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Differential Equations question

2025 · 29 Jan · Shift 1 · Q36
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  5. /2025 · 29 Jan · Shift 1 · Q36

Differential Equations question

2025 · 29 Jan · Shift 1 · Q36

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation : cos⁡x(log⁡e(cos⁡x))2dy+(sin⁡x−3ysin⁡xlog⁡e(cos⁡x))dx=0\cos x\left(\log _e(\cos x)\right)^2 d y+\left(\sin x-3 y \sin x \log _e(\cos x)\right) d x=0cosx(loge​(cosx))2dy+(sinx−3ysinxloge​(cosx))dx=0, x ∈ (0, π2\frac{\pi}{2}2π​). If y(π4)y(\frac{\pi}{4})y(4π​)=−1log⁡e2-\frac{1}{\log_{e}2}−loge​21​, then y(π6)y(\frac{\pi}{6})y(6π​) is equal to :
  1. A
    2log⁡e(3)−log⁡e(4)\frac{2}{\log_{e}(3)−\log_{e}(4)}loge​(3)−loge​(4)2​
  2. B
    −1log⁡e(4)-\frac{1}{\log_{e}(4)}−loge​(4)1​
  3. C
    1log⁡e(4)−log⁡e(3)\frac{1}{\log_{e}(4)−\log_{e}(3)}loge​(4)−loge​(3)1​
  4. D
    1log⁡e(3)−log⁡e(4)\frac{1}{\log_{e}(3)−\log_{e}(4)}loge​(3)−loge​(4)1​
View written solutionFree

Correct answer: D

  1. Rewrite the differential equation in standard form

Given

cos⁡x (ln⁡(cos⁡x))2 dy+(sin⁡x−3ysin⁡xln⁡(cos⁡x))dx=0.\cos x\,(\ln(\cos x))^2\,dy+\left(\sin x-3y\sin x\ln(\cos x)\right)dx=0.cosx(ln(cosx))2dy+(sinx−3ysinxln(cosx))dx=0.

Divide throughout by dxdxdx:

cos⁡x (ln⁡(cos⁡x))2 dydx+sin⁡x−3ysin⁡xln⁡(cos⁡x)=0.\cos x\,(\ln(\cos x))^2\,\frac{dy}{dx}+\sin x-3y\sin x\ln(\cos x)=0.cosx(ln(cosx))2dxdy​+sinx−3ysinxln(cosx)=0.

So,

dydx+sin⁡x−3ysin⁡xln⁡(cos⁡x)cos⁡x(ln⁡(cos⁡x))2=0.\frac{dy}{dx}+\frac{\sin x-3y\sin x\ln(\cos x)}{\cos x(\ln(\cos x))^2}=0.dxdy​+cosx(ln(cosx))2sinx−3ysinxln(cosx)​=0.

Simplify:

dydx+tan⁡x(ln⁡(cos⁡x))2−3ytan⁡xln⁡(cos⁡x)=0.\frac{dy}{dx}+\frac{\tan x}{(\ln(\cos x))^2}-\frac{3y\tan x}{\ln(\cos x)}=0.dxdy​+(ln(cosx))2tanx​−ln(cosx)3ytanx​=0.

Hence,

dydx−3tan⁡xln⁡(cos⁡x)y=−tan⁡x(ln⁡(cos⁡x))2.\frac{dy}{dx}-\frac{3\tan x}{\ln(\cos x)}y=-\frac{\tan x}{(\ln(\cos x))^2}.dxdy​−ln(cosx)3tanx​y=−(ln(cosx))2tanx​.

This is a linear differential equation:

dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x)dxdy​+P(x)y=Q(x)

with

P(x)=−3tan⁡xln⁡(cos⁡x),Q(x)=−tan⁡x(ln⁡(cos⁡x))2.P(x)=-\frac{3\tan x}{\ln(\cos x)},\qquad Q(x)=-\frac{\tan x}{(\ln(\cos x))^2}.P(x)=−ln(cosx)3tanx​,Q(x)=−(ln(cosx))2tanx​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x)dx=e∫−3tan⁡xln⁡(cos⁡x)dx.\text{I.F.}=e^{\int P(x)dx}=e^{\int -\frac{3\tan x}{\ln(\cos x)}dx}.I.F.=e∫P(x)dx=e∫−ln(cosx)3tanx​dx.

Let

t=ln⁡(cos⁡x).t=\ln(\cos x).t=ln(cosx).

Then

dtdx=−tan⁡x⇒dt=−tan⁡x dx.\frac{dt}{dx}=-\tan x \quad\Rightarrow\quad dt=-\tan x\,dx.dxdt​=−tanx⇒dt=−tanxdx.

Therefore,

∫−3tan⁡xln⁡(cos⁡x)dx=∫3tdt=3ln⁡∣t∣.\int -\frac{3\tan x}{\ln(\cos x)}dx=\int \frac{3}{t}dt=3\ln|t|.∫−ln(cosx)3tanx​dx=∫t3​dt=3ln∣t∣.

So,

I.F.=e3ln⁡∣ln⁡(cos⁡x)∣=(ln⁡(cos⁡x))3.\text{I.F.}=e^{3\ln|\ln(\cos x)|}=(\ln(\cos x))^3.I.F.=e3ln∣ln(cosx)∣=(ln(cosx))3.

(Here x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2), so cos⁡x∈(0,1)\cos x\in(0,1)cosx∈(0,1) and ln⁡(cos⁡x)<0\ln(\cos x)<0ln(cosx)<0, but the power 333 works consistently.)

  1. Solve the equation

Multiplying the differential equation by (ln⁡(cos⁡x))3(\ln(\cos x))^3(ln(cosx))3,

(ln⁡(cos⁡x))3dydx−3tan⁡x(ln⁡(cos⁡x))2y=−tan⁡xln⁡(cos⁡x).(\ln(\cos x))^3\frac{dy}{dx}-3\tan x(\ln(\cos x))^2y=-\tan x\ln(\cos x).(ln(cosx))3dxdy​−3tanx(ln(cosx))2y=−tanxln(cosx).

The left side is

ddx[y(ln⁡(cos⁡x))3],\frac{d}{dx}\left[y(\ln(\cos x))^3\right],dxd​[y(ln(cosx))3],

because

ddx(ln⁡(cos⁡x))3=3(ln⁡(cos⁡x))2⋅(−tan⁡x).\frac{d}{dx}(\ln(\cos x))^3=3(\ln(\cos x))^2\cdot(-\tan x).dxd​(ln(cosx))3=3(ln(cosx))2⋅(−tanx).

Hence,

ddx[y(ln⁡(cos⁡x))3]=−tan⁡xln⁡(cos⁡x).\frac{d}{dx}\left[y(\ln(\cos x))^3\right]=-\tan x\ln(\cos x).dxd​[y(ln(cosx))3]=−tanxln(cosx).

Integrate both sides:

y(ln⁡(cos⁡x))3=∫−tan⁡xln⁡(cos⁡x) dx+C.y(\ln(\cos x))^3=\int -\tan x\ln(\cos x)\,dx+C.y(ln(cosx))3=∫−tanxln(cosx)dx+C.

Again use t=ln⁡(cos⁡x)t=\ln(\cos x)t=ln(cosx), dt=−tan⁡x dxdt=-\tan x\,dxdt=−tanxdx:

\int -\tan x\ln(\cos x)\,dx=\int t\,dt=\frac{t^2}{2}+C= rac{(\ln(\cos x))^2}{2}+C.

Thus,

y(ln⁡(cos⁡x))3=(ln⁡(cos⁡x))22+C.y(\ln(\cos x))^3=\frac{(\ln(\cos x))^2}{2}+C.y(ln(cosx))3=2(ln(cosx))2​+C.

So,

y=12ln⁡(cos⁡x)+C(ln⁡(cos⁡x))3.y=\frac{1}{2\ln(\cos x)}+\frac{C}{(\ln(\cos x))^3}.y=2ln(cosx)1​+(ln(cosx))3C​.
  1. Use the initial condition

Given

y(π4)=−1ln⁡2.y\left(\frac{\pi}{4}\right)=-\frac{1}{\ln 2}.y(4π​)=−ln21​.

Now,

cos⁡π4=12\cos\frac{\pi}{4}=\frac{1}{\sqrt2}cos4π​=2​1​

so

ln⁡(cos⁡π4)=ln⁡(12)=−12ln⁡2.\ln\left(\cos\frac{\pi}{4}\right)=\ln\left(\frac{1}{\sqrt2}\right)=-\frac{1}{2}\ln2.ln(cos4π​)=ln(2​1​)=−21​ln2.

Substitute into the general solution:

−1ln⁡2=12(−12ln⁡2)+C(−12ln⁡2)3.-\frac{1}{\ln2}=\frac{1}{2\left(-\frac12\ln2\right)}+\frac{C}{\left(-\frac12\ln2\right)^3}.−ln21​=2(−21​ln2)1​+(−21​ln2)3C​.

But

12(−12ln⁡2)=−1ln⁡2.\frac{1}{2\left(-\frac12\ln2\right)}=-\frac{1}{\ln2}.2(−21​ln2)1​=−ln21​.

So,

−1ln⁡2=−1ln⁡2+C(−12ln⁡2)3.-\frac{1}{\ln2}=-\frac{1}{\ln2}+\frac{C}{\left(-\frac12\ln2\right)^3}.−ln21​=−ln21​+(−21​ln2)3C​.

Hence,

C=0.C=0.C=0.

Therefore,

y=12ln⁡(cos⁡x).y=\frac{1}{2\ln(\cos x)}.y=2ln(cosx)1​.
  1. Find y(π6)y\left(\frac{\pi}{6}\right)y(6π​)

Since

cos⁡π6=32,\cos\frac{\pi}{6}=\frac{\sqrt3}{2},cos6π​=23​​,

we get

y(π6)=12ln⁡(32).y\left(\frac{\pi}{6}\right)=\frac{1}{2\ln\left(\frac{\sqrt3}{2}\right)}.y(6π​)=2ln(23​​)1​.

Now,

ln⁡(32)=ln⁡(3)−ln⁡2=12ln⁡3−ln⁡2.\ln\left(\frac{\sqrt3}{2}\right)=\ln(\sqrt3)-\ln 2=\frac12\ln3-\ln2.ln(23​​)=ln(3​)−ln2=21​ln3−ln2.

So,

2ln⁡(32)=ln⁡3−2ln⁡2=ln⁡3−ln⁡4.2\ln\left(\frac{\sqrt3}{2}\right)=\ln3-2\ln2=\ln3-\ln4.2ln(23​​)=ln3−2ln2=ln3−ln4.

Therefore,

y(π6)=1ln⁡3−ln⁡4.y\left(\frac{\pi}{6}\right)=\frac{1}{\ln3-\ln4}.y(6π​)=ln3−ln41​.
  1. Compare with options
1ln⁡3−ln⁡4\frac{1}{\ln3-\ln4}ln3−ln41​

matches Option D.

  1. Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

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