Separable Equations, differential equations lesson

Let $t$ be the independent variable and write $y' = dy/dt$ . Suppose a quantity changes at a rate that is the product of a factor depending only on time and a factor depending only on the current state, as in $y' = 2ty$ . Then each variable can be gathered on its own side of the equation and integrated. This lesson explains why that step is legitimate, which solutions the division step can lose, when a relation between the variables is the natural answer, and how to find the interval on which a solution is valid.

Key terms

Separable equation

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

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

where $a$ is continuous on an open interval $I$ and $b$ is continuous on an open interval $J$ . The right side is a product of a function of $t$ alone and a function of $y$ alone. An autonomous equation $y' = f(y)$ with $f$ continuous is the special case $a(t) = 1$ .

In plain words Time and state affect the rate through separate factors that are multiplied together. Neither factor contains the other variable.

Example In $y' = 2ty$ the time factor is $a(t) = 2t$ and the state factor is $b(y) = y$ . The equation $y' = e^{t + y}$ is separable too, because $e^{t + y} = e^t e^y$ . The equation $y' = t + y$ is not. If $t + y = a(t)b(y)$ held on a rectangle, then for two fixed times $t_1 \ne t_2$ the ratio $(t_1 + y)/(t_2 + y)$ would equal the constant $a(t_1)/a(t_2)$ wherever $t_2 + y \ne 0$ . But that ratio is $1 + (t_1 - t_2)/(t_2 + y)$ , which changes with $y$ .

Separation of variables

Definition Let $y' = a(t)b(y)$ be separable as above, and let $J_0 \subseteq J$ be an open interval on which $b(y) \ne 0$ . Let $A$ be an antiderivative of $a$ on $I$ and $H$ an antiderivative of $1/b$ on $J_0$ . A differentiable function $y$ on an interval contained in $I$ , with values in $J_0$ , is a solution if and only if

H\bigl(y(t)\bigr) = A(t) + C

for some constant $C$ . The reason is the chain rule,

\frac{d}{dt}H\bigl(y(t)\bigr) = \frac{y'(t)}{b\bigl(y(t)\bigr)},

which equals $a(t)$ at every point exactly when $y$ solves the equation, and two functions with the same derivative on an interval differ by a constant. The usual shorthand for this relation is

\int \frac{dy}{b(y)} = \int a(t)\,dt.

In plain words Divide by the state factor where it is not zero, then integrate each side with respect to its own variable. Writing $dy$ and $dt$ on separate sides is a compact record of the chain rule argument, not a separate rule.

Example Solve $y' = 2ty$ on a branch where $y \ne 0$ . The function $1/b(y) = 1/y$ has the antiderivative $\ln\lvert y \rvert$ on each of the intervals $y > 0$ and $y < 0$ , so

\ln\lvert y \rvert = t^2 + C_0.

On the branch $y > 0$ this gives $y = e^{C_0}e^{t^2}$ , and on the branch $y < 0$ it gives $y = -e^{C_0}e^{t^2}$ . Both are of the form $y = Ce^{t^2}$ with $C \ne 0$ , where the sign of $C$ is the sign of the branch.

Equilibria lost by division

Definition If $b(c) = 0$ for some $c$ in $J$ , the constant function $y(t) = c$ solves $y' = a(t)b(y)$ on all of $I$ , because both sides are zero. Such a constant solution is an equilibrium. Separation divides by $b(y)$ , so it only produces solutions that avoid the zeros of $b$ , and the equilibria have to be found before dividing and checked on their own. If in addition $b$ is continuously differentiable on $J$ , the uniqueness theorem applies to $f(t, y) = a(t)b(y)$ , so a solution that equals $c$ at one time equals $c$ at every time. Every solution is then either an equilibrium or never reaches one, and in the second case its values stay inside a single interval $J_0$ where separation applies. Without that hypothesis a solution can reach an equilibrium in finite time, as $y = t^3$ does for $y' = 3y^{2/3}$ at $t = 0$ .

