Solving separable differential equations

A separable differential equation is a first-order equation whose variables can be rearranged so that every expression involving $y$ is with $dy$ and every expression involving $x$ is with $dx$. Integration then gives a family of solution curves.

The standard form is

$$\frac{dy}{dx}=f(x)g(y).$$

Where $g(y)\ne0$, rearrange it as

$$\frac1{g(y)}\,dy=f(x)\,dx$$

and integrate both sides:

$$\boxed{\int\frac1{g(y)}\,dy=\int f(x)\,dx.}$$

This lesson is about solving the equation. For what a differential equation says about a real situation, continue to interpreting differential-equation models.

Prerequisites

You should be able to:

• integrate powers, exponentials and reciprocal functions using integration basics
• use integration by substitution
• rearrange equations involving logarithms and exponentials
• apply an initial condition to find a constant
• differentiate with the chain rule to check a solution

What does solving mean?

An equation such as

$$\frac{dy}{dx}=2xy$$

does not ask for one number. It asks for functions $y(x)$ whose gradient at each point $(x,y)$ equals $2xy$. There is usually a family of such functions, distinguished by an arbitrary constant.

An initial condition, such as $y=3$ when $x=0$, selects one member of the family. It is often written $y(0)=3$.

The method

For $\frac{dy}{dx}=f(x)g(y)$:

1. Separate the variables, putting the $y$ terms with $dy$ and the $x$ terms with $dx$.
2. Integrate both sides.
3. Write just one arbitrary constant, usually on the right.
4. Rearrange for $y$ if an explicit solution is required and possible.
5. Use any initial condition to find the constant.
6. Check by differentiating and substituting into the original equation.

The rearrangement with $dy$ and $dx$ is a compact way to perform a valid chain-rule substitution. It is not simply cancelling decorative symbols.

Worked example 1: separate and integrate

Solve

$$\frac{dy}{dx}=3x^2y.$$

For $y\ne0$, divide by $y$ and multiply by $dx$:

$$\frac1y\,dy=3x^2\,dx.$$

Integrate:

$$\begin{aligned} \int\frac1y\,dy&=\int3x^2\,dx\\ \ln|y|&=x^3+C. \end{aligned}$$

Exponentiating gives

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

Rather than carrying the positive quantity $e^C$ and a separate sign, absorb both into a new non-zero constant $A$:

$$y=Ae^{x^3}.$$

The division by $y$ excluded $y=0$, but $y=0$ also satisfies the original equation. Including it amounts to allowing $A=0$. Thus the complete family is

$$\boxed{y=Ae^{x^3}},\qquad A\in\mathbb R.$$

Check:

$$\frac{dy}{dx}=Ae^{x^3}(3x^2)=3x^2y.$$

Worked example 2: use an initial condition

Solve

$$\frac{dy}{dx}=\frac{2x}{y}, \qquad y=3\text{ when }x=2.$$

Separate the variables:

$$y\,dy=2x\,dx.$$

Integrate both sides:

$$\frac12y^2=x^2+C.$$

It is convenient to multiply through by $2$ and rename the arbitrary constant:

$$y^2=2x^2+C.$$

Apply $(x,y)=(2,3)$:

$$9=2(2^2)+C,$$

so $C=1$. Hence

$$y^2=2x^2+1.$$

Taking a square root creates two algebraic branches, but the initial condition $y=3>0$ selects the positive branch:

$$\boxed{y=\sqrt{2x^2+1}}.$$

Check by implicit differentiation:

$$2y\frac{dy}{dx}=4x \quad\Longrightarrow\quad \frac{dy}{dx}=\frac{2x}{y}.$$

Implicit and explicit solutions

An explicit solution has $y$ isolated, such as

$$y=\sqrt{2x^2+1}.$$

An implicit solution gives a relationship between $x$ and $y$, such as

$$y^2-2x^2=1.$$

Both forms can be valid. Do not force an implicit equation into an awkward or impossible explicit form unless the question asks for $y$ in terms of $x$.

Worked example 3: an implicit solution

Find the solution of

$$\frac{dy}{dx}=\frac{x^2+1}{y^2+1}$$

that passes through $(0,1)$. Give your answer implicitly.

Separate:

$$(y^2+1)\,dy=(x^2+1)\,dx.$$

Integrate:

$$\frac{y^3}{3}+y=\frac{x^3}{3}+x+C.$$

Use $(0,1)$:

$$\frac13+1=C,$$

so $C=\frac43$. Multiplying by $3$ gives the cleaner implicit answer

$$\boxed{y^3+3y=x^3+3x+4}.$$

There is no benefit in trying to solve this cubic for $y$.

Worked example 4: logarithms and algebra

Solve

$$\frac{dy}{dx}=x(1+y), \qquad y(0)=2.$$

For $y\ne-1$,

$$\frac1{1+y}\,dy=x\,dx.$$

Therefore

$$\ln|1+y|=\frac{x^2}{2}+C.$$

Using the initial condition before exponentiating is efficient:

$$\ln3=C.$$

Thus

$$\ln|1+y|=\frac{x^2}{2}+\ln3.$$

Since $1+y=3>0$ at $x=0$, the relevant branch is positive:

$$1+y=3e^{x^2/2}.$$

Hence

$$\boxed{y=3e^{x^2/2}-1}.$$

The excluded constant function $y=-1$ also solves the differential equation, but it cannot satisfy $y(0)=2$.

