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Separable differential equations

Introduction

In Sections and , we have seen several ways to approximate the solution to an initial value problem. Given the frequency with which differential equations arise in the world around us, we would like to have some techniques for finding explicit algebraic solutions of certain initial value problems. In this section, we focus on a particular class of differential equations (called separable) and develop a method for finding algebraic formulas for their solutions.

A separable differential equation is a differential equation whose algebraic structure allows the variables to be separated in a particular way. For instance, consider the equation \[\begin{aligned}\end{aligned}\].

We would like to separate the variables \(t\) and \(y\) so that all occurrences of \(t\) appear on the right-hand side, and all occurrences of \(y\) appear on the left, multiplied by \(dy/dt\). For this example, we divide both sides by \(y\) so that \[\begin{aligned}\end{aligned}\].

Note that when we attempt to separate the variables in a differential equation, we require that one side is a product in which the derivative \(dy/dt\) is one factor and the other factor is solely an expression involving \(y\).

Not every differential equation is separable. For example, if we consider the equation \[\begin{aligned}\end{aligned}\], it may seem natural to separate it by writing \[\begin{aligned}\end{aligned}\].

As we will see, this turns out not to be helpful, since the left-hand side is not a product of a function of \(y\) with \(\frac{dy}{dt}\).

Exploration

Solving separable differential equations

Before we discuss a general approach to solving a separable differential equation, it is instructive to consider an example.

The strategy of Example may be applied to any differential equation of the form \[\begin{aligned}\end{aligned}\], and any differential equation of this form is said to be separable. We work to solve a separable differential equation by writing \[\begin{aligned}\end{aligned}\], and then integrating both sides with respect to \(t\). After integrating, we try to solve algebraically for \(y\) in order to write \(y\) as a function of \(t\).

Example

Solve the differential equation \[\begin{aligned}\end{aligned}\].

Solution

Following the same strategy as in Example, we have \[\begin{aligned}\end{aligned}\].

Integrating both sides with respect to \(t\), \[\begin{aligned}\end{aligned}\], and thus \[\begin{aligned}\end{aligned}\].

Antidifferentiating and including the integration constant, we find that \[\begin{aligned}\end{aligned}\].

Finally, we need to solve for \(y\). Here, one point deserves careful attention. By the definition of the natural logarithm function, it follows that \[\begin{aligned}\end{aligned}\].

Since \(C\) is an unknown constant, \(e^C\) is as well, though we do know that it is positive (because \(e^x\) is positive for any \(x\)). When we remove the absolute value in order to solve for \(y\), however, this constant may be either positive or negative. To account for a possible \(+\) or \(-\), we denote this updated constant by \(C\) to obtain \[\begin{aligned}\end{aligned}\].

There are two more technical points to make. First, notice that \(y=0\) is an equilibrium solution to this differential equation. In solving the equation above, we begin by dividing both sides by \(y\), which is not allowed if \(y=0\). To be perfectly careful, therefore, we should consider the equilibrium solutions separately. In this case, notice that the final form of our solution captures the equilibrium solution by allowing \(C=0\).

Condensed — the full section is in Boelkins, Active Calculus.

Summary

A separable differential equation is one that may be rewritten with all occurrences of the dependent variable multiplying the derivative and all occurrences of the independent variable on the other side of the equation.
We may find the solutions to certain separable differential equations by separating variables, integrating with respect to \(t\), and ultimately solving the resulting algebraic equation for \(y\).
This technique allows us to solve many important differential equations that arise in the world around us. For instance, questions of growth and decay and Newton's Law of Cooling give rise to separable differential equations. Later, we will learn in Section that the important logistic differential equation is also separable.

