Separable Differential Equations

A rate that depends on the quantity

An equation that contains a derivative is a differential equation, and a solution of one is a function. $\frac{dy}{dx}=ky$dy/dx = ky says that y changes at a rate proportional to y itself. With k > 0, the larger y is, the faster it grows: that is growth. With k < 0, y shrinks, and shrinks more slowly as it gets smaller: that is decay.

For $\frac{dy}{dx}=0.5y$dy/dx = 0.5y, the gradient at height 1 is 0.5, at height 2 it is 1, and at height 3 it is 1.5. The gradient depends only on the height, so a solution steepens as it climbs.

Short segments at whole-number points, each drawn at the gradient $\frac{dy}{dx}=0.5y$dy/dx = 0.5y: 0.5 along the line y = 1, 1 along y = 2, 1.5 along y = 3, and flat along the x-axis. The gold curve $y=e^{0.5x}$y = e^(0.5x) passes through (0, 1) and runs along every segment it meets; at (2, e) its gradient is $\frac{e}{2}\approx 1.36$e/2 ≈ 1.36.

Separate, then integrate

Treat dy and dx as the two sides of a rate, and move every y to the side with dy and every x to the side with dx. Dividing $\frac{dy}{dx}=ky$dy/dx = ky by y gives $\left(\frac{1}{y}\right)dy=kdx$(1/y) dy = k dx.

Now each side is in one letter, so each can be integrated in its own letter: $\int \left(\frac{1}{y}\right)dy=\int kdx$∫ (1/y) dy = ∫ k dx, which gives ln|y| = kx + c.

Each integral has its own constant, but two constants on opposite sides combine into one. Write a single c, on the side with x.

Undo the log

Take e to the power of both sides: $|y|=e^{kx+c}$|y| = e^(kx + c). A power of a sum splits into a product, so $|y|=e^c\cdot e^{kx}$|y| = e^c · e^(kx). The number $e^c$e^c is a constant like any other, so call it A, and drop the absolute value by letting A be negative as well as positive: $y=A{e}^{kx}$y = Ae^(kx).

A = 0 gives y = 0, which also solves $\frac{dy}{dx}=ky$dy/dx = ky, since both sides are 0. Dividing by y at the start had quietly set that solution aside.

Check: $A{e}^{kx}$Ae^(kx) differentiates to $k\times A{e}^{kx}$k × Ae^(kx), which is k times y.

A starting value fixes A

The equation alone has a whole family of solutions, one for each A. A known value, called an initial condition, picks one of them. Solve $\frac{dy}{dx}=2y$dy/dx = 2y with y = 6 when x = 0. Separating and integrating gives $y=A{e}^{2x}$y = Ae^(2x). At x = 0, $e^0=1$e^0 = 1, so 6 = A, and the solution is $y=6e^{2x}$y = 6e^(2x).

Check it at a point. At x = 0.5, $y=6e\approx 16.31$y = 6e ≈ 16.31. A difference quotient across a tiny step gives the derivative there as 32.62, which is 2 × 16.31, as the equation says it must be.

In the same way, P = 100 at t = 0 with $\frac{dP}{dt}=\left(\ln 2\right)P$dP/dt = (ln 2)P gives $P=100e^{\left(\ln 2\right)t}=100\times 2^t$P = 100e^((ln 2)t) = 100 × 2ᵗ. It doubles each unit of time, and reaches 800 at t = 3.

Other separable equations

Solve $\frac{dy}{dx}=xy$dy/dx = xy with y = 1 when x = 0. Separating gives $\left(\frac{1}{y}\right)dy=xdx$(1/y) dy = x dx, and integrating gives $\ln y=\frac{x^2}{2}+c$ln y = x²/2 + c. At x = 0, ln 1 = 0, so c = 0, and the solution is $y=e^{\frac{x^2}{2}}$y = e^(x²/2). At x = 1 it is $e^{0.5}\approx 1.649$e^(0.5) ≈ 1.649, and the gradient there, measured by a difference quotient, is also 1.649, which is x times y with x = 1.

Solve $\frac{dy}{dx}=\frac{x}{y}$dy/dx = x/y with y = 3 when x = 0. Separating gives y dy = x dx, and integrating gives $\frac{y^2}{2}=\frac{x^2}{2}+c$y²/2 = x²/2 + c. At x = 0, $\frac{9}{2}=c$9/2 = c. Multiplying by 2, $y^2=x^2+9$y² = x² + 9, and since y is positive at the start, $y=\sqrt{x^2+9}$y = √(x² + 9). At x = 4, y = 5, and the equation asks for a gradient of $\frac{4}{5}=0.8$4/5 = 0.8 there, which the curve has.

Use the condition after integrating, not before: the constant only exists once both sides have been integrated.

The slope field of $\frac{dy}{dx}=xy$dy/dx = xy: flat along the y-axis, gradient 1 at (1, 1), 2 at (1, 2) and at (2, 1), and −1 at (−1, 1). The gold curve $y=e^{\frac{x^2}{2}}$y = e^(x²/2) is the solution through (0, 1); the dashed curve $y=2e^{\frac{x^2}{2}}$y = 2e^(x²/2) is the one through (0, 2). Both run along the segments.

Decay

With a negative k the same steps give decay. $\frac{dy}{dx}=-0.5y$dy/dx = −0.5y with y = 8 at x = 0 gives $y=8e^{-0.5x}$y = 8e^(−0.5x). Each $2\ln 2\approx 1.386$2 ln 2 ≈ 1.386 units of x halves y, whatever y is: at x = 2, $y=8e^{-1}\approx 2.943$y = 8e^(−1) ≈ 2.943, and the gradient there is $-0.5\times 2.943\approx -1.472$−0.5 × 2.943 ≈ −1.472.

The usual mistakes

Integrating without separating. $\frac{dy}{dx}=ky$dy/dx = ky does not integrate to y = kx + c: that solves $\frac{dy}{dx}=k$dy/dx = k, a constant rate, and ignores the y on the right.

Adding the constant after exponentiating. From ln y = kx + c the next line is $y=e^{kx+c}=A{e}^{kx}$y = e^(kx + c) = Ae^(kx), not $e^{kx}+c$e^(kx) + c.

Leaving A at 1. With y = 6 at x = 0, the solution of $\frac{dy}{dx}=2y$dy/dx = 2y is $6e^{2x}$6e^(2x), not $e^{2x}$e^(2x).

Swapping the rate and the starting value. $\frac{dy}{dx}=2y$dy/dx = 2y with y = 6 at x = 0 is $6e^{2x}$6e^(2x), not $2e^{6x}$2e^(6x): the rate goes in the exponent and the starting value in front.

A water butt emptying

In the application below, the depth of water falls at a rate proportional to its square root. Separating and integrating gives $\sqrt{h}$√h as a straight line in time, and two readings fix its two constants.