In plain words Before dividing by $b(y)$ , list every value $c$ with $b(c) = 0$ . Each one is a constant solution, and the division may have removed it from the family that separation produces.

Example For $y' = 2ty$ the zero of $b(y) = y$ gives the equilibrium $y = 0$ , which does solve the equation, since both sides are $0$ . The separated family $y = Ce^{t^2}$ with $C \ne 0$ happens to recover it at $C = 0$ , and $b(y) = y$ is continuously differentiable, so every solution is $y = Ce^{t^2}$ for some real $C$ . That last step was checked in the original equation, and it does not always work. For $y' = 2ty^2$ , separation gives

\begin{gathered} -\frac{1}{y} = t^2 + C, \\[0.4em] y = -\frac{1}{t^2 + C}, \end{gathered}

and no value of $C$ produces the equilibrium $y = 0$ , because the fraction is never zero.

Implicit solution

Definition A relation $F(t, y) = C$ is an implicit solution on an open interval if it defines there a differentiable function $y = \varphi(t)$ that solves the equation and satisfies $F\bigl(t, \varphi(t)\bigr) = C$ for every $t$ in the interval. By the implicit function theorem, if $F$ has continuous first partial derivatives near a point $(t_0, y_0)$ of the relation and $F_y(t_0, y_0) \ne 0$ , then on some open interval around $t_0$ the part of the relation near $(t_0, y_0)$ is the graph of a differentiable function $\varphi$ with $\varphi(t_0) = y_0$ , and

\varphi'(t) = -\frac{F_t\bigl(t, \varphi(t)\bigr)}{F_y\bigl(t, \varphi(t)\bigr)}.

The separated relation $H(y) - A(t) = C$ has $F_y = 1/b(y)$ , which is never zero on $J_0$ , so it always defines $y$ locally there.

In plain words Separation often ends with an equation that links the two variables rather than a formula for $y$ . That equation is a valid answer wherever it defines $y$ as a differentiable function, and the initial value picks out the piece of its graph that the solution follows.

Example Take $y' = -x/y$ with $y(0) = 2$ , where now $x$ is the independent variable and $y' = dy/dx$ . The equation itself requires $y \ne 0$ , so multiplying by $y$ loses nothing and gives $yy' = -x$ . The left side is the derivative of $\tfrac{1}{2}y^2$ , so $\tfrac{1}{2}y^2 = -\tfrac{1}{2}x^2 + C_0$ , and the initial point gives $C_0 = 2$ , that is

x^2 + y^2 = 4.

The whole circle is not the graph of one function. The initial value $y(0) = 2 > 0$ selects the upper half,

y = \sqrt{4 - x^2},

and differentiating gives $y' = -x/\sqrt{4 - x^2} = -x/y$ , as required.

Solution interval

Definition A solution of an initial-value problem is a function on an open interval containing the initial time that takes the initial value, is differentiable and satisfies the equation at every point of that interval. The solution interval is the largest such interval for the selected solution, its maximal interval of existence. A formula can be defined on a larger set than the solution interval, or on a set that is not an interval, so the interval has to be read off separately. For a separated formula, look for the times where $t$ leaves $I$ , where $y$ becomes unbounded, and where the values $y(t)$ approach an end of $J_0$ , such as a point where $b$ is zero or undefined. In the last case the separated formula stops, and whether the solution can be continued needs a separate check in the original equation.

In plain words After finding a formula, ask how far it can be followed from the initial time in each direction while it stays differentiable and keeps satisfying the equation.

Example The function $y = 3e^{t^2}$ solves $y' = 2ty$ with $y(0) = 3$ , since $y' = 6te^{t^2} = 2t \cdot 3e^{t^2}$ for every real $t$ , so its solution interval is the whole line. Its derivative is negative for $t < 0$ and positive for $t > 0$ , so it falls to its minimum value $3$ at $t = 0$ and rises afterward. For $y' = 2ty^2$ with $y(0) = 1$ , the family above gives $C = -1$ and

y = \frac{1}{1 - t^2}.

