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CBSE Board Examination 2024 · Section A · 1 mark

Q8: Integrating factor of (1−x2)dydx+xy=ax(1 - x^2)\frac{dy}{dx} + xy = ax

Board 2024Ch 9 · Differential Equations
The question

The integrating factor of the differential equation (1−x2)dydx+xy=ax(1 - x^2)\frac{dy}{dx} + xy = ax⁠, −1<x<1-1 < x < 1⁠, is :

(A) 1x2−1\dfrac{1}{x^2 - 1}

(B) 1x2−1\dfrac{1}{\sqrt{x^2 - 1}}

(C) 11−x2\dfrac{1}{1 - x^2}

(D) 11−x2\dfrac{1}{\sqrt{1 - x^2}}

Exam answer

Dividing by 1−x21 - x^2 (which is positive for −1<x<1-1 < x < 1⁠):

dydx+x1−x2 y=ax1−x2\frac{dy}{dx} + \frac{x}{1 - x^2}\,y = \frac{ax}{1 - x^2}
Why?

The integrating factor is read from the standard linear form, where the coefficient of dy/dxdy/dx is 11⁠:

dydx+Py=Q,I.F.=e∫P dx\frac{dy}{dx} + Py = Q,\quad \text{I.F.} = e^{\int P\,dx}

So we first divide every term by 1−x21 - x^2⁠.

This is linear with P=x1−x2P = \frac{x}{1 - x^2}⁠.

I.F.=e∫x1−x2 dx=e−12log⁡(1−x2)=(1−x2)−12=11−x2\begin{aligned} \text{I.F.} &= e^{\int \frac{x}{1 - x^2}\,dx} \\ &= e^{-\frac{1}{2}\log(1 - x^2)} \\ &= (1 - x^2)^{-\frac{1}{2}} \\ &= \frac{1}{\sqrt{1 - x^2}} \end{aligned}
Why?

The integral of PP comes from the substitution t=1−x2t = 1 - x^2⁠:

t=1−x2dtdx=0−2x=−2xdt=−2x dxx dx=−12 dt(divide by −2)∫x dx1−x2=∫1t(−12 dt)=−12∫dtt=−12log⁡t=−12log⁡(1−x2)\begin{aligned} t &= 1 - x^2 \\ \frac{dt}{dx} &= 0 - 2x = -2x \\ dt &= -2x\,dx \\ x\,dx &= -\frac{1}{2}\,dt \quad (\text{divide by } {-2}) \\[4pt] \int \frac{x\,dx}{1 - x^2} &= \int \frac{1}{t}\Big(-\frac{1}{2}\,dt\Big) \\ &= -\frac{1}{2}\int\frac{dt}{t} \\ &= -\frac{1}{2}\log t \\ &= -\frac{1}{2}\log(1 - x^2) \end{aligned}

No modulus is needed, because 1−x2>01 - x^2 > 0 for −1<x<1-1 < x < 1⁠.

Then eklog⁡u=uke^{k\log u} = u^k⁠, here with k=−1/2k = -1/2 and u=1−x2u = 1 - x^2⁠.

e−12log⁡(1−x2)=elog⁡(1−x2)−12=(1−x2)−12=1(1−x2)12=11−x2\begin{aligned} e^{-\frac12\log(1 - x^2)} &= e^{\log(1 - x^2)^{-\frac12}} \\ &= (1 - x^2)^{-\frac12} \\ &= \frac{1}{(1 - x^2)^{\frac12}} \\ &= \frac{1}{\sqrt{1 - x^2}} \end{aligned}

(First klog⁡u=log⁡ukk\log u = \log u^k⁠, then elog⁡w=we^{\log w} = w⁠. A power of −12-\frac12 means one over the square root.)

Of the other options, (B) is not real for −1<x<1-1 < x < 1⁠; (C) is the square of this I.F., and (A) is the negative of (C).

Answer: (D) 11−x2\dfrac{1}{\sqrt{1 - x^2}}

Where this comes from

Chapter 9, Differential Equations: Linear differential equations: integrating factor.

Question type: Linear equations

How to spot it: click to see

Type 7. Linear equations: the integrating factor

Spot it: dydx+Py=Q\frac{dy}{dx} + Py = Q⁠, with PP and QQ in xx only, and yy only to power 1.

Here: Divide by 1−x21 - x^2 so that dydx\frac{dy}{dx} stands alone: P=x1−x2P = \frac{x}{1 - x^2}⁠. Then ∫P dx=−12log⁡(1−x2)\int P\,dx = -\frac12\log(1 - x^2)⁠, so I.F.=11−x2\text{I.F.} = \frac{1}{\sqrt{1 - x^2}}⁠.

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