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the initial value problem \[\begin{aligned}\end{aligned}\]
1. Find the solution of the initial value problem and sketch its graph.
2. For what values of \(t\) is the solution defined?
3. What is the value of \(y\) at the last time that the solution is defined?
4. By looking at the differential equation, explain why we should not expect to find solutions with the value of \(y\) you noted in (c).
အဖြေကို ဖော်ပြပါ
1. Separating the variables, we find that \[\begin{aligned}\end{aligned}\] and integrating with respect to \(t\) \[\begin{aligned}\end{aligned}\]. Evaluating the integrals and solving for \(y\), we see that \(\frac{1}{2}y^2 = -\frac{1}{2}t^2 + C\) and therefore \[\begin{aligned}\end{aligned}\], so \(y = \pm \sqrt{C - t^2}\). Applying the initial condition \(y(0) = 8\), we find that \(8 = \sqrt{C}\) (we choose the \(+\) from \(\pm\) since \(8 \gt 0\)), so \(C = 64\). Hence the solution to the IVP is \[\begin{aligned}\end{aligned}\].
2. The solution to the IVP is defined for \(-8 \le t \le 8\).
3. The last time that the solution is defined is \(t = 8\) and the corresponding value of \(y\) at this point is \(y(8) = 0\).
4. Recall that the differential equation is \(\frac{dy}{dt} = -\frac ty\), and the righthand expression is undefined when \(y = 0\). Hence the solution value is technically not defined at the point we noted in (c); this is connected to there being a vertical tangent line at this point in the direction field.
Suppose that a cylindrical water tank with a hole in the bottom is filled with water. The water, of course, will leak out and the height of the water will decrease. Let \(h(t)\) denote the height of the water. A physical principle called Torricelli's Law implies that the height decreases at a rate proportional to the square root of the height.
1. Express this fact using \(k\) as the constant of proportionality.
2. Suppose you have two tanks, one with \(k=-1\) and another with \(k=-10\). What physical differences would you expect to find?
3. Suppose you have a tank for which the height decreases at \(20\) inches per minute when the water is filled to a depth of \(100\) inches. Find the value of \(k\).
4. Solve the initial value problem for the tank in part (c), and graph the solution you determine.
5. How long does it take for the water to run out of the tank?
6. Is the solution that you found valid for all time \(t\)? If so, explain how you know this. If not, explain why not.
အဖြေကို ဖော်ပြပါ
1. Since \(\frac{dh}{dt}\) is proportional to the square root of the height of the water, we have \[\begin{aligned}\end{aligned}\].
2. If one tank has \(k=-1\) and the other has \(k=-10\), the tank with \(k = -10\) will have water leaving the tank much more rapidly, which should correspond to that tank having a larger hole.
3. Since height decreases at \(20\) inches per minute when the water is filled to a depth of \(100\) inches, this tells us that \(\frac{dh}{dt}\vert_{h=100} = -20\). Applying this information to the differential equation, \[\begin{aligned}\end{aligned}\] and thus \(k = -2\).
4. To solve the differential equation \(\frac{dh}{dt} = -2 \sqrt{h}\) with initial condition \(h(0) = 100\), we first separate the variables and write \[\begin{aligned}\end{aligned}\]. Rewriting the first fraction and integrating with respect to \(t\), we have \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\], or \(h = \left( C - t \right)^2\). Knowing that \(h(0) = 100\), we see that \(100 = C^2\), and thus \(C = 10\). This shows that the solution to the IVP is \[\begin{aligned}\end{aligned}\]. A plot of this function shows it to be a concave up parabola with \(y\)-intercept \((0,100)\) and a repeated zero at \((10,0)\).
5. It takes \(10\) minutes for the tank to empty since \(h(10) = 0\).
6. No, this solution is not valid for all time \(t\). The physical constraints of the problem tell us that the solution is only valid for \(0 \lt t \lt 10\) since at \(t = 10\) the tank is empty.
The Gompertz equation is a model that is used to describe the growth of certain populations. Suppose that \(P(t)\) is the population of some organism and that \[\begin{aligned}\end{aligned}\].
1. Sketch a slope field for \(P(t)\) over the range \(0\leq P\leq 6\).
2. Identify any equilibrium solutions and determine whether they are stable or unstable.
3. Find the population \(P(t)\) assuming that \(P(0) = 1\) and sketch its graph. What happens to \(P(t)\) after a very long time?
4. Find the population \(P(t)\) assuming that \(P(0) = 6\) and sketch its graph. What happens to \(P(t)\) after a very long time?
