Differential Equations: Separable, Homogeneous or Linear?

You look at a differential equation, recognise every symbol, and still wonder: which method should I use?

That hesitation is often the hardest part. Once you know whether to separate the variables, put y = vx, or find an integrating factor, the question becomes a sequence of familiar steps.

A separable equation lets you split the variables. A homogeneous equation depends on a ratio such as y/x. A first-order linear equation can be arranged as dy/dx + P(x)y = Q(x). The useful skill is spotting those structures after a little rearrangement.

This guide focuses on first-order equations with the derivative appearing to the first power. We will use y' as shorthand for dy/dx, and ln for the natural logarithm.

The Three-Question Method Checklist

Start by simplifying brackets and collecting terms. Then ask:

1. Can I write the right side as a product of an x-only expression and a y-only expression? Try separation.
2. Can I arrange the equation as y' + P(x)y = Q(x)? Try the linear method.
3. Does the right side become a function of y/x alone? Try the homogeneous substitution y = vx.

This is a practical checking order, not a rule that the categories cannot overlap. Sometimes two or even all three methods work. Choose the shortest valid route unless the question specifies a method.

Type	What to recognise	First useful move
Separable	y' = f(x)g(y)	Put y terms with dy and x terms with dx
Linear in y	y' + P(x)y = Q(x)	Find an integrating factor
Homogeneous ratio form	y' = F(y/x)	Substitute y = vx

The standard forms and the ratio substitution are covered in the official NCERT chapter on differential equations. The examples below are a separate practice workshop designed to make the choice visible.

1. Separable: Can the Variables Live on Opposite Sides?

Look at:

y' = 3x(y + 2)

The right side is already a product. The expression 3x depends only on x, and y + 2 depends only on y. For y != -2, rearrange:

dy/(y + 2) = 3x dx

Each integral now has only its own variable. Compare that with:

y' = x + y^2

A sum of an x term and a y term does not automatically separate. Writing dy/y^2 = x dx would simply throw away part of the equation.

Brackets can hide a separable product. For example, xy + 2x + y + 2 factors as (x + 1)(y + 2). Factor before deciding that an equation is difficult.

Worked Example 1: Separation With an Initial Condition

Solve:

y' = 2x(y - 3),     y(0) = 5

Identify: the right side is an x-only factor multiplied by a y-only factor.

Before dividing by y - 3, check y = 3. Its derivative is zero, and the right side is zero too, so it is a constant solution of the differential equation. It does not meet our initial condition.

For the other solutions:

dy/(y - 3) = 2x dx
ln|y - 3| = x^2 + C1
y - 3 = A e^(x^2)

The constant A may be positive or negative. Including A = 0 also restores the constant solution we checked separately.

Use y(0) = 5:

5 - 3 = A e^0
A = 2

Answer: y = 3 + 2e^(x^2)

Check: differentiating gives y' = 4x e^(x^2). Substituting into 2x(y - 3) gives the same expression. At x = 0, y = 5.

For more on this method and its solution families, see OpenStax’s separable equations lesson.

2. Homogeneous: Does Only the Ratio Matter?

Here, homogeneous means that an equation written as y' = F(x,y) has a right side depending only on the ratio y/x, on the domain being considered.

For example:

y' = (2x^2 + xy + y^2)/x^2
   = 2 + y/x + (y/x)^2

Scaling both x and y by the same nonzero factor leaves this expression unchanged. That is why a ratio is useful: the overall size disappears.

For polynomial quotients, a quick clue is that every term in the numerator and denominator has the same total degree. The total degree of xy is 2, just like x^2 or y^2.

Do not look only at the highest power. The quotient (x^2 + y + 1)/x^2 contains terms of different degrees and is not a function of y/x alone.

Also, this test is about the degree of expressions in x and y. It is different from the degree of a differential equation, which concerns powers of derivatives.

Why the Derivative Has Two Terms

Set:

y = vx

The new variable v usually changes with x. Applying the product rule gives:

y' = v + x dv/dx

Substituting into y' = F(y/x) produces:

v + x dv/dx = F(v)
x dv/dx = F(v) - v

This is separable wherever the required divisions are valid. If F(v) - v = 0 for some constant v, check the corresponding straight-line solution y = vx separately.

Worked Example 2: A Clear Homogeneous Pattern

Solve:

y' = (2x^2 + xy + y^2)/x^2,     x != 0

Identify: dividing through by x^2 leaves 2 + y/x + (y/x)^2.

Put y = vx and y' = v + x dv/dx:

v + x dv/dx = 2 + v + v^2
x dv/dx = 2 + v^2

dv/(v^2 + 2) = dx/x

Integrate:

(1/sqrt(2)) arctan(v/sqrt(2)) = ln|x| + C

Replace v with y/x:

Answer:
(1/sqrt(2)) arctan(y/(sqrt(2)x))
= ln|x| + C

This implicit answer is sufficient. It describes solutions on suitable intervals excluding x = 0. There are no real constant-v solutions lost through division because v^2 + 2 is always positive.

