§2 First-Order Equations

A first-order scalar ODE has one unknown real-valued function $y$ and no derivative higher than $y'$. Here $t$ is the independent variable, $y=y(t)$ is the unknown function, and

$$y'=\frac{dy}{dt}.$$

The equation can often be written as

$$y'=f(t,y).$$

There is no single method for every first-order equation. In this section, two recognizable forms are especially important:

The hard part is usually recognizing the form. Once that is done, the remaining steps are mostly algebra, differentiation, and integration.

First-order linear equations

A first-order linear equation may first appear as

$$a_0(t)y'(t)+a_1(t)y(t)=h(t).$$

On any interval where $a_0(t)\ne0$, divide the entire equation by $a_0(t)$:

$$y'(t)+P(t)y(t)=g(t),$$

where

$$P(t)=\frac{a_1(t)}{a_0(t)}, \qquad g(t)=\frac{h(t)}{a_0(t)}.$$

This is the standard form. Always reach this form before finding an integrating factor. In particular, the coefficient of $y'$ must be $1$.

The idea behind the integrating factor

We want to turn the two terms on the left,

$$y'+P(t)y,$$

into the derivative of one product. Introduce an unknown function $\mu(t)$, read “mu of $t$,” and multiply the equation by it:

$$\mu y'+\mu P y=\mu g.$$

The product rule says

$$\frac{d}{dt}(\mu y)=\mu y'+\mu' y.$$

The $\mu y'$ terms already match. To make the other terms match as well, choose $\mu$ so that

$$\mu' y=\mu P y.$$

It is enough to require

$$\mu'=P\mu.$$

For a nonzero $\mu$,

$$\frac{\mu'}{\mu}=P(t).$$

Integrating gives

$$\ln|\mu(t)|=\int P(t)\,dt,$$

so we may choose the integrating factor

$$\boxed{\mu(t)=e^{\int P(t)\,dt}}.$$

There is no $+C$ in the usual formula for $\mu$. Including one would only multiply $\mu$ by a nonzero constant, and that constant cancels from the final answer. We simply choose it to be $1$.

With this choice,

$$\frac{d}{dt}(\mu y)=\mu g.$$

Some WeBWorK questions use $u(t)$ for the reciprocal of the integrating factor:

$$u(t)=\frac{1}{\mu(t)}.$$

Under that convention, $\mu$ is still the factor multiplied into the ODE, while $u$ is the factor that appears after solving for $y$. Read the requested symbol carefully; $u$ and $\mu$ may be reciprocals rather than competing answers.

Integrating and then dividing by $\mu$ gives

$$\boxed{ y(t)=\frac{1}{\mu(t)} \left(\int \mu(t)g(t)\,dt+C\right). }$$

The formula is useful, but the idea is more important: the integrating factor is chosen to create a product derivative.

A reliable procedure

For

$$a_0(t)y'+a_1(t)y=h(t),$$

use these steps:

1. Divide by $a_0(t)$ to obtain $y'+P(t)y=g(t)$.
2. Calculate $\mu(t)=e^{\int P(t)\,dt}$.
3. Multiply every term by $\mu(t)$.
4. Rewrite the left side as $(\mu y)'$.
5. Integrate both sides and include $+C$ here.
6. Divide by $\mu(t)$.
7. If an initial condition is given, use it to determine $C$.
8. Differentiate and substitute to check the result.

Example: a variable coefficient

Consider

$$y'+\frac{2t}{4+t^2}y=\frac{4t}{4+t^2}.$$

This is already in standard form, with

$$P(t)=\frac{2t}{4+t^2}.$$

The integrating factor is

$$\begin{aligned} \mu(t) &=e^{\int \frac{2t}{4+t^2}\,dt}\\ &=e^{\ln(4+t^2)}\\ &=4+t^2. \end{aligned}$$

Here $4+t^2$ is always positive, so no interval split is needed. Multiply the equation by $4+t^2$:

$$(4+t^2)y'+2ty=4t.$$

By the product rule,

$$\frac{d}{dt}\bigl((4+t^2)y\bigr)=4t.$$

Integrate:

$$(4+t^2)y=2t^2+C.$$

Therefore,

$$\boxed{y(t)=\frac{2t^2+C}{4+t^2}}.$$

This is the general solution: different values of $C$ give different solutions.

If $y(0)=0$, then

$$0=\frac{C}{4},$$

so $C=0$ and

$$y(t)=\frac{2t^2}{4+t^2}.$$

If instead $y(0)=1$, then $C=4$, so

$$y(t)=\frac{2t^2+4}{4+t^2}.$$

An initial condition selects one member of the general family.

Example: an exponential right-hand side

Consider

$$y'+\frac12y=\frac12e^{t/3}.$$

The integrating factor is

$$\mu(t)=e^{\int 1/2\,dt}=e^{t/2}.$$

Multiplying the equation gives

$$\frac{d}{dt}\left(e^{t/2}y\right) =\frac12e^{t/2}e^{t/3} =\frac12e^{5t/6}.$$

Integrate:

$$e^{t/2}y =\frac12\int e^{5t/6}\,dt+C =\frac35e^{5t/6}+C.$$

Divide by $e^{t/2}$:

$$\boxed{y(t)=\frac35e^{t/3}+Ce^{-t/2}}.$$

The two terms have different roles. The first responds to the right-hand side of the ODE. The term $Ce^{-t/2}$ solves the corresponding homogeneous equation, obtained by replacing the right side with zero:

$$y'+\frac12y=0.$$

Separable equations

A first-order ODE is separable if it can be written as

$$\frac{dy}{dt}=f_1(t)f_2(y).$$

The right side must be a product of:

• a function depending only on $t$;
• a function depending only on $y$.