5. Verify that the long-term behavior of your solutions agrees with what you predicted by looking at the slope field.
အဖြေကို ဖော်ပြပါ
2. The only equilibrium solution is \(P = 3\), since \(\ln(1) = 0\) (note: \(\ln(0)\) is undefined, so the differential equation is not defined for \(P = 0\)). From the slope field pictured in (a), the equilibrium solution \(P = 3\) is stable.
3. To solve the IVP with \(\frac{dP}{dt} = -P\ln\left(\frac P3\right)\) and \(P(0) = 1\), we use separation of variables. Observe that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\] On the left, using the substitution \(u = \ln (P/3)\), it follows \(du = \frac{3}{P} \cdot \frac{1}{3} \, dP\), we can rewrite the most recent equation as \[\begin{aligned}\end{aligned}\]. Therefore, \[\begin{aligned}\end{aligned}\], so \(|u| = e^{C-t}\), or \(u = Ke^{-t}\). Next, we recall that \(u = \ln \left(\frac P3\right)\), so \[\begin{aligned}\end{aligned}\] and \(P = 3e^{Ke^{-t}}\). Applying the initial condition, \(1 = 3e^K\), so \(K = \ln \left(\frac{1}{3} \right)\), and thus \[\begin{aligned}\end{aligned}\]. This function's plot is shown in blue in the image above in (a) and we see (both graphically and algebraically) that \(P(t) \to 3\) as \(t \to \infty\) since \(e^{-t} \to 0\).
4. Changing the initial condition to \(P(0) = 6\), we can use all of our preceding work in (c) up to where we established \(P = 3e^{Ke^{-t}}\). Applying this initial condition, \(6 = 3e^K\), so \(K = \ln \left(2 \right)\), and thus \[\begin{aligned}\end{aligned}\]. This function's plot is shown in red in the image above in (a) and we see (both graphically and algebraically) that \(P(t) \to 3\) as \(t \to \infty\) since \(e^{-t} \to 0\).
5. Our work in (c) and (d) suggests that regardless of the initial condition with \(P(0) \gt 0\), the long-term behavior of \(P\) will be for it to tend to \(3\) as \(t\) increases without bound. This matches our initial observations based on the slope field.
Consider the initial value problem \[\begin{aligned}\end{aligned}\]
1. Find the solution of the initial value problem and sketch its graph.
2. For what values of \(t\) is the solution defined?
3. What is the value of \(y\) at the last time that the solution is defined?
4. By looking at the differential equation, explain why we should not expect to find solutions with the value of \(y\) you noted in (c).
အဖြေကို ဖော်ပြပါ
1. Separating the variables, we find that \[\begin{aligned}\end{aligned}\] and integrating with respect to \(t\) \[\begin{aligned}\end{aligned}\]. Evaluating the integrals and solving for \(y\), we see that \(\frac{1}{2}y^2 = -\frac{1}{2}t^2 + C\) and therefore \[\begin{aligned}\end{aligned}\], so \(y = \pm \sqrt{C - t^2}\). Applying the initial condition \(y(0) = 8\), we find that \(8 = \sqrt{C}\) (we choose the \(+\) from \(\pm\) since \(8 \gt 0\)), so \(C = 64\). Hence the solution to the IVP is \[\begin{aligned}\end{aligned}\].
2. The solution to the IVP is defined for \(-8 \le t \le 8\).
3. The last time that the solution is defined is \(t = 8\) and the corresponding value of \(y\) at this point is \(y(8) = 0\).
4. Recall that the differential equation is \(\frac{dy}{dt} = -\frac ty\), and the righthand expression is undefined when \(y = 0\). Hence the solution value is technically not defined at the point we noted in (c); this is connected to there being a vertical tangent line at this point in the direction field.
Suppose that a cylindrical water tank with a hole in the bottom is filled with water. The water, of course, will leak out and the height of the water will decrease. Let \(h(t)\) denote the height of the water. A physical principle called Torricelli's Law implies that the height decreases at a rate proportional to the square root of the height.
1. Express this fact using \(k\) as the constant of proportionality.
2. Suppose you have two tanks, one with \(k=-1\) and another with \(k=-10\). What physical differences would you expect to find?
3. Suppose you have a tank for which the height decreases at \(20\) inches per minute when the water is filled to a depth of \(100\) inches. Find the value of \(k\).
4. Solve the initial value problem for the tank in part (c), and graph the solution you determine.
5. How long does it take for the water to run out of the tank?
6. Is the solution that you found valid for all time \(t\)? If so, explain how you know this. If not, explain why not.
အဖြေကို ဖော်ပြပါ
1. Since \(\frac{dh}{dt}\) is proportional to the square root of the height of the water, we have \[\begin{aligned}\end{aligned}\].
