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Differential Equations question

2023 · 8 Apr · Shift 2 · Q40
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  5. /2023 · 8 Apr · Shift 2 · Q40

Differential Equations question

2023 · 8 Apr · Shift 2 · Q40

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let the solution curve x=x(y),0<y<π2x=x(y), 0 \lt y \lt \frac{\pi}{2}x=x(y),0<y<2π​, of the differential equation (log⁡e(cos⁡y))2cos⁡y dx−(1+3xlog⁡e(cos⁡y))sin⁡ydy=0\left(\log _{e}(\cos y)\right)^{2} \cos y \mathrm{~d} x-\left(1+3 x \log _{e}(\cos y)\right) \sin \mathrm{y} d y=0(loge​(cosy))2cosy dx−(1+3xloge​(cosy))sinydy=0 satisfy x(π3)=12log⁡e2x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _{e} 2}x(3π​)=2loge​21​. If x(π6)=1log⁡em−log⁡enx\left(\frac{\pi}{6}\right)=\frac{1}{\log _{e} m-\log _{e} n}x(6π​)=loge​m−loge​n1​, where mmm and nnn are coprime, then mnm nmn is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Rewrite the differential equation in terms of x(y)x(y)x(y)

Given

(ln⁡(cos⁡y))2cos⁡y dx−(1+3xln⁡(cos⁡y))sin⁡y dy=0.(\ln(\cos y))^2 \cos y\, dx-(1+3x\ln(\cos y))\sin y\, dy=0.(ln(cosy))2cosydx−(1+3xln(cosy))sinydy=0.

Divide by dydydy:

(ln⁡(cos⁡y))2cos⁡y dxdy−(1+3xln⁡(cos⁡y))sin⁡y=0.(\ln(\cos y))^2 \cos y\, \frac{dx}{dy}-(1+3x\ln(\cos y))\sin y=0.(ln(cosy))2cosydydx​−(1+3xln(cosy))siny=0.

So

dxdy=(1+3xln⁡(cos⁡y))sin⁡y(ln⁡(cos⁡y))2cos⁡y.\frac{dx}{dy}=\frac{(1+3x\ln(\cos y))\sin y}{(\ln(\cos y))^2\cos y}.dydx​=(ln(cosy))2cosy(1+3xln(cosy))siny​.

Using tan⁡y=sin⁡ycos⁡y\tan y=\dfrac{\sin y}{\cos y}tany=cosysiny​,

dxdy=tan⁡y(1(ln⁡(cos⁡y))2+3xln⁡(cos⁡y)).\frac{dx}{dy}=\tan y\left(\frac{1}{(\ln(\cos y))^2}+\frac{3x}{\ln(\cos y)}\right).dydx​=tany((ln(cosy))21​+ln(cosy)3x​).

Hence

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

This is a linear differential equation in xxx.

  1. Simplify the coefficient using substitution

Let

t=ln⁡(cos⁡y).t=\ln(\cos y).t=ln(cosy).

Then

dtdy=−tan⁡y.\frac{dt}{dy}=-\tan y.dydt​=−tany.

So tan⁡y dy=−dt\tan y\,dy=-dttanydy=−dt.

The differential equation becomes

dxdy−3tan⁡ytx=tan⁡yt2.\frac{dx}{dy}-\frac{3\tan y}{t}x=\frac{\tan y}{t^2}.dydx​−t3tany​x=t2tany​.

Convert derivatives using dxdy=dxdtdtdy=−tan⁡ydxdt\dfrac{dx}{dy}=\dfrac{dx}{dt}\dfