The Power of Substitutions

Substitutions are inevitable in useful mathematics, since they shift our perspective from challenging problems to trivial ones.

Problem 1. Let $H$ be a continuous function such that for any $x,y$ , $H(x,y) = H(1, y/x)$ . Prove that the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = H(x,y)$

can be solved using the method of separable variables. Obviously, we assume $x \neq 0$ . This is known as a homogeneous differential equation.

(Click for Solution)

Solution. Make the substitution $y = ux$ . On the right-hand side,

$H(x,y) = H(1,y/x) = H(1,u).$

On the left-hand side, the product rule yields

$\displaystyle \begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx}(ux) \\ & = x \cdot \frac{\mathrm du}{\mathrm dx} + u \cdot 1 \\ & = x \cdot \frac{\mathrm du}{\mathrm dx} + u. \end{aligned}$

Then the original differential equation simplifies to

$\displaystyle \begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= H(x,y)\\ x \cdot \frac{\mathrm du}{\mathrm dx} + u &= H(1, u) \\ \frac{\mathrm du}{\mathrm dx} &= \frac 1x \cdot (H(1, u) - u), \end{aligned}$

which can be solved by the method of separable variables. In fact,

$\begin{aligned} \int \frac{1}{H(1,u) - u}\, \mathrm du &= \int \frac 1x\, \mathrm dx = \ln|x| + C. \end{aligned}$

Problem 2. Let $P,Q$ be continuous functions and $n$ be a real number. Prove that the Bernoulli ordinary differential equation given by

$\displaystyle \frac{\mathrm dy}{\mathrm dx} + P(x) y = Q(x) y^n$

can be solved using existing techniques.

(Click for Solution)

Solution. Dividing by $y^n$ on both sides,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} \cdot y^{-n} + P(x) \cdot y^{1-n} = Q(x).$

This motivates the substitution $u = y^{1-n}$ , with hopes that the term $\displaystyle \frac{\mathrm dy}{\mathrm dx} \cdot y^{-n}$ simplifies to $\displaystyle \frac{\mathrm du}{\mathrm dx}$ . Indeed, differentiating,

$\displaystyle \begin{aligned} \frac{\mathrm du}{\mathrm dx} &= (1-n) \cdot y^{(1-n) - 1} \cdot \frac{\mathrm dy}{dx} \\ \frac{\mathrm dy}{dx} \cdot y^{-n} &= \frac 1{1-n} \cdot \frac{\mathrm du}{\mathrm dx}\end{aligned}$

Thus, making the substitutions,

$\displaystyle \begin{aligned} \frac 1{1-n} \cdot \frac{\mathrm du}{\mathrm dx} + P(x) u &= Q(x) \\ \frac{\mathrm du}{\mathrm dx} + (1-n)P(x) u &= (1-n)Q(x).\end{aligned}$

Thus, the equation can be solved via the method of integrating factors. In fact, using the integrating factor

$\mu(x) = e^{(1-n)\int P(x)\, \mathrm dx},$

the general solution to the Bernoulli ordinary differential equation is given by

$\displaystyle y = \frac{(1-n) x}{e^{\int (1-n)P(x)\, \mathrm dx}} \int Q(x)e^{(1-n)\int P(x)\, \mathrm dx}\, \mathrm dx.$

after back-substitution.

—Joel Kindiak, 28 Jan 25, 1727H

KindiakMath

The Power of Substitutions

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The Power of Substitutions

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