Separable Differential Equations and Exponential Models

Q: What is a separable differential equation?

One that can be written dy/dx = g(x) h(y), so that all the y's can be moved to one side and all the x's to the other.

Q: How do I solve a separable equation?

Write dy/h(y) = g(x) dx, integrate both sides, add one constant, and solve for y. Use an initial condition to find the constant.

Q: What is the solution of dy/dt = ky?

y = y₀e^(kt), where y₀ is the value at t = 0. It grows when k > 0 and decays when k < 0.

Q: How are half-life and k related?

For decay y₀e^(kt), the half-life is ln 2 / |k|: the time for the amount to fall to half, whatever amount you start with.

Q: Why is only one constant needed?

Constants from both sides can be combined into one: ∫ dy/h(y) + C₁ = ∫ g(x) dx + C₂ is the same as one equation with C = C₂ − C₁.

Quick answer

A differential equation is separable when its right side factors as a function of x times a function of y. Move every y to one side and every x to the other, integrate both sides, and solve for y, using an initial condition to fix the constant. The most important case is dy/dt = ky, whose solutions are y = y₀e^(kt): exponential growth when k > 0 and decay when k < 0, with a fixed doubling time or half-life. Newton's law of cooling separates the same way.

What you'll learn

Recognize and solve separable differential equations
Find particular solutions from initial conditions
Model exponential growth and decay with dy/dt = ky
Solve problems about half-life, doubling time and cooling

Separating the variables

The equation $\tfrac{dy}{dx} = 2xy$ has a right side that factors into an $x$ part and a $y$ part. Such an equation is separable. Divide by the $y$ part and multiply by $dx$ , so each variable has its own side:

\frac{dy}{y} = 2x\,dx

Integrate both sides:

\int \frac{dy}{y} = \int 2x\,dx \quad\Longrightarrow\quad \ln\lvert y \rvert = x^2 + C

and solve for $y$ : $\lvert y \rvert = e^{x^2 + C} = e^C e^{x^2}$ , so $y = Ae^{x^2}$ , with $A$ any constant.

Why separating is legitimate

Moving $dx$ around looks like treating $\tfrac{dy}{dx}$ as a fraction. What is really happening is the chain rule.

Suppose $y(x)$ solves $\tfrac{dy}{dx} = g(x)\,h(y)$ . Let $H$ be an antiderivative of $\tfrac{1}{h}$ and $G$ one of $g$ . By the chain rule,

\frac{d}{dx} H\big(y(x)\big) = \frac{1}{h(y)} \cdot \frac{dy}{dx} = g(x) = \frac{d}{dx} G(x)

Two functions with the same derivative differ by a constant, so $H(y) = G(x) + C$ — exactly what integrating both sides produced. Separation of variables is the chain rule applied to an unknown solution.

Exponential growth and decay

When a quantity changes at a rate proportional to its size,

\frac{dy}{dt} = ky \quad\Longrightarrow\quad y = y_0 e^{kt}

where $y_0 = y(0)$ . Separating gives $\ln\lvert y \rvert = kt + C$ , and the initial value fixes the constant. For $k > 0$ this is growth, with a fixed doubling time $\tfrac{\ln 2}{k}$ . For $k < 0$ it is decay, with a fixed half-life $\tfrac{\ln 2}{\lvert k \rvert}$ .

The half-life is the same whatever amount is left. Each 5-year step below halves the amount, from $100$ to $50$ and from $12.5$ to $6.25$ alike.

y = 100 · (1/2)^(t/5)

Every 5 years, half of what is left decays

Newton’s law of cooling, $\tfrac{dT}{dt} = -k(T - T_{\text{room}})$ , separates the same way and gives $T = T_{\text{room}} + (T_0 - T_{\text{room}})e^{-kt}$ : the difference from room temperature decays exponentially.

Worked examples

Common mistakes

Practice problems

Solve $\tfrac{dy}{dx} = 3y$ with $y(0) = 4$ .

Answer

$y = 4e^{3x}$

Full solution

It has the form $\tfrac{dy}{dx} = ky$ with $k = 3$ and $y_0 = 4$ .
Find the general solution of $\tfrac{dy}{dx} = \tfrac{x^2}{y^2}$ .

Answer

$y = \sqrt[3]{x^3 + C}$

Full solution

$y^2\,dy = x^2\,dx$ , so $\tfrac{y^3}{3} = \tfrac{x^3}{3} + C_1$ and $y^3 = x^3 + C$ .
Solve $\tfrac{dy}{dx} = y\cos x$ with $y(0) = 2$ .

Answer

$y = 2e^{\sin x}$

Full solution

$\tfrac{dy}{y} = \cos x\,dx$ gives $\ln\lvert y \rvert = \sin x + C$ , so $y = Ae^{\sin x}$ , and $y(0) = A = 2$ .
Solve $\tfrac{dy}{dx} = e^{x - y}$ with $y(0) = 0$ .

Answer

$y = x$

Full solution

$e^{x - y} = e^x e^{-y}$ , so $e^y\,dy = e^x\,dx$ . Then $e^y = e^x + C$ , and $y(0) = 0$ gives $1 = 1 + C$ , so $C = 0$ and $y = x$ .
A substance decays with half-life $8$ days. Find $k$ in $y = y_0 e^{kt}$ .

Answer

$k = -\tfrac{\ln 2}{8} \approx -0.0866$ per day

Full solution

$\tfrac{1}{2} = e^{8k}$ gives $8k = -\ln 2$ .
A bacteria culture starts with $500$ cells and triples every $2$ hours. How many cells after $5$ hours?

Answer

About $7794$

Full solution

$y = 500 \cdot 3^{t/2}$ , so $y(5) = 500 \cdot 3^{2.5} \approx 7794$ .
Coffee at $85°$ C sits in a $25°$ C room, with $k = 0.05$ per minute. What is its temperature after $10$ minutes?

Answer

About $61.4°$ C

Full solution

$T = 25 + 60e^{-0.05 \cdot 10} = 25 + 60e^{-0.5} \approx 25 + 36.4 = 61.4$ .
Is $\tfrac{dy}{dx} = xy + x$ separable? If so, solve it.

Answer

Yes: $y = Ae^{x^2/2} - 1$

Full solution

$xy + x = x(y + 1)$ . Then $\tfrac{dy}{y + 1} = x\,dx$ , so $\ln\lvert y + 1 \rvert = \tfrac{x^2}{2} + C$ and $y + 1 = Ae^{x^2/2}$ .
How long does an investment growing continuously at $7\%$ a year take to double?

Answer

About $9.9$ years

Full solution

$t = \tfrac{\ln 2}{0.07} \approx 9.90$ .
A student solves $\tfrac{dy}{dx} = 2y$ , $y(0) = 3$ as $\ln y = 2x$ , then $y = e^{2x} + 3$ . What went wrong?

Hint

Where should the constant go, and when?

Answer

The constant was dropped at integration and added at the end instead. The solution is $y = 3e^{2x}$ .

Full solution

Integrating gives $\ln\lvert y \rvert = 2x + C$ , so $y = Ae^{2x}$ . The initial condition gives $A = 3$ .

The student’s function has $y(0) = 4$ and does not satisfy $y' = 2y$ : its derivative is $2e^{2x}$ , but $2y = 2e^{2x} + 6$ .

Frequently asked questions

What is a separable differential equation?