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Separable differential equations

Solve separable differential equations by grouping variables and integrating. Understand arbitrary constants, general and particular solutions, equilibrium solutions and checking the original equation.

Before you startIntegration, logarithms, exponentials and the chain rule.

01 / An equation for a function and its rate of change

A solution must satisfy the derivative relationship on an interval.

dy/dx = f(x)g(y)

Separable means the right-hand side is a product of a function of x and a function of y.

The unknown is a function y(x), not just a number. One differential equation often has a family of solutions. An initial condition supplies a value of y at one x and can select a particular member.

One equation, a family of solutionsExplore

Constant A = y(0)Inspect at x

At x = 0, y = 1.

Derivative from the formula = 0; required xy = 0.

For A ≠ 0, division by y is valid along this entire solution.

Changing A chooses another solution of the same equation. The initial condition y(0) = A selects one curve. The zero solution must be retained even though logarithmic separation divides by y.

02 / Put each variable with its own differential

Check any division before integrating.

Solve dy/dx = xy for nonzero y.Worked example

(1/y) dy = x dx

Divide by y only for the nonzero solutions.

∫(1/y) dy = ∫x dx

Integrate the separated sides.

ln|y| = x²/2 + C

Use one arbitrary constant after combining the constants from both sides.

y = A e^(x²/2), A ≠ 0

Exponentiation gives a nonzero magnitude; the sign is absorbed into A.

Watch: a constant selects a solution curve

Pause, replay or seek freely. The notes explain the same idea and stay in view.

01 · Separate a new equation

Separate dy/dx = 2xy for y ≠ 0.

Hint

Divide by y.

Worked solution

(1/y) dy = 2x dx.

03 / Integrate after separation, not before

The y-integral is with respect to y.

Solve dy/dx = 2xy, including constant solutions.Worked example

ln|y| = x² + C for y ≠ 0

Integrate both separated sides.

y = A e^(x²), A ≠ 0

The new arbitrary constant absorbs sign and exponential magnitude.

y = 0 also satisfies the original equation

Check the case excluded by division.

Thus y = A e^(x²), A any real constant

The same expression includes zero when A = 0.

02 · Recover the derivative

Differentiate y = A e^(x²).

Hint

A is a constant; use the chain rule.

Worked solution

y′ = 2x A e^(x²) = 2xy.

03 · Keep the variable in the integral

Why is ∫(1/y) dy not y⁻¹x?

Hint

The variable of integration is y.

Worked solution

It equals ln|y| + C on an interval with y ≠ 0. Treating y as constant while integrating it is invalid.

04 / Use one arbitrary constant for the family

Changing its name or scale does not restrict it.

Why is one constant sufficient after integrating both sides?Worked example

F(y) + C₁ = G(x) + C₂

Each side could have its own constant.

F(y) = G(x) + (C₂ − C₁)

Move the constants together.

Write C = C₂ − C₁

Their difference is one arbitrary constant.

Be careful when changing the form of the constant. eᶜ is positive, so a sign must be handled when passing from ln|y| to y. A = 0 is not produced by exponentiating a finite C; it is added only after checking the original equation.

04 · Absorb a scale

If y²/2 = x³/3 + C, can the equation be written 3y² = 2x³ + K?

Hint

Multiply by 6.

Worked solution

Yes. K = 6C is another arbitrary real constant.

05 · Do not add a second independent constant

Why is adding an independent B to y = A e^(x²/2) generally invalid?

Hint

Substitute the modified expression into y′ = xy.

Worked solution

The derivative has no added B term, but xy gains xB. The equation would fail unless B = 0.

05 / Recover solutions lost by dividing

Zeros of g(y) may be constant solutions.

Solve dy/dx = x(1 − y).Worked example

First check y = 1

Both sides are zero, so y = 1 is a solution.

For y ≠ 1, dy/(1 − y) = x dx

Now divide on the non-equilibrium branches.

−ln|1 − y| = x²/2 + C

The negative sign comes from differentiating 1 − y.

y = 1 + A e^(−x²/2)

Absorb signs and constants.

Allow A = 0 after checking y = 1

The full family includes the equilibrium solution.

06 · An excluded zero

For y′ = y², does y = 0 satisfy the equation?

Hint

Substitute a constant zero function.

Worked solution

Yes. It is lost if you divide by y² without recording it.

