Separable Differential Equations, calculus lesson

Some rates of change split neatly into a part that depends only on time and a part that depends only on the current amount. Radioactive decay, simple growth and many cooling laws look like this. For such equations you can put each variable on its own side and integrate, which is the first general method for solving differential equations.

Key terms

Separation of variables

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

\frac{dy}{dx} = g(x)\,h(y),

with $g$ continuous on an open interval $I$ and $h$ continuous on an open interval $J$ . Let $J_0$ be a subinterval of $J$ on which $h(y) \neq 0$ , let $H$ be an antiderivative of $\frac{1}{h}$ on $J_0$ and $G$ an antiderivative of $g$ on $I$ . Then a differentiable function $y$ on a subinterval of $I$ with values in $J_0$ is a solution if and only if

H\left(y(x)\right) = G(x) + C

for some constant $C$ . In the usual shorthand this is written

\int \frac{dy}{h(y)} = \int g(x)\,dx + C.

In plain words Move everything with $y$ to the left, everything with $x$ to the right, and integrate both sides. The chain rule is what makes this legal, since

\frac{d}{dx}H\left(y(x)\right) = \frac{y'(x)}{h\left(y(x)\right)}.

Example Solve $y' = 2x\left(1 + y^2\right)$ . Here $h(y) = 1 + y^2$ is never zero, so

\int \frac{dy}{1 + y^2} = \int 2x\,dx \quad\Longrightarrow\quad \arctan y = x^2 + C \quad\Longrightarrow\quad y = \tan\left(x^2 + C\right).

Solving for $y$ is allowed on any interval where $x^2 + C$ stays strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ , the range of $\arctan$ .

Explicit and implicit solutions

Definition The relation $H(y) = G(x) + C$ is an implicit solution. If it can be solved for $y$ as a differentiable function $y = \varphi(x)$ on an interval, that function is an explicit solution there. By the implicit function theorem the relation defines such a $\varphi$ near any point $(x_0, y_0)$ on it with $H'(y_0) = \frac{1}{h(y_0)} \neq 0$ , even when no formula for $\varphi$ exists.

In plain words Often you can stop at an equation linking $x$ and $y$ . That equation still defines the solution, even if you cannot write $y = \ldots$ .

Example For $y' = \frac{x}{y + e^y}$ , separating gives

\int \left(y + e^y\right)dy = \int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} + e^y = \frac{x^2}{2} + C.

There is no elementary formula for $y$ , so this implicit form is the answer. With $y(0) = 0$ we get $0 + 1 = 0 + C$ , so $C = 1$ and the solution satisfies $y^2 + 2e^y = x^2 + 2$ .

Lost equilibrium solutions

Definition If $h(c) = 0$ , the constant function $y = c$ solves $y' = g(x)h(y)$ on all of $I$ . Dividing by $h(y)$ assumes $h(y) \neq 0$ , so these constant solutions must be checked separately. They may or may not appear in the family obtained by separation.

In plain words Before dividing by $h(y)$ , write down every value where $h(y) = 0$ . Each gives a solution that the division may have thrown away.

Example Solve $y' = y^2$ . The constant $y = 0$ is a solution. For $y \neq 0$ ,

\int y^{-2}\,dy = \int dx \quad\Longrightarrow\quad -\frac{1}{y} = x + C \quad\Longrightarrow\quad y = -\frac{1}{x + C}.

No value of $C$ gives $y = 0$ , so the complete answer is the family together with the lost solution $y = 0$ . Compare $y' = xy$ , where separation gives $y = Ae^{x^2/2}$ with $A \neq 0$ and the lost solution $y = 0$ comes back as $A = 0$ .

Maximal intervals

Definition The maximal interval of a solution of an initial-value problem is the largest open interval containing the initial point $x_0$ on which the solution is defined and satisfies the equation. For a separable problem look for the places where $x$ reaches an end of $I$ , where $y$ becomes unbounded, or where $y(x)$ reaches an end of $J_0$ . In the last case the separated formula stops being valid, and whether the solution continues past that point needs a separate look.

In plain words After finding the formula, ask how far you can go left and right from the starting point before something goes wrong.

Example For $y' = 2x\left(1 + y^2\right)$ , $y(0) = 0$ , the formula above gives $\arctan 0 = C$ , so $C = 0$ and $y = \tan\left(x^2\right)$ . This requires $x^2 < \frac{\pi}{2}$ , so the maximal interval is

\left(-\sqrt{\pi/2},\ \sqrt{\pi/2}\right) \approx (-1.2533,\ 1.2533).

At the ends the solution goes to $+\infty$ . Beyond them the formula is defined again in places, but those pieces belong to other solutions.

Orthogonal trajectories

Definition Let a one-parameter family of curves satisfy the differential equation $y' = f(x, y)$ , obtained by differentiating and eliminating the parameter. Its orthogonal trajectories are the curves that meet every curve of the family at right angles. Where $f(x, y) \neq 0$ , they are the solution curves of

y' = -\frac{1}{f(x, y)}.

In plain words Find the slope of the family at each point, take the negative reciprocal, and solve the new equation.

Example Take the parabolas $y = cx^2$ . Then $y' = 2cx$ , and with $c = \frac{y}{x^2}$ this is $y' = \frac{2y}{x}$ for $x \neq 0$ . The orthogonal trajectories satisfy $y' = -\frac{x}{2y}$ , which is separable, and separating the variables gives

\int 2y\,dy = -\int x\,dx \quad\Longrightarrow\quad y^2 = -\frac{x^2}{2} + K \quad\Longrightarrow\quad x^2 + 2y^2 = 2K.

So the orthogonal trajectories are the ellipses $x^2 + 2y^2 = k$ with $k > 0$ . At $(1, 1)$ the parabola $y = x^2$ has slope $2$ and the ellipse $x^2 + 2y^2 = 3$ has slope $-\frac{1}{2}$ , and $2 \cdot \left(-\frac{1}{2}\right) = -1$ .

Check yourself

Solve $y' = \frac{x^2}{y}$ with $y(0) = 2$ and give the maximal interval.

Show the answer

Separating the variables gives

\int y\,dy = \int x^2\,dx, \qquad\text{so}\qquad \frac{y^2}{2} = \frac{x^3}{3} + C,

and $y(0) = 2$ gives $C = 2$ . Since $y > 0$ at the start,

y = \sqrt{\frac{2x^3}{3} + 4}.

This formula needs $\frac{2x^3}{3} + 4 > 0$ , so $x^3 > -6$ , and the maximal interval is $\left(-\sqrt[3]{6}, \infty\right)$ . At the left end $y \to 0$ and $y'$ blows up.

Find all solutions of $y' = y\cos x$ .

Show the answer

The constant $y = 0$ is a solution. For $y \neq 0$ , separating gives $\ln\lvert y \rvert = \sin x + C$ , so $y = Ae^{\sin x}$ with $A \neq 0$ . Taking $A = 0$ brings back $y = 0$ , so all solutions are $y = Ae^{\sin x}$ with $A$ any real number.

Why is the constant solution $y = 1$ of $y' = x(y - 1)^2$ easy to miss?

Show the answer

Separation divides by $(y - 1)^2$ and gives $-\frac{1}{y - 1} = \frac{x^2}{2} + C$ , so

y = 1 - \frac{2}{x^2 + 2C}.

No choice of $C$ gives $y = 1$ , yet $y = 1$ makes both sides zero.

Separable Differential Equations

Key terms

Separation of variables

Explicit and implicit solutions

Lost equilibrium solutions

Maximal intervals

Orthogonal trajectories

Check yourself

How this lesson connects

Builds on

Used later in