Worked example 5: a less obvious separation

Solve

$$(1+x^2)\frac{dy}{dx}=2xy^2, \qquad y(0)=1.$$

First isolate the derivative:

$$\frac{dy}{dx}=\frac{2x}{1+x^2}y^2.$$

For $y\ne0$, separate:

$$y^{-2}\,dy=\frac{2x}{1+x^2}\,dx.$$

Integrate. The right side uses the reverse chain rule:

$$-\frac1y=\ln(1+x^2)+C.$$

Use $x=0$, $y=1$:

$$-1=\ln1+C,$$

so $C=-1$. Therefore

$$-\frac1y=\ln(1+x^2)-1.$$

Rearranging,

$$\boxed{y=\frac1{1-\ln(1+x^2)}}.$$

This formula is valid only where its denominator is non-zero. The interval containing the initial point ends when

$$1-\ln(1+x^2)=0 \quad\Longrightarrow\quad x=\pm\sqrt{e-1}.$$

So the solution through $(0,1)$ is defined on

$$-\sqrt{e-1}<x<\sqrt{e-1}.$$

This is a useful A* detail: a solution formula and its interval of validity are not always the same thing.

Do not lose constant solutions

Dividing by an expression involving $y$ may discard solutions for which that expression is zero.

For example, consider

$$\frac{dy}{dx}=x(y-2)(y+1).$$

Separating by division gives

$$\frac1{(y-2)(y+1)}\,dy=x\,dx,$$

but this assumes $(y-2)(y+1)\ne0$. Before dividing, test its zeros in the original equation:

$$\boxed{y=2\quad\text{and}\quad y=-1}$$

are constant solutions because each makes both sides zero. They must be recorded separately if the complete solution set is required.

The same warning applies when cancelling a factor. Always ask: what values made the factor zero?

Common misconceptions

Integrating $\frac{dy}{dx}$ term by term

From

$$\frac{dy}{dx}=xy,$$

it is wrong to write $y=\frac12x^2\cdot\frac12y^2+C$. The variables must first be separated:

$$\frac1y\,dy=x\,dx.$$

Adding two independent constants

Writing

$$F(y)+C_1=G(x)+C_2$$

is not incorrect, but only the difference $C_2-C_1$ matters. Replace it with one constant:

$$F(y)=G(x)+C.$$

Forgetting the absolute value

The integral is

$$\int\frac1y\,dy=\ln|y|+C,$$

not automatically $\ln y+C$. An initial condition may later determine whether the relevant branch is positive or negative.

Choosing both square-root branches

If $y^2=h(x)$, then $y=\pm\sqrt{h(x)}$ describes two branches. A non-zero initial value selects one branch. Do not leave $\pm$ when the condition determines the sign.

Applying the condition too early

First integrate to obtain the family, then substitute the given $x$ and $y$ values. Never put the initial values into the differential equation and expect to determine the integration constant, because the constant has not appeared yet.

Self-check

1. Recognise separable equations

Which equations are separable?

$$\text{(a) }\frac{dy}{dx}=x^2(1-y), \qquad \text{(b) }\frac{dy}{dx}=x+y, \qquad \text{(c) }y\frac{dy}{dx}=\cos x.$$

Answer

(a) and (c) are separable:

$$\frac1{1-y}\,dy=x^2\,dx, \qquad y\,dy=\cos x\,dx.$$

Equation (b) is not separable in its given form because $x+y$ cannot be written as a product of a function of $x$ and a function of $y$.

2. Find a general solution

Solve

$$\frac{dy}{dx}=4xy.$$

Answer

For $y\ne0$,

$$\frac1y\,dy=4x\,dx.$$

Hence

$$\ln|y|=2x^2+C$$

and

$$\boxed{y=Ae^{2x^2}}.$$

Allowing $A=0$ includes the constant solution $y=0$.

3. Apply an initial condition

Given

$$\frac{dy}{dx}=\frac{x}{y^2}, \qquad y(0)=2,$$

find $y$ explicitly.

Answer

Separate and integrate:

$$y^2\,dy=x\,dx \quad\Longrightarrow\quad \frac{y^3}{3}=\frac{x^2}{2}+C.$$

Using $y(0)=2$ gives $C=\frac83$. Therefore

$$2y^3=3x^2+16,$$

so

$$\boxed{y=\sqrt[3]{\frac{3x^2+16}{2}}}.$$

4. Spot the missing solutions

Which solutions could be lost when separating

$$\frac{dy}{dx}=e^x y(y-5)?$$

Answer

Division by $y(y-5)$ excludes $y=0$ and $y=5$. Both are constant solutions of the original equation:

$$\boxed{y=0\quad\text{and}\quad y=5}.$$

5. Verify a solution

Show that

$$y=\frac{2}{3-x^2}$$

satisfies $\frac{dy}{dx}=xy^2$ wherever it is defined.

Answer

Differentiate using the chain rule:

$$\frac{dy}{dx} =2(-1)(3-x^2)^{-2}(-2x) =\frac{4x}{(3-x^2)^2}.$$

Also,

$$xy^2=x\left(\frac2{3-x^2}\right)^2 =\frac{4x}{(3-x^2)^2}.$$

The expressions agree. The solution is not defined at $x=\pm\sqrt3$.

Exam checklist

• Confirm that the equation can be written as $\frac{dy}{dx}=f(x)g(y)$.
• Separate completely before integrating.
• Use one arbitrary constant.
• Check values excluded by division for constant solutions.
• Apply the initial condition to the integrated equation.
• Select the correct branch after taking roots or exponentiating.
• Give an implicit answer unless an explicit form is requested.
• Differentiate your result to check it against the original equation.

Next steps

Use interpreting differential-equation models to connect solution families and initial conditions to growth, decay and other real processes. Revisit integration by substitution if either side of a separated equation is difficult to integrate, and use checking answers to strengthen substitution, differentiation and domain checks.