07 · A nonzero equilibrium

Find the constant solution of y′ = (x + 1)(y − 4).

Hint

Set the y-factor to zero.

Worked solution

y = 4.

06 / A power integral may give an implicit solution

Do not choose a square-root sign without information.

Solve y′ = 3x²/(2y), where y ≠ 0.Worked example

2y dy = 3x² dx

Multiply by 2y.

y² = x³ + C

Integrate.

y = ±√(x³ + C) where x³ + C > 0

Each sign defines a branch on an appropriate interval.

y = 0 is not an extra solution here

The original equation is undefined at y = 0.

08 · Another implicit family

Solve y′ = x/y where y ≠ 0.

Hint

Use y dy = x dx.

Worked solution

y² = x² + C, with nonzero branches on intervals where x² + C > 0.

09 · Multiplication does not repair the original domain

May y = 0 be accepted in y′ = x/y just because multiplying by y gives yy′ = x?

Hint

The original quotient must remain defined.

Worked solution

No. A solution must satisfy the original equation, which excludes y = 0.

07 / Solve a reciprocal-power separation

Retain equilibrium solutions separately when necessary.

Solve y′ = y².Worked example

For y ≠ 0: y⁻² dy = dx

Divide by y².

−1/y = x + C

Integrate the negative power.

y = −1/(x + C)

Use a connected interval avoiding x = −C.

Also y = 0

This equilibrium is not represented by any finite C in the reciprocal family.

10 · Check the reciprocal solution

Verify y = −1/(x + C).

Hint

Differentiate and square the original expression.

Worked solution

y′ = 1/(x + C)² = y² wherever x + C ≠ 0.

08 / Use an initial condition to select a member

Substitute the condition into the integrated family.

Solve y′ = xy with y(0) = 2.Worked example

General family y = A e^(x²/2)

Includes all real A.

2 = A e⁰, so A = 2

Use x = 0 and y = 2.

y = 2e^(x²/2)

This satisfies both the equation and the initial value.

11 · Negative initial value

For the same equation, find the solution with y(0) = −3.

Hint

A equals y(0).

Worked solution

y = −3e^(x²/2).

12 · Zero initial value

Which family member has y(0) = 0?

Hint

Use A = 0.

Worked solution

The equilibrium solution y = 0.

09 / Check that the variables really separate

An additive mixture is not automatically a product.

For y′ = x(1 + y²), separate as dy/(1 + y²) = x dx, giving arctan y = x²/2 + C. If you solve explicitly with tangent, restrict to an interval on which the expression is finite and the branch is valid.

The equation y′ = x + y does not separate by dividing both sides by y: that produces x/y + 1, which still mixes the variables. It requires a different method, beyond this lesson.

13 · Recognise a separable product

Is y′ = (x² + 1)(y − 2) separable?

Hint

The x-part and y-part are multiplied.

Worked solution

Yes. First record y = 2, then use dy/(y − 2) = (x² + 1) dx for other branches.

10 / Check the original equation and any condition

An integrated expression alone is not enough.

Differentiate your proposed solution, compare its derivative with the original right-hand side, and check every denominator and logarithm. Then substitute the initial condition. Keep a connected interval on which the solution and equation are defined.

14 · Reject a wrong sign

Does y = A e^(−x²/2), A ≠ 0, solve y′ = xy?

Hint

Differentiate carefully.

Worked solution

No. Its derivative is −xy. It solves y′ = −xy instead.

15 · Where is the family valid?

For y = −1/(x + 2), what point must be excluded?

Hint

Locate the zero denominator.

Worked solution

x = −2. Use an interval wholly on one side of this point.

11 / Separate, integrate, restore, verify

Record excluded cases before any division.

Recognise a product of an x-function and a y-function.
Check constant solutions before dividing by a y-factor.
Group variables with their own differentials.
Integrate with one arbitrary constant.
Handle signs, branches and original domains.
Use initial data when supplied.
Differentiate back and verify the original equation.

16 · A final separated equation

Solve y′ = 2x(y − 3), including equilibrium.

Hint

Use ln|y − 3| = x² + C for nonzero y − 3.

Worked solution

y = 3 + A e^(x²), A any real constant. A = 0 gives y = 3.

Section 1 of 11 · An equation for a function and its rate of change