Separable: which two parts?
If the rate of change splits into a part that depends only on x times a part that depends only on y, you can handle the two sides separately.
If a first-order equation can be written as
it is called a separable differential equation. y′=y fits with f(x)=1 and g(y)=y; y′=2xy fits with f(x)=2x and g(y)=y. The rate of change depends on both x and y, yet it splits into two factors that each contain only one variable.
y′=x+y has no such product form, and you can’t “split x+y across the two sides” and call it separation. Some equations become separable after a substitution, but that is a separate step that comes later.
Calculus reminder: the chain rule
If the input of H is y(x), differentiating with respect to x takes both the outer and the inner derivative: dH(y(x))/dx=H′(y(x))y′(x). For example, d(y²/2)/dx=y y′, and d ln|y|/dx=y′/y, though the second holds only where y≠0.
Separation comes from the chain rule
Moving dy and dx around looks like working with fractions, but the real reason it works is the chain rule. This section fills in that reason.
Assume that on some interval g(y(x))≠0, and that f and g are continuous where needed. Choose an antiderivative H with H′(u)=1/g(u); u is just a placeholder variable for the integration.
Divide by the nonzero factor
y′/g(y) = f(x)This step already restricts us to stretches of solution where g(y) is not zero. Don’t forget that condition.
Recognize the left side as the derivative of a composite function
dH(y(x))/dx = H′(y(x))y′(x) = f(x)That is the chain rule: the outer derivative of H is 1/g, and the inner derivative is y′.
Integrate both sides with respect to the same x
H(y(x)) = ∫ f(x) dx + CBoth sides get the same operation. On the left, the fundamental theorem of calculus gives back H; C is an arbitrary constant.
Write it in the usual notation
∫ dy/g(y) = ∫ f(x) dx + CWriting dy is a compact way to express integration by substitution. There is no need to treat dy/dx as an ordinary fraction and cancel.
If you write a constant of integration on each side, moving them to one side leaves only their difference, which is still one arbitrary constant. You may not be able to solve explicitly for y at the end; reaching H(y)=F(x)+C can already be a useful implicit description.
Put back any solutions you may have lost. Every constant c with g(c)=0 is worth checking: if the equation is defined there, y(x)=c has derivative zero and the right side is zero too, so it is an equilibrium solution. If the uniqueness conditions fail, there may also be solutions that rest at an equilibrium value and then leave, so listing the constant solutions does not prove you have found them all.
Worked example: y′=2xy, y(0)=2
Go all the way through once: separate, integrate, use the starting value to fix the constant, then check. The problem is solved only when all four steps are done.
Compared with y′=y, the coefficient 2x now changes with position. For the first time, we actually derive the function from the differential equation, instead of knowing the answer and checking it.
Spot the form and list what division would exclude
f(x)=2x, g(y)=y; y=0 is a solutionThe starting value 2 is not on the zero solution, so first work on the nonzero stretch of solution that contains x=0.
Separate and integrate
y′/y=2x ⇒ ln|y|=x²+C₁The left side is the derivative of ln|y(x)|; the right side integrates to x². This is not integrating with y treated as a constant.
Undo the logarithm and restore the sign
|y|=eC₁ex² ⇒ y=Aex²e^(C₁)>0. On a connected nonzero stretch, y cannot suddenly change sign. Absorbing the sign, A can be any nonzero real number.
Use the starting value to choose the constant
2=Ae⁰ ⇒ A=2This gives y=2e^(x²). The zero solution does not meet this starting value, but it still belongs to the equation’s family of solutions.
Check the equation and the starting value separately
y′=4xex² = 2x(2ex²) = 2xyThe chain rule brings out the 2x; putting in x=0 gives y(0)=2. Neither check can be skipped.
State the interval and the whole family
y=Aex², A∈ℝ; x∈ℝPut back A=0 for the zero solution. f=2xy and fᵧ=2x are continuous everywhere, so every starting value has exactly one solution, and A reaches every y(0). So in this example no solutions that cross zero or wait at zero have been missed.
The 2e≈5.436563 on the plot is the value at x=1; the decimal is a rounded display. The formula 2e^(x²) itself is exact. For x<0 the positive solution has negative slope 2xy, and for x>0 its slope is positive, so its lowest point is at x=0.
Not every right side separates
Before you divide by y, ask whether y can be 0. The curve that division throws away is often exactly the missing solution.
y′=x+y looks like it has only two terms, yet it cannot be separated directly into y′=f(x)g(y). If you write dy/(x+y)=dx and then “integrate in y,” the denominator still contains x, which keeps changing; you have not built the H(y) described above.
y′=y(1−y), on the other hand, really is separable, but you can’t declare that the nonzero formula from integration is the whole answer. Before dividing by y(1−y), list y=0 and y=1 and check both. On the page about autonomous equations, these two equilibrium values explain the overall behavior.
A reminder: being separable does not promise an easy integral, or that you can solve for y with elementary functions afterward. The method gives structure; it still has to be combined with integration, inverse functions and a check of the interval of validity.
From spotting the form to checking the answer
Spot the form → separate → integrate → fix the constant → check. Write down any condition that any step needs.
1. y′=2xy with y(0)=−1. What is the solution, and on what interval?
See the full reasoning
Integrating in the same way gives ln|y|=x²+C₁, and the nonzero solutions are Ae^(x²). The starting value gives A=−1, so y=−e^(x²), valid on the whole real line. Its derivative is −2xe^(x²)=2xy, and y(0)=−1. Taking an exponential does not make the minus sign disappear.
2. Solve y′=y, y(0)=3 again, this time by separation.
See the full reasoning
First list y=0. For nonzero solutions, y′/y=1 integrates to ln|y|=x+C₁, and combining the signs gives y=Aeˣ. y(0)=3 gives A=3, so y=3eˣ. Differentiating gives 3eˣ again, the starting value is right, and the interval is ℝ. Now you know where the earlier family Ceˣ comes from.
3. y′=(1+x²)(y−3), y(0)=3. Can you divide by y−3 and then find the constant?
See the full reasoning
That division cannot cover this solution. Try y=3 first: its derivative is 0, the right side is 0 too, and it meets the starting value. Integrating gives the other solutions, y=3+A e^(x+x³/3) with A≠0; putting back A=0 includes the branch where the factor is zero. Since fᵧ=1+x² is continuous, the starting value y(0)=3 picks out only the equilibrium solution.
A formula is not the end
Having a formula does not finish the job: the formula may not even be defined at some x.
We now have a first complete method: spot the product form, keep the zero factors, separate and integrate using the chain rule, put in the starting value, and check against the original equation. The next page takes up the two questions left over: does an equation you haven’t solved for y count as a solution? And how do you find the largest interval on which the formula works?
Sources and scope
Definitions and the conditions of theorems were checked against the authors and university materials below. The explanations, worked examples and interactive plots here are written independently. Decimals on the plots are rounded displays; the mathematical conclusions rest on derivation and substitution. This page does not prove the general existence and uniqueness theorems it cites.
- Jiří Lebl · Separable equations: separating and integrating, the chain rule, zero solutions and implicit solutions.