Substitutions and Bernoulli equations

What you will be able to do

• turn $\dfrac{dy}{dx} = f(ax + by + c)$ into a separable equation with one substitution;
• solve $\dfrac{dy}{dx} = \dfrac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$ in both cases, with parallel lines and with crossing lines;
• recognise a Bernoulli equation and reduce it to a linear equation.

When the equation is not standard

In the free courses you met three types: separable, homogeneous and linear. Many equations are none of these. The idea of this lesson is simple. Replace $y$ by a new unknown $v$ so that the new equation is one of the three types. Solve it, and then put $y$ back.

A good substitution comes from looking at the equation. If the same combination appears again and again, call it $v$.

The equation $\dfrac{dy}{dx} = f(ax + by + c)$

Here $y$ appears only inside one linear expression. Put $v = ax + by + c$. Differentiate: $\dfrac{dv}{dx} = a + b\dfrac{dy}{dx} = a + b\,f(v)$. This equation has only $v$ and $x$, and the variables separate.

The left graph shows the solutions of the first example below.

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

1. Let $v = x + y + 1$. Then $\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx} = 1 + v^2$.
2. Separate: $\dfrac{dv}{1 + v^2} = dx$, so $\tan^{-1}v = x + C$.
3. Return to $y$: $x + y + 1 = \tan(x + C)$, so $y = \tan(x + C) - x - 1$.

Example 2. Solve $\dfrac{dy}{dx} = \dfrac{1}{x + y}$.

1. Let $v = x + y$. Then $\dfrac{dv}{dx} = 1 + \dfrac{1}{v} = \dfrac{v + 1}{v}$.
2. Separate: $\dfrac{v}{v + 1}\,dv = dx$. Write $\dfrac{v}{v + 1} = 1 - \dfrac{1}{v + 1}$, so $v - \ln|v + 1| = x + C$.
3. Return to $y$: $x + y - \ln|x + y + 1| = x + C$, which simplifies to $y - \ln|x + y + 1| = C$.

A fraction of two linear expressions

Now take $\dfrac{dy}{dx} = \dfrac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$. Each of $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is a straight line. There are two cases.

The lines are parallel. Then $a_1b_2 = a_2b_1$, and the top is a multiple of $a_2x + b_2y$ plus a constant. Put $v = a_2x + b_2y$ (or $a_1x + b_1y$). The equation becomes the type $f(ax + by + c)$ above.

Example 3. Solve $\dfrac{dy}{dx} = \dfrac{x + y + 1}{2x + 2y + 3}$.

1. The lines are parallel. Let $v = x + y$. Then $\dfrac{dv}{dx} = 1 + \dfrac{v + 1}{2v + 3} = \dfrac{3v + 4}{2v + 3}$.
2. Separate: $\dfrac{2v + 3}{3v + 4}\,dv = dx$. Write $\dfrac{2v + 3}{3v + 4} = \dfrac{2}{3} + \dfrac{1}{3(3v + 4)}$. Integrating gives $\dfrac{2v}{3} + \dfrac{1}{9}\ln|3v + 4| = x + C$.
3. Return to $y$ and multiply by $9$: $6(x + y) + \ln|3x + 3y + 4| = 9x + C$, so $6y - 3x + \ln|3x + 3y + 4| = C$.

The lines cross at a point $(h, k)$. Shift the origin to the crossing point: put $x = X + h$ and $y = Y + k$. The constants $c_1$ and $c_2$ disappear, and the equation becomes $\dfrac{dY}{dX} = \dfrac{a_1X + b_1Y}{a_2X + b_2Y}$. This is homogeneous. Use $Y = VX$ from the free course.

Example 4. Solve $\dfrac{dy}{dx} = \dfrac{x + y - 3}{x - y - 1}$.

1. The lines $x + y = 3$ and $x - y = 1$ cross at $(2, 1)$. Put $x = X + 2$ and $y = Y + 1$. The equation is $\dfrac{dY}{dX} = \dfrac{X + Y}{X - Y}$.
2. Let $Y = VX$. Then $V + X\dfrac{dV}{dX} = \dfrac{1 + V}{1 - V}$, so $X\dfrac{dV}{dX} = \dfrac{1 + V^2}{1 - V}$.
3. Separate: $\dfrac{1 - V}{1 + V^2}\,dV = \dfrac{dX}{X}$. Integrating gives $\tan^{-1}V - \tfrac{1}{2}\ln(1 + V^2) = \ln|X| + C$.
4. Return to $x$ and $y$: $\tan^{-1}\dfrac{y - 1}{x - 2} = \tfrac{1}{2}\ln\left[(x - 2)^2 + (y - 1)^2\right] + C$.

In polar form about the point $(2, 1)$ this says $\theta = \ln r + C$. The solution is a spiral, and the right graph of the picture shows it.

Bernoulli equations

A Bernoulli equation has the form

where $n$ is a number other than $0$ and $1$. (For $n = 0$ it is linear already, and for $n = 1$ it is separable.) The substitution $v = y^{1-n}$ makes it linear.

1. Divide by $y^n$: $y^{-n}\dfrac{dy}{dx} + P\,y^{1-n} = Q$.
2. Put $v = y^{1-n}$. Then $\dfrac{dv}{dx} = (1 - n)\,y^{-n}\dfrac{dy}{dx}$.
3. The equation becomes $\dfrac{dv}{dx} + (1 - n)P\,v = (1 - n)Q$. This is linear in $v$, and the integrating factor $e^{\int (1 - n)P\,dx}$ solves it.

Example 5. Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = xy^2$ for $x > 0$.

1. Here $n = 2$, so $v = y^{-1}$ and $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$.
2. Divide by $y^2$: $y^{-2}\dfrac{dy}{dx} + \dfrac{1}{x}y^{-1} = x$. This is $-\dfrac{dv}{dx} + \dfrac{v}{x} = x$, or $\dfrac{dv}{dx} - \dfrac{v}{x} = -x$.
3. The integrating factor is $e^{-\ln x} = \dfrac{1}{x}$. So $\dfrac{d}{dx}\left(\dfrac{v}{x}\right) = -1$, which gives $\dfrac{v}{x} = -x + C$.
4. Return to $y$: $v = Cx - x^2$, so $y = \dfrac{1}{Cx - x^2}$.

Example 6: $x$ as the unknown. Solve $\dfrac{dy}{dx} = \dfrac{y}{x - x^2y}$. Turn the equation over: $\dfrac{dx}{dy} = \dfrac{x}{y} - x^2$. Now $x$ is the unknown and $y$ is the variable, and this is a Bernoulli equation with $n = 2$.

1. Let $v = \dfrac{1}{x}$. Then $\dfrac{dv}{dy} = -\dfrac{1}{x^2}\dfrac{dx}{dy} = -\dfrac{v}{y} + 1$.
2. So $\dfrac{dv}{dy} + \dfrac{v}{y} = 1$. The integrating factor is $y$, and $\dfrac{d}{dy}(yv) = y$.
3. Then $yv = \dfrac{y^2}{2} + C$, so $\dfrac{y}{x} = \dfrac{y^2}{2} + C$ and $x = \dfrac{2y}{y^2 + 2C}$.

Other useful substitutions

If $xy$ appears in the equation, try $v = xy$. Then $\dfrac{dv}{dx} = y + x\dfrac{dy}{dx}$. If $\dfrac{y}{x^n}$ appears, try $v = \dfrac{y}{x^n}$. The test for a good substitution is always the same: after it, the new equation must be separable, homogeneous or linear.

What the syllabi ask

JEE Main and JEE Advanced expect you to find the substitution yourself. A-Level Further Maths and IB Higher Level usually give the substitution in the question ("show that the substitution $z = y^{-1}$ reduces the equation to..."). The method is the same.

Practise with new numbers

Try it in Python

The program checks that $y = \dfrac{1}{Cx - x^2}$ solves the Bernoulli equation of Example 5 for two values of $C$. Then it checks the linear equation for $v = \dfrac{1}{y}$.

Common mistakes

• Forgetting the new letter. After $v = x + y$ the equation must contain only $v$ and $x$. If a $y$ is left, the substitution is not finished.
• Treating $\dfrac{dv}{dx}$ as $\dfrac{dy}{dx}$. For $v = ax + by + c$, $\dfrac{dv}{dx} = a + b\dfrac{dy}{dx}$. The $a$ is easy to lose.
• Shifting the origin for parallel lines. Parallel lines never cross, so there is no point to shift to. Use $v = a_2x + b_2y$.
• Wrong power in a Bernoulli equation. The substitution is $v = y^{1-n}$. For $n = 2$ that is $y^{-1}$.
• Forgetting the factor $(1 - n)$. It multiplies both $P$ and $Q$ in the linear equation.

Remember these

• $\dfrac{dy}{dx} = f(ax + by + c)$: put $v = ax + by + c$, then $\dfrac{dv}{dx} = a + b\,f(v)$.
• Fraction of two lines, parallel: put $v = a_2x + b_2y$. Crossing: shift the origin to the crossing point, then put $Y = VX$.
• Bernoulli: $\dfrac{dy}{dx} + Py = Qy^n$. Put $v = y^{1-n}$ and get $\dfrac{dv}{dx} + (1 - n)Pv = (1 - n)Q$.
• If $x$ is easier as the unknown, turn the equation over and work with $\dfrac{dx}{dy}$.

Check yourself

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

1. Solve $\dfrac{dy}{dx} = \cos(x + y)$.

2. The equation $\dfrac{dy}{dx} + \dfrac{2y}{x} = \dfrac{y^3}{x^2}$ is a Bernoulli equation. Write the linear equation for $v = y^{-2}$.

3. Solve $\dfrac{dy}{dx} + y = xy^3$.

4. For $\dfrac{dy}{dx} = \dfrac{2x + y - 5}{x - y - 1}$, find the point to which the origin must be shifted, and write the equation in $X$ and $Y$.

5. Can you solve $\dfrac{dy}{dx} = \dfrac{4x + 2y + 1}{2x + y + 3}$ by shifting the origin? What do you do instead?

References

• OpenStax, Calculus Volume 2, Chapter 4, sections 4.3 Separable Differential Equations and 4.5 First-order Linear Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on homogeneous and linear differential equations.