KindiakMath

Un-abusing the Product Rule

Problem 1. Given f, solve the differential equation

\displaystyle \frac{\mathrm d}{\mathrm dx}(y \cdot f(x)) = \frac{\mathrm dy}{\mathrm dx} \cdot f'(x).

(Click for Solution)

Solution. Using the product rule correctly to expand the left-hand side,

\displaystyle \frac{\mathrm d}{\mathrm dx}(y \cdot f(x)) = \frac{\mathrm dy}{\mathrm dx} \cdot f(x) + y \cdot f'(x).

Therefore,

\displaystyle \frac{\mathrm dy}{\mathrm dx} \cdot f(x) + y \cdot f'(x) = \frac{\mathrm dy}{\mathrm dx} \cdot f'(x).

Making \frac{\mathrm dy}{\mathrm dx} the subject,

\begin{aligned} \frac 1y \cdot \frac{\mathrm dy}{\mathrm dx} = \frac{ f'(x) }{ f'(x) - f(x) }. \end{aligned}

Therefore, integrating with respect to x on both sides,

\displaystyle \ln|y| = \int \frac{f'(x)}{f(x) - f'(x)}\, \mathrm dx.

Problem 2. Deduce the solution to the differential equation

\displaystyle \frac{\mathrm d}{\mathrm dx}((y^2+1) \cdot \sin(x)) = 2y \cdot \frac{\mathrm dy}{\mathrm dx} \cdot \cos(x).

(Click for Solution)

Solution. Firstly, since \frac{\mathrm d}{\mathrm dx}(y^2 + 1) = 2y \cdot \frac{\mathrm dy}{\mathrm dx}, setting f(x) = \sin(x) in Problem 1 yields the solution

\begin{aligned}\ln(y^2 + 1) &= \int \frac{\cos(x)}{ \cos(x) - \sin(x) }\, \mathrm dx. \end{aligned}

Use trigonometric identities to simplify the integrand as follows:

\begin{aligned} \frac{ \cos(x) }{ \cos(x) - \sin(x) } &= \frac{ \cos(x) }{ \cos(x) - \sin(x) } \cdot \frac{ \cos(x) + \sin(x) }{ \cos(x) + \sin(x) } \\ &= \frac{ \cos^2(x) + \sin(x) \cos(x) }{ \cos^2(x) - \sin^2(x) } \\ &= \frac{ \frac 12 (\cos(2x)+1) + \frac 12 \sin(2x) }{ \cos(2x) } \\ &= \frac 12 \cdot \left( 1 + 2\sec(2x) + \tan(2x) \right). \end{aligned}

Therefore, integrating the right-hand side,

\begin{aligned} \int \frac{ \cos(x) }{ \cos(x) - \sin(x) } \, \mathrm dx &= \frac 12 \cdot \int \left( 1 + 2\sec(2x) + \tan(2x) \right)\, \mathrm dx \\ &= \frac 12 \left( x + \ln|{\sec(2x) + \tan(2x)}| + \frac 12 \ln |{ \sec(2x) }| \right) + C \\ &= \frac 12 x + \frac 12 \ln|{\sec(2x) + \tan(2x)}| + \frac 14 \ln |{ \sec(2x) }| + C. \end{aligned}

Therefore, the required solution is

\displaystyle \ln(y^2 + 1) = \frac 12 x + \frac 12 \ln|{\sec(2x) + \tan(2x)}| + \frac 14 \ln |{ \sec(2x) }| + C.

—Joel Kindiak, 1 Sept 2025, 1645H

,

Published by

joelkindiak

Leave a comment