A removable extension of the differential equation · source reference

Printed question

(i) Find the general solution of the differential equation $\left(\frac{x^2+1}{x}\right)\frac{dy}{dx}=2$. [2]

(ii) Sketch the particular solution of the differential equation in (i) for which $y=0$ and $x=0$. [2]

(iii) Describe the gradient of every solution curve as $x\to+\infty$. [1]

[2010/MI/Prelim/I/11(b) (modified)]

Printed working

$(x^2+1)y'/x=2\Rightarrow y'=2x/(x^2+1)$.

General solution: $y=\ln(x^2+1)+c$. When $x=0,y=0$, $c=0$. Particular solution: $y=\ln(x^2+1)$.

As $x\to+\infty$, $y'=2x/(x^2+1)\to0$, i.e. the gradient tends towards zero.

Printed answer key

(i) $y=\ln(x^2+1)+c$. (ii) The native sketch is given below. (iii) The gradient tends towards zero as $x\to+\infty$.

Teacher note 1

Given DE of form $dy/dx=f(x)$. Integrate with respect to $x$: $y=\int f(x)dx$.

Teacher note 2

Note: gradient at $x=0$ is $dy/dx=0/(0+1)=0$.

Diagram 10-01-solution

Mathematical review

(i) For $x\ne0$, $y'=2x/(1+x^2)$, so $y=\ln(1+x^2)+C$ on an interval excluding zero.

(ii) The intended extension through $(0,0)$ sets $C=0$. It is even, nonnegative, with a horizontal tangent and minimum at the origin, and increases for positive x. Inflections are at $x=\pm1$, since $y''=2(1-x^2)/(1+x^2)^2$.

(iii) Every member has $y'=2x/(1+x^2)\to0$ as $x\to+\infty$.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 1. Working page(s): 1. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.