only solution through $x = 0$ is the constant $y = 1$ . The equation at $x = 0$ reads $0 \cdot y'(0) + y(0) = 1$ , so the problem with $y(0) = 1$ has exactly one solution and the problem with $y(0) = y_0 \ne 1$ has none. The members with $C \ne 0$ become unbounded as $x \to 0$ .

Definite-integral solution formula

Definition Under the hypotheses of the previous theorem, choose the integrating factor that equals $1$ at $t_0$ , and accumulate the input from $t_0$ onward,

\begin{gathered} \mu(t) = \exp\left(\int_{t_0}^{t} p(s)\,ds\right), \\[0.4em] R(t) = \int_{t_0}^{t} \mu(s)\,q(s)\,ds. \end{gathered}

Integrating $(\mu y)' = \mu q$ from $t_0$ to $t$ and using $\mu(t_0) = 1$ gives the unique solution

y(t) = \frac{y_0 + R(t)}{\mu(t)},

valid for every $t$ in $J$ . The initial value enters the formula directly, so no arbitrary constant has to be fitted afterward. The term $y_0/\mu(t)$ is the response to the initial value alone, and $R(t)/\mu(t)$ is the response to the input when the solution starts from $0$ .

In plain words Start the clock at $t_0$ . The initial value is carried forward by the factor $1/\mu$ , and the input is added up from $t_0$ onward.

Example Solve $y' + 2ty = 1$ with $y(0) = 0$ . Here $\mu(t) = e^{t^2}$ , and since $e^{s^2}$ has no elementary antiderivative, the definite integral is the natural answer,

y(t) = e^{-t^2}\int_0^t e^{s^2}\,ds.

It holds for every real $t$ , because $p(t) = 2t$ and $q(t) = 1$ are continuous everywhere. By the fundamental theorem of calculus, $y' = -2ty + e^{-t^2}e^{t^2} = 1 - 2ty$ , which confirms the solution. Numerically, $y(1) \approx 0.5381$ .

Check yourself

Solve $y' - y = e^{2t}$ with $y(0) = 3$ .

Show the answer

Here $p(t) = -1$ , so $\mu = e^{-t}$ and $(e^{-t}y)' = e^{t}$ . Integrating gives $e^{-t}y = e^{t} + C$ , so $y = e^{2t} + Ce^{t}$ . The initial value gives $1 + C = 3$ , so

y = e^{2t} + 2e^{t}.

As a check, $y' - y = 2e^{2t} + 2e^{t} - e^{2t} - 2e^{t} = e^{2t}$ . The coefficients are continuous everywhere, so the solution holds for all real $t$ .

A classmate multiplies $ty' + 2y = 6t^2$ by $e^{2t}$ , reasoning that the coefficient of $y$ is $2$ . Explain why this fails, then solve the equation with $y(1) = 2$ and state the interval on which the solution is guaranteed.

Show the answer

The rule $\mu = e^{\int p\,dt}$ uses the coefficient $p$ of the normalized form, in which $y'$ has coefficient $1$ . Here the coefficient of $y'$ is $t$ , and the multiplied left side $te^{2t}y' + 2e^{2t}y$ is not $(e^{2t}y)' = e^{2t}y' + 2e^{2t}y$ except at $t = 1$ . On $t > 0$ the normalized form is

y' + \frac{2}{t}\,y = 6t,

so $\mu = e^{2\ln t} = t^2$ and $(t^2y)' = 6t^3$ . Then $t^2y = \tfrac{3}{2}t^4 + C$ , and $y(1) = 2$ gives $C = \tfrac{1}{2}$ , so

y = \frac{3}{2}t^2 + \frac{1}{2t^2}.

The solution is guaranteed on $(0, \infty)$ , the largest interval containing $t_0 = 1$ on which $p(t) = 2/t$ and $q(t) = 6t$ are continuous. It becomes unbounded as $t \to 0^{+}$ , so it cannot be continued past $0$ .

For $ty' - y = t^2$ , find the solution with $y(1) = 3$ . Then decide for which values $y_0$ the problem with $y(0) = y_0$ has a solution, and how many.

Show the answer

On $t > 0$ the normalized form is $y' - y/t = t$ , with $\mu = e^{-\ln t} = 1/t$ . Then $(y/t)' = 1$ , so $y = t^2 + Ct$ , and $y(1) = 3$ gives $C = 2$ and $y = t^2 + 2t$ . The theorem guarantees this solution on $(0, \infty)$ . In fact every $t^2 + Ct$ satisfies the original equation for all real $t$ , since $t(2t + C) - (t^2 + Ct) = t^2$ . A solution on $t < 0$ has the form $t^2 + C't$ , so differentiability at $0$ forces $C' = 2$ , and the solution through $(1, 3)$ continues across $0$ only as $t^2 + 2t$ .

At $t = 0$ the equation reads $0 \cdot y'(0) - y(0) = 0$ , so $y(0) = 0$ is necessary, and there is no solution when $y_0 \ne 0$ . When $y_0 = 0$ , every function $y = t^2 + Ct$ is a solution, so there are infinitely many. The point $t = 0$ is singular, and the existence and uniqueness theorem makes no claim there.

Linear First-Order Equations and Integrating Factors

Key terms

Linear first-order equation

Normalized linear form

Derivative of a product

Integrating factor

Continuous coefficients

Definite-integral solution formula

Check yourself

How this lesson connects

Builds on

Used later in