Variables separable

What you will be able to do

• recognise a separable equation and pull its two variables apart;
• integrate both sides, add one constant, and fix it from a point on the curve;
• find the constant solutions that the division loses.

Pulling the letters apart

A bank pays interest on the money that is in the account. The more money there is, the faster it grows. If $y$ is the amount and $x$ is the time, the rule is $\dfrac{dy}{dx} = ky$. The rate depends on $y$ and on nothing else. Can you find $y$?

Move every $y$ to one side and every $x$ to the other: $\dfrac{dy}{y} = k\,dx$. Now each side can be integrated by itself. This is the whole idea of this lesson.

An equation of the form

is called separable. NCERT calls it a differential equation of the "variable separable" type. A-Level, IB and AP texts say "separable". The two names mean the same thing.

The steps

1. Check. The right side must be a product of a function of $x$ alone and a function of $y$ alone.
2. Separate. Write $\dfrac{dy}{g(y)} = f(x)\,dx$. Here we divide by $g(y)$, so we need $g(y) \ne 0$.
3. Integrate both sides. $\displaystyle\int \dfrac{dy}{g(y)} = \int f(x)\,dx + C$. One constant is enough.
4. Use the condition. A point $(x_0, y_0)$ on the curve fixes $C$.
5. Look back at the division. If $g(y) = 0$ has a solution $y = y_1$, then the constant function $y = y_1$ also satisfies the equation. Check whether the general solution already contains it.

A convenient form of the constant

Take $\dfrac{dy}{dx} = 4xy$. Then $\dfrac{dy}{y} = 4x\,dx$, so $\ln|y| = 2x^2 + c$. The constant $c$ is any real number. Raise $e$ to both sides: $|y| = e^{c}e^{2x^2}$. Write $\pm e^{c}$ as one letter $C$. Then

Here $C$ is any non-zero number. The value $C = 0$ gives $y = 0$, which is also a solution (it is the lost one, because $g(y) = y$ is zero there). So $C$ may be any real number.

You may also write the constant as $\ln C$ from the start. Then $\ln|y| = 2x^2 + \ln C$ becomes $|y| = C e^{2x^2}$ with $C > 0$. Both ways are correct.

Worked examples

Example 1. Find the curve with $\dfrac{dy}{dx} = 4xy$ through $(0, 3)$.

1. The general solution is $y = Ce^{2x^2}$.
2. Put in the point: $3 = Ce^{0} = C$.
3. The curve is $y = 3e^{2x^2}$.

Example 2. Solve $(1 + x^2)\dfrac{dy}{dx} = 1 + y^2$ with $y = 1$ at $x = 0$.

1. Separate: $\dfrac{dy}{1 + y^2} = \dfrac{dx}{1 + x^2}$.
2. Integrate: $\tan^{-1}y = \tan^{-1}x + c$.
3. At $(0, 1)$: $\dfrac{\pi}{4} = 0 + c$. So $\tan^{-1}y = \tan^{-1}x + \dfrac{\pi}{4}$.
4. Take the tangent of both sides and use $\tan(A + B)$ with $\tan\dfrac{\pi}{4} = 1$: $y = \dfrac{x + 1}{1 - x\cdot 1}$. So $y = \dfrac{1 + x}{1 - x}$ for $x < 1$.

Example 3. Solve $\dfrac{dy}{dx} = e^{x + y}$ with $y = 0$ at $x = 0$.

1. Write $e^{x + y} = e^x e^y$ and separate: $e^{-y}\,dy = e^x\,dx$.
2. Integrate: $-e^{-y} = e^x + c$, so $e^x + e^{-y} = C$.
3. At $(0, 0)$: $1 + 1 = C = 2$. So $e^{-y} = 2 - e^x$ and $y = -\ln(2 - e^x)$ for $x < \ln 2$.

When division loses a solution

Take $\dfrac{dy}{dx} = x(1 - y^2)$. Here $g(y) = 1 - y^2$ is zero at $y = 1$ and $y = -1$. Both constant functions satisfy the equation, because their slope is $0$ and so is the right side.

Now divide where $g(y) \ne 0$. The integral of $\dfrac{dy}{1 - y^2}$ is $\tfrac{1}{2}\ln\left|\dfrac{1 + y}{1 - y}\right|$. So

For $|y| < 1$ this gives $y = \tanh\left(\dfrac{x^2}{2} + c\right)$. For $|y| > 1$ it gives $y = \coth\left(\dfrac{x^2}{2} + c\right)$. No choice of $c$ gives $y = 1$ exactly, so the two constant solutions are extra.

The curves never touch the lines $y = \pm 1$. This is not an accident. Two different solutions of this equation do not cross.

Practise with new numbers

Try it in Python

The program takes $\dfrac{dy}{dx} = xy$ with $y = 1$ at $x = 0$. Separation gives $\ln y = \dfrac{x^2}{2}$. The program checks the answer by its slope, and by the two sides of the separated equation.

Common mistakes

• Writing $\int \dfrac{dy}{y} = \dfrac{y^2}{2}$ or $\int y\,dx$ on the left. The $y$ side must have $dy$, and the $x$ side must have $dx$.
• Adding a constant on each side. One constant on one side is enough.
• Losing a constant solution. After dividing by $g(y)$, ask what happens when $g(y) = 0$.
• Dropping the absolute value in $\ln|y|$ before the constant is fixed. Fix the sign by the point, not by habit.
• Trying to separate a sum. In $\dfrac{dy}{dx} = x + y$ the right side is not a product, so this method does not apply. You need the linear method of the third lesson.

Remember these

• $\dfrac{dy}{dx} = f(x)g(y)$ separates into $\dfrac{dy}{g(y)} = f(x)\,dx$.
• Integrate both sides, with one constant, and use the point to fix it.
• The zeros of $g$ give constant solutions. Check whether the general solution has them.
• NCERT says "variable separable". A-Level, IB and AP say "separable".

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$ for $x > 0$ and $y > 0$.

2. Solve $e^{y}\,\dfrac{dy}{dx} = 1$.

3. Solve $\dfrac{dy}{dx} = x^2(y + 3)$.

4. Find the constant solutions of $\dfrac{dy}{dx} = (y - 2)(y + 1)$. What is lost if you divide by $(y - 2)(y + 1)$?

5. Solve $\dfrac{dy}{dx} = y\sin x$ with $y = 1$ at $x = 0$.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.3 Separable Differential Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the section on differential equations with variables separable.