Check the structure: differentiating the implicit answer gives x y' - y = 2x + y^2/x, or equivalently y' = y/x + 2 + y^2/x^2, exactly the original equation.

Worked Example 3: The Straight Line You Could Lose

Solve:

y' = y/x + (y/x - 2)^2,     x != 0

Identify: every occurrence of x and y on the right is through y/x.

After y = vx:

v + x dv/dx = v + (v - 2)^2
x dv/dx = (v - 2)^2

Before dividing, try v = 2. This gives the solution y = 2x. It really works: its derivative is 2, and the right side is 2 + 0.

For v != 2:

dv/(v - 2)^2 = dx/x
-1/(v - 2) = ln|x| + C

Answer for the remaining family:
-x/(y - 2x) = ln|x| + C

You may also write this family as:

y = 2x - x/(ln|x| + C)

Use intervals where x is nonzero and the displayed denominator is nonzero. State y = 2x separately. It cannot be obtained by assigning an ordinary finite value to C in that family.

3. Linear: Normalise the Derivative First

A first-order equation linear in y has the form:

y' + P(x)y = Q(x)

The coefficient P and the right side Q contain only x, or are constants. Terms such as y^2, sin(y) and y y' prevent this form unless valid simplification removes them.

The word linear does not mean x must occur only to the first power. For instance, y' + x^4 y = cos(x) is linear in y.

A suitable integrating factor is:

IF = e^(integral P(x) dx)

Multiplication by IF turns the left side into the derivative of y IF. Integrating then gives:

y IF = integral [Q(x) IF] dx + C

The product-rule explanation and the standard procedure appear in OpenStax’s first-order linear equations lesson.

Worked Example 4: Do Not Read P Too Early

Solve:

2x y' + 6y = 8x^2,     x > 0

A rushed solution might take P = 6. But the coefficient of y' is not yet 1.

Divide every term by 2x:

y' + (3/x)y = 4x
P(x) = 3/x
Q(x) = 4x

Now:

IF = e^(integral 3/x dx)
   = e^(3 ln x)
   = x^3

Multiply the normalised equation by x^3:

x^3 y' + 3x^2 y = 4x^4
d(x^3 y)/dx = 4x^4
x^3 y = (4/5)x^5 + C

Answer: y = (4/5)x^2 + C/x^3

Check: y' = (8/5)x - 3C/x^4. Substituting in 2x y' + 6y leaves 8x^2; the terms containing C cancel.

The condition x > 0 makes the logarithm step simple. Without it, solve on an interval avoiding zero and handle the absolute value in ln|x| correctly.

4. Sometimes x Is the Dependent Variable

An equation may be awkward as an equation for y but linear when x is viewed as a function of y. The corresponding form is:

dx/dy + P(y)x = Q(y)

Now the integrating factor and every integral use dy.

Worked Example 5: Switch the Viewpoint

Solve:

(y + 2x) dy = dx

Reading the equation in terms of dx/dy gives:

dx/dy - 2x = y

This is linear in x, with P(y) = -2 and Q(y) = y. Therefore:

IF = e^(-2y)
d[x e^(-2y)]/dy = y e^(-2y)

Integrating by parts:

x e^(-2y) = -(y/2)e^(-2y)
           -(1/4)e^(-2y) + C

Answer: x = Ce^(2y) - y/2 - 1/4

Check: differentiating with respect to y gives dx/dy = 2Ce^(2y) - 1/2. The expression y + 2x gives exactly the same result.

If you express the equation as dy/dx = 1/(y + 2x), that representation requires y + 2x != 0. The x-as-a-function-of-y form avoids dividing by that expression.

5. What If More Than One Method Fits?

Worked Example 6: Three Labels, One Easy Choice

Consider:

y' = 3y/x,     x != 0

It is separable because the right side is (3/x)y. It is linear because it can be written y' - (3/x)y = 0. It is also homogeneous in the ratio sense because the right side is 3(y/x).

Separation is a short route:

dy/y = 3 dx/x
ln|y| = 3 ln|x| + C1

Answer: y = A x^3

The last form works on an interval on either side of zero. Checking y = 0 before division lets us include A = 0. Differentiation gives y' = 3Ax^2, matching 3y/x.

If asked only to identify the type, name all applicable types. If asked to solve by a particular method, use that method even when another looks quicker.

Two Meanings of Homogeneous

There is another use of the word: a homogeneous linear equation has zero on the right of its linear standard form, such as y' + 5y = 0.

