Separable Differential Equations

A separable differential equation is one you can write as dy/dx = g(x)·h(y), so all the y’s can move to one side and all the x’s to the other. To solve it, separate the variables, integrate both sides, add one constant C, and solve for y. An initial condition then gives the value of C.

Before this lesson: Verifying Solutions to Differential Equations, Euler’s Method and U-Substitution

The idea: sort the x’s and the y’s

A differential equation like dy/dx = 4x/y mixes x and y. You can’t integrate the right side with respect to x, because y is an unknown function of x.

The fix is to sort. Move every y, along with dy, to the left side. Move every x, along with dx, to the right side. Now each side has only one variable, so you can integrate each side on its own.

This works because of the chain rule. Integrating (1/h(y)) dy on the left is the same as integrating (1/h(y))·(dy/dx) dx, so both sides are really integrals with respect to x.

How to solve separable differential equations

\dfrac{dy}{dx} = g(x)\,h(y) \quad\Longrightarrow\quad \int \dfrac{1}{h(y)}\,dy = \int g(x)\,dx

1. Separate: divide both sides by h(y) and multiply both sides by dx, so dy is with the y’s and dx is with the x’s.
2. Integrate both sides.
3. Add one constant of integration, C, on the x side. (Constants from both sides combine into one.)
4. Solve for y if you can. If you can’t, leave the answer as an equation in x and y, called an implicit solution.
5. Use the initial condition to find C, and use it to choose a sign if a square root appears.

Watch for lost solutions. Dividing by h(y) is only allowed when h(y) ≠ 0. If h(c) = 0 for some number c, then the constant function y = c is also a solution, called an equilibrium solution.

How to tell if an equation is separable

The right side must factor into a function of x times a function of y. Factoring or using exponent rules sometimes shows the split.

Separable differential equations examples

Example 1: Square root solution

Solve dy/dx = 4x/y with y(1) = 3.

y\,dy = 4x\,dx

\int y\,dy = \int 4x\,dx

\dfrac{y^2}{2} = 2x^2 + C

y^2 = 4x^2 + 2C

y^2 = 4x^2 + K

3^2 = 4(1)^2 + K

9 = 4 + K

K = 5

y^2 = 4x^2 + 5

y = \sqrt{4x^2 + 5}

Answer: y = √(4x² + 5). Check: y′ = 4x/√(4x² + 5) = 4x/y, and y(1) = √9 = 3.

Example 2: Exponential solution

Solve dy/dx = 3x²y with y(0) = 5.

\dfrac{1}{y}\,dy = 3x^2\,dx

\int \dfrac{1}{y}\,dy = \int 3x^2\,dx

\ln|y| = x^3 + C

|y| = e^{x^3 + C}

|y| = e^{C}e^{x^3}

y = A\,e^{x^3}

5 = A\,e^{0}

A = 5

Answer: y = 5e^(x³). The constant C became a multiplier A, which is how exponential solutions usually look.

Example 3: Splitting an exponential

Solve dy/dx = e^(2x − y) with y(0) = 0.

\dfrac{dy}{dx} = e^{2x}e^{-y}

e^{y}\,dy = e^{2x}\,dx

\int e^{y}\,dy = \int e^{2x}\,dx

e^{y} = \tfrac{1}{2}e^{2x} + C

e^{0} = \tfrac{1}{2}e^{0} + C

1 = \tfrac{1}{2} + C

C = \tfrac{1}{2}

e^{y} = \tfrac{1}{2}e^{2x} + \tfrac{1}{2}

y = \ln\left(\tfrac{1}{2}e^{2x} + \tfrac{1}{2}\right)

Answer: y = ln((e^(2x) + 1)/2).

Example 4: Factor first

Solve dy/dx = xy² + x with y(0) = 1.

\dfrac{dy}{dx} = x(y^2 + 1)

\dfrac{1}{y^2 + 1}\,dy = x\,dx

\int \dfrac{1}{1 + y^2}\,dy = \int x\,dx

\arctan y = \dfrac{x^2}{2} + C

\arctan 1 = 0 + C

C = \dfrac{\pi}{4}

\arctan y = \dfrac{x^2}{2} + \dfrac{\pi}{4}

y = \tan\left(\dfrac{x^2}{2} + \dfrac{\pi}{4}\right)

Answer: y = tan(x²/2 + π/4). This solution is valid only while x²/2 + π/4 stays below π/2, so for |x| less than √(π/2) ≈ 1.25.

Example 5: Draining a tank (word problem)

Water drains from a tank through a hole in the bottom. The depth h (in feet) after t minutes follows dh/dt = −0.2√h, and the tank starts 16 feet deep. Find the depth after 10 minutes and when the tank is empty.

h^{-1/2}\,dh = -0.2\,dt

\int h^{-1/2}\,dh = \int -0.2\,dt

2h^{1/2} = -0.2t + C

2\sqrt{16} = -0.2(0) + C

C = 8

2\sqrt{h} = 8 - 0.2t

\sqrt{h} = 4 - 0.1t

h = (4 - 0.1t)^2

h(10) = (4 - 1)^2 = 9

4 - 0.1t = 0

t = 40

Answer: 9 feet deep after 10 minutes, and empty after 40 minutes. The depth drops fastest at the start, when the water pressure is highest.