This formula is defined for every $t \ne \pm 1$ , but the solution interval is $(-1, 1)$ , the interval containing $0$ , because $y$ becomes unbounded at both ends. The upper half circle $y = \sqrt{4 - x^2}$ has solution interval $(-2, 2)$ . Its formula is defined on $[-2, 2]$ , but at $x = \pm 2$ the value $y = 0$ makes the right side $-x/y$ undefined.

Check yourself

Decide which of the equations $y' = ty + t$ , $y' = e^{t - y}$ and $y' = t + y^2$ are separable, and name the two factors when they are.

Show the answer

The first is separable, since $ty + t = t(y + 1)$ with $a(t) = t$ and $b(y) = y + 1$ , and it has the equilibrium $y = -1$ . The second is separable, since $e^{t - y} = e^t e^{-y}$ , and it has no equilibrium because $e^{-y}$ is never zero. The third is not separable. If $t + y^2 = a(t)b(y)$ held on a rectangle, then for two fixed times $t_1 \ne t_2$ the ratio $(t_1 + y^2)/(t_2 + y^2)$ would equal the constant $a(t_1)/a(t_2)$ wherever $t_2 + y^2 \ne 0$ . But the ratio is $1 + (t_1 - t_2)/(t_2 + y^2)$ , which changes as $y^2$ changes.

Solve the problem below, leaving the answer as a relation between $t$ and $y$ , and explain why that relation defines $y$ as a differentiable function of $t$ for every real $t$ .

\begin{gathered} y' = \frac{2t}{3y^2 + 1}, \\[0.4em] y(0) = 1 \end{gathered}

Show the answer

The state factor $1/(3y^2 + 1)$ is never zero, so there are no equilibria to set aside. Separating the variables gives

\int (3y^2 + 1)\,dy = \int 2t\,dt,

so $y^3 + y = t^2 + C$ , and $y(0) = 1$ gives $C = 2$ . The solution satisfies

y^3 + y = t^2 + 2.

The function $g(y) = y^3 + y$ has $g'(y) = 3y^2 + 1 > 0$ and takes every real value, so for each $t$ there is exactly one $y$ with $g(y) = t^2 + 2$ . The derivative of the relation in $y$ , which is $3y^2 + 1$ , is never zero, so by the implicit function theorem this $y$ is a differentiable function of $t$ . Differentiating the relation gives back $(3y^2 + 1)y' = 2t$ . The solution interval is the whole line, and for example $y(1)$ is the real root of $y^3 + y = 3$ , about $1.2134$ .

Solve $y' = (y - 2)^2\cos t$ with $y(0) = 1$ and give the solution interval. Which solution has $y(0) = 2$ , and why does the separated formula not show it?

Show the answer

The zero of $(y - 2)^2$ gives the equilibrium $y = 2$ . For $y \ne 2$ , separating the variables gives

-\frac{1}{y - 2} = \sin t + C.

The initial value gives $1 = 0 + C$ , so $C = 1$ and

y = 2 - \frac{1}{1 + \sin t}.

As a check, $y' = \cos t/(1 + \sin t)^2$ , which equals $(y - 2)^2\cos t$ . The denominator vanishes where $\sin t = -1$ , and the zeros nearest to $0$ are $t = -\pi/2$ and $t = 3\pi/2$ , where $y \to -\infty$ . The solution interval is $(-\pi/2, 3\pi/2)$ .

The problem with $y(0) = 2$ is solved by the equilibrium $y = 2$ , and since $(y - 2)^2$ is continuously differentiable it is the only solution. No value of $C$ in $2 - 1/(\sin t + C)$ gives the constant $2$ , because the fraction is never zero. The equilibrium was lost when the equation was divided by $(y - 2)^2$ .

Separable Equations

Key terms

Separable equation

Separation of variables

Equilibria lost by division

Implicit solution

Solution interval

Check yourself

How this lesson connects

Builds on

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