That does not automatically make it homogeneous in the ratio sense. Here y' = -5y is not a function of y/x alone.

Whenever you read the word homogeneous, ask whether the question means the ratio test or zero forcing in a linear equation. The surrounding equation will usually settle it.

A Small Change Can Change the Method

Compare these three questions:

A. y' = (x + y)/x
B. y' = (x + y + 1)/x
C. y' = x + y^2

A becomes 1 + y/x, so it is homogeneous. It is also linear: y' - y/x = 1.

B becomes 1 + y/x + 1/x. The extra 1/x breaks the ratio-only pattern, but it remains linear: y' - y/x = 1 + 1/x.

C is not immediately separable, not linear in y, and not homogeneous in the ratio sense. These three methods do not directly cover every first-order equation. If a question does not fit, reread it and check your algebra rather than forcing a formula.

A Reliable Answer Layout

A clear written solution needs five things:

1. The rearranged equation and the reason your method fits.
2. The separation, substitution or integrating factor.
3. The integration, with a constant.
4. The answer in the original variables and any relevant restrictions.
5. The initial-condition substitution, if one is given.

When possible, finish with a quick differentiation check. It catches sign errors more reliably than rereading the same integration repeatedly.

For extra help with that final condition, read general versus particular solutions of differential equations.

Try These Before Looking at the Answers

For questions 1 to 4, identify a suitable method and write the first useful step. For questions 5 and 6, solve fully.

1. y' = (x + 2)(y - 4)
2. y' = (x^2 + 3y^2)/(xy)
3. 3x y' - 6y = 9x^4
4. dx/dy + x/y = y^2
5. y' = 3x^2(y + 1), y(0) = 2
6. y' + 2y/x = 5x^2, x > 0

Answers and Reasons

1. Separable: write dy/(y - 4) = (x + 2) dx, checking y = 4 separately. It is also linear after expanding the right side.

2. Homogeneous: write y' = x/y + 3y/x, then set y = vx. This gives x dv/dx = 1/v + 2v, which separates. The original equation requires both x and y to be nonzero.

3. Linear: divide by 3x to get y' - 2y/x = 3x^3, on an interval excluding x = 0. An integrating factor is x^(-2).

4. Linear in x: integrate with respect to y, on an interval where y != 0. The convenient integrating factor y gives d(xy)/dy = y^3. Thus x = y^3/4 + C/y.

5. Separable: integration gives ln|y + 1| = x^3 + C1. The initial condition selects y = 3e^(x^3) - 1. The separate constant solution y = -1 does not meet that condition.

6. Linear: IF = x^2, so d(x^2 y)/dx = 5x^4. Hence x^2 y = x^5 + C, giving y = x^3 + C/x^2.

Frequently Asked Questions

How do I identify a separable differential equation quickly?

Try to factor its right side into an x-only expression multiplied by a y-only expression. If that works, move each variable’s terms to the side containing its own differential. Check any solutions that division might exclude.

Is every equation containing y/x homogeneous?

No. The entire right side, after isolating dy/dx, must depend only on the ratio. An extra standalone x term usually breaks that pattern. For example, y' = y/x + x is linear but not homogeneous in the ratio sense.

Why is y = vx followed by y’ = v + x dv/dx?

Because v is a function of x. Differentiating the product vx requires the product rule. Treating v as a constant removes the very change you are trying to find.

How can I recognise a linear equation when it has brackets?

Expand or rearrange enough to make the coefficient of dy/dx equal to 1. Then check whether the remaining y term is exactly a function of x multiplied by y, with an x-only expression on the other side.

Can a separable equation also be linear?

Yes. For instance, y' = 2xy is separable and becomes the linear equation y' - 2xy = 0. Use whichever allowed method leads to simpler calculations.

Can I use the integrating factor before dividing by the derivative’s coefficient?

First put the equation in standard form. In a(x)y' + b(x)y = c(x), the coefficient for the integrating factor is b(x)/a(x), wherever a(x) is nonzero, rather than b(x) alone.

What happens when the equation contains dx/dy?

Treat x as the dependent variable and y as the independent variable. If the equation is linear in x, calculate e^(integral P(y) dy) and integrate with respect to y throughout.

Why should I check constant solutions before dividing?

Division by an expression such as y - 3 assumes it is nonzero. The excluded value y = 3 might itself satisfy the original equation. Check it directly so a valid solution is not silently lost.

Must I always make y the subject of my final answer?

No. A correct implicit relation is often an acceptable solution, especially when solving explicitly would make the answer less clear. Return to the original variables and state the domain restrictions that your working needs.

What is the best way to practise choosing the method?

Take a mixed set and write only the method plus the first transformed line for each equation. Then solve a few fully and differentiate your answers. This separates the skill of recognising a pattern from the skill of carrying out the integration.