Common mistakes

Related methods

Slope fields and Euler’s method

A slope field shows the shape of the solutions, and Euler’s method estimates them with numbers. Separation of variables gives the exact formula, so you can use it to check both.

Checking your answer

Differentiate your solution and substitute it back into the equation, as in verifying solutions to differential equations. Also check that it meets the initial condition.

Growth, cooling and logistic models

The rest of this unit applies separation to dy/dt = ky (Exponential Models with Differential Equations), to Newton’s law of cooling and to logistic growth. All three are separable. Compare the algebra version in exponential growth and decay.

Equations that aren’t separable

Equations like dy/dx = x + y need other methods, such as an integrating factor. Those are taught in the Differential Equations course.

Practice problems

1. Find the general solution of dy/dx = 2x/y².
2. Solve dy/dx = 5y with y(0) = 2.
3. Solve dy/dx = x²/y with y(0) = −2.
4. Solve dy/dx = y cos x with y(0) = 3.
5. Solve dy/dx = e^(x + y) with y(0) = 0.
6. Solve dy/dx = (1 + y²)/(1 + x²) with y(0) = 1. Write y without inverse trig functions.
7. Solve dy/dx = xy − 2y with y(0) = 4.
8. Which are separable? (a) dy/dx = x − xy (b) dy/dx = sin(xy) (c) dy/dx = eˣy²
9. Solve dy/dx = 2xy² with y(0) = 1.
10. A chemical breaks down so that its mass A (in grams) after t hours follows dA/dt = −0.5A², with A(0) = 4. Find a formula for A and the mass after 6 hours.

y^2\,dy = 2x\,dx

\int y^2\,dy = \int 2x\,dx

\dfrac{y^3}{3} = x^2 + C

y^3 = 3x^2 + K

y = \sqrt[3]{3x^2 + K}

\dfrac{1}{y}\,dy = 5\,dx

\ln|y| = 5x + C

y = A\,e^{5x}

2 = A\,e^{0}

A = 2

y\,dy = x^2\,dx

\dfrac{y^2}{2} = \dfrac{x^3}{3} + C

y^2 = \dfrac{2x^3}{3} + K

(-2)^2 = 0 + K

K = 4

y = -\sqrt{\dfrac{2x^3}{3} + 4}

\dfrac{1}{y}\,dy = \cos x\,dx

\ln|y| = \sin x + C

y = A\,e^{\sin x}

3 = A\,e^{\sin 0} = A

\dfrac{dy}{dx} = e^{x}e^{y}

e^{-y}\,dy = e^{x}\,dx

-e^{-y} = e^{x} + C

-e^{0} = e^{0} + C

C = -2

e^{-y} = 2 - e^{x}

-y = \ln(2 - e^{x})

y = -\ln(2 - e^{x})

\dfrac{1}{1 + y^2}\,dy = \dfrac{1}{1 + x^2}\,dx

\arctan y = \arctan x + C

\arctan 1 = \arctan 0 + C

C = \dfrac{\pi}{4}

y = \tan\left(\arctan x + \dfrac{\pi}{4}\right)

y = \dfrac{\tan(\arctan x) + \tan\frac{\pi}{4}}{1 - \tan(\arctan x)\tan\frac{\pi}{4}}

y = \dfrac{x + 1}{1 - x}

\dfrac{dy}{dx} = y(x - 2)

\dfrac{1}{y}\,dy = (x - 2)\,dx

\ln|y| = \dfrac{x^2}{2} - 2x + C

y = A\,e^{x^2/2 - 2x}

4 = A\,e^{0}

x - xy = x(1 - y)

\sin(xy)

e^{x}y^2

y^{-2}\,dy = 2x\,dx

-y^{-1} = x^2 + C

-\dfrac{1}{1} = 0 + C

C = -1

-\dfrac{1}{y} = x^2 - 1

\dfrac{1}{y} = 1 - x^2

y = \dfrac{1}{1 - x^2}

A^{-2}\,dA = -0.5\,dt

-A^{-1} = -0.5t + C

-\dfrac{1}{4} = 0 + C

\dfrac{1}{A} = 0.5t + \dfrac{1}{4}

\dfrac{1}{A} = \dfrac{2t + 1}{4}

A = \dfrac{4}{2t + 1}

A(6) = \dfrac{4}{13} \approx 0.308

Separable differential equations FAQ

What makes a differential equation separable?

It is separable when dy/dx can be written as a function of x times a function of y. Then every y can go with dy and every x with dx.

How do you solve a separable differential equation?

Separate the variables, integrate both sides, add one constant C, and solve for y. Use the initial condition, if one is given, to find C.

Why is there only one constant of integration?

Each side does produce a constant, but you can subtract one from the other and combine them into a single constant. Putting it on the x side is the usual choice.

Can every first-order differential equation be solved by separation of variables?

No. Equations like dy/dx = x + y or dy/dx = sin(xy) don’t split into an x part times a y part. They need other methods or a numerical method like Euler’s method.

Summary

• A separable equation can be written as dy/dx = g(x)·h(y).
• Separate: (1/h(y)) dy = g(x) dx, then integrate both sides.
• Add one constant C right after integrating, before solving for y.
• Exponential solutions come out as y = Ae^(…), where the constant is a multiplier.
• Use the initial condition to find C and to pick the sign of a square root.
• Values where h(y) = 0 give constant (equilibrium) solutions.

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Next lesson: Exponential Models with Differential Equations

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