For example,

$$y'=e^{t+y}=e^t e^y$$

is separable. By contrast,

$$y'=t^2+y^2$$

is not separable in this form because the right side is a sum that cannot generally be factored into $f_1(t)f_2(y)$.

Assuming $f_2(y)\ne0$, divide by $f_2(y)$ and integrate:

$$\boxed{ \int\frac{1}{f_2(y)}\,dy = \int f_1(t)\,dt+C. }$$

The result may define $y$ implicitly, meaning that $y$ is not isolated. That is still a valid solution.

Do not lose equilibrium solutions

Before dividing by $f_2(y)$, solve

$$f_2(y)=0.$$

Any constant value $y=y_*$ satisfying this equation gives

$$y'(t)=0=f_1(t)f_2(y_*),$$

so $y(t)\equiv y_*$ is an equilibrium solution. Dividing by $f_2(y)$ would hide it, so check these values separately.

Why separation works

Writing $dy$ and $dt$ on opposite sides is a useful mnemonic, but the chain rule is the real justification.

Starting from

$$y'=f_1(t)f_2(y),$$

rearrange it as

$$-f_1(t)+\frac{1}{f_2(y)}y'=0.$$

Define

$$M(t)=-f_1(t), \qquad N(y)=\frac{1}{f_2(y)}.$$

Choose antiderivatives $H_1$ and $H_2$ such that

$$H_1'(t)=M(t), \qquad H_2'(y)=N(y).$$

Because $y$ depends on $t$, the chain rule gives

$$\frac{d}{dt}H_2(y(t)) =H_2'(y)y' =N(y)y'.$$

Therefore,

$$\begin{aligned} M(t)+N(y)y' &=\frac{d}{dt}H_1(t) +\frac{d}{dt}H_2(y(t))\\ &=\frac{d}{dt}\bigl(H_1(t)+H_2(y(t))\bigr). \end{aligned}$$

The ODE says this derivative is zero, so the quantity inside must be constant:

$$\boxed{H_1(t)+H_2(y)=C}.$$

This is exactly the implicit equation obtained by separating and integrating.

Potential functions and level curves

Define

$$\psi(t,y)=H_1(t)+H_2(y).$$

The function $\psi$, read “psi,” is called a potential function. A solution follows a curve on which $\psi$ is constant:

$$\psi(t,y)=C.$$

Such a curve is called a level curve. Different values of $C$ give different curves in the solution family.

Example: circles as implicit solutions

Consider

$$\frac{dy}{dx}=-\frac{x}{y}.$$

Separate and integrate:

$$\begin{aligned} y\,dy&=-x\,dx,\\ \frac{y^2}{2}&=-\frac{x^2}{2}+C. \end{aligned}$$

Move both squared terms to the same side:

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

The level curves are circles centered at the origin. If the initial condition is

$$y(0)=1,$$

then

$$C=\frac{0^2}{2}+\frac{1^2}{2}=\frac12.$$

Thus

$$x^2+y^2=1.$$

Solving for $y$ produces two branches:

$$y=\pm\sqrt{1-x^2}.$$

The initial value $y(0)=1$ selects the upper branch:

$$\boxed{y(x)=\sqrt{1-x^2}}.$$

Its maximal solution interval—the largest interval containing the initial point on which the formula satisfies the original ODE—is

$$-1 < x < 1.$$

Although the formula is real at $x=\pm1$, it has $y=0$ there, and the original ODE contains division by $y$. The ODE itself is therefore undefined at those endpoints.

Finite-time blowup

A solution has finite-time blowup when its magnitude becomes unbounded at a finite endpoint of its interval of existence.

Consider

$$y'=y^2, \qquad y(0)=y_0>0.$$

First note that $y\equiv0$ is an equilibrium solution. For a nonzero solution, separate variables:

$$\frac{dy}{y^2}=dt.$$

Integrate:

$$-\frac1y=t+C.$$

The initial condition gives

$$C=-\frac1{y_0}.$$

Therefore,

$$\boxed{ y(t)=\frac{1}{\frac1{y_0}-t} =\frac{y_0}{1-y_0t}. }$$

The denominator becomes zero at

$$t=\frac1{y_0}.$$

As $t$ approaches this time from the left,

$$\lim_{t\to(1/y_0)^-}y(t)=+\infty.$$

The maximal interval containing $t=0$ is therefore

$$\boxed{\left(-\infty,\frac1{y_0}\right)}.$$

This is different from ordinary exponential growth. A solution of $y'=y$ grows like $e^t$ and becomes unbounded only as $t\to\infty$. The solution of $y'=y^2$ becomes unbounded at a finite time.

Method check

When you see a first-order equation, ask:

1. Can it be written as $y'+P(t)y=g(t)$? Use an integrating factor.
2. Can it be written as $y'=f_1(t)f_2(y)$? Separate the variables.
3. Before dividing by a function of $y$, does that function have zeros? Check the corresponding equilibrium solutions.
4. Does the final formula contain a denominator, logarithm, or square root? Use it to determine the actual solution interval.
5. Does an initial condition select a constant, a branch, or both?