2. If one tank has \(k=-1\) and the other has \(k=-10\), the tank with \(k = -10\) will have water leaving the tank much more rapidly, which should correspond to that tank having a larger hole.
3. Since height decreases at \(20\) inches per minute when the water is filled to a depth of \(100\) inches, this tells us that \(\frac{dh}{dt}\vert_{h=100} = -20\). Applying this information to the differential equation, \[\begin{aligned}\end{aligned}\] and thus \(k = -2\).
4. To solve the differential equation \(\frac{dh}{dt} = -2 \sqrt{h}\) with initial condition \(h(0) = 100\), we first separate the variables and write \[\begin{aligned}\end{aligned}\]. Rewriting the first fraction and integrating with respect to \(t\), we have \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\], or \(h = \left( C - t \right)^2\). Knowing that \(h(0) = 100\), we see that \(100 = C^2\), and thus \(C = 10\). This shows that the solution to the IVP is \[\begin{aligned}\end{aligned}\]. A plot of this function shows it to be a concave up parabola with \(y\)-intercept \((0,100)\) and a repeated zero at \((10,0)\).
5. It takes \(10\) minutes for the tank to empty since \(h(10) = 0\).
6. No, this solution is not valid for all time \(t\). The physical constraints of the problem tell us that the solution is only valid for \(0 \lt t \lt 10\) since at \(t = 10\) the tank is empty.
The Gompertz equation is a model that is used to describe the growth of certain populations. Suppose that \(P(t)\) is the population of some organism and that \[\begin{aligned}\end{aligned}\].
1. Sketch a slope field for \(P(t)\) over the range \(0\leq P\leq 6\).
2. Identify any equilibrium solutions and determine whether they are stable or unstable.
3. Find the population \(P(t)\) assuming that \(P(0) = 1\) and sketch its graph. What happens to \(P(t)\) after a very long time?
4. Find the population \(P(t)\) assuming that \(P(0) = 6\) and sketch its graph. What happens to \(P(t)\) after a very long time?
5. Verify that the long-term behavior of your solutions agrees with what you predicted by looking at the slope field.
အဖြေကို ဖော်ပြပါ
2. The only equilibrium solution is \(P = 3\), since \(\ln(1) = 0\) (note: \(\ln(0)\) is undefined, so the differential equation is not defined for \(P = 0\)). From the slope field pictured in (a), the equilibrium solution \(P = 3\) is stable.
3. To solve the IVP with \(\frac{dP}{dt} = -P\ln\left(\frac P3\right)\) and \(P(0) = 1\), we use separation of variables. Observe that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\] On the left, using the substitution \(u = \ln (P/3)\), it follows \(du = \frac{3}{P} \cdot \frac{1}{3} \, dP\), we can rewrite the most recent equation as \[\begin{aligned}\end{aligned}\]. Therefore, \[\begin{aligned}\end{aligned}\], so \(|u| = e^{C-t}\), or \(u = Ke^{-t}\). Next, we recall that \(u = \ln \left(\frac P3\right)\), so \[\begin{aligned}\end{aligned}\] and \(P = 3e^{Ke^{-t}}\). Applying the initial condition, \(1 = 3e^K\), so \(K = \ln \left(\frac{1}{3} \right)\), and thus \[\begin{aligned}\end{aligned}\]. This function's plot is shown in blue in the image above in (a) and we see (both graphically and algebraically) that \(P(t) \to 3\) as \(t \to \infty\) since \(e^{-t} \to 0\).
4. Changing the initial condition to \(P(0) = 6\), we can use all of our preceding work in (c) up to where we established \(P = 3e^{Ke^{-t}}\). Applying this initial condition, \(6 = 3e^K\), so \(K = \ln \left(2 \right)\), and thus \[\begin{aligned}\end{aligned}\]. This function's plot is shown in red in the image above in (a) and we see (both graphically and algebraically) that \(P(t) \to 3\) as \(t \to \infty\) since \(e^{-t} \to 0\).
5. Our work in (c) and (d) suggests that regardless of the initial condition with \(P(0) \gt 0\), the long-term behavior of \(P\) will be for it to tend to \(3\) as \(t\) increases without bound. This matches our initial observations based on the slope field.

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

\frac{\partial u}{\partial t},\ \nabla^2 u

partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Separable differential equations

What is a separable differential equation?
How can we find solutions to a separable differential equation?
Are some of the differential equations that arise in applications separable?

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

သင့်ရဲ့ကိုယ်ပိုင်စမ်းသပ်

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

ပိုပြီး Differential Equations

Separable equations First-order linear equations Second-order, constant coefficients Nonhomogeneous equations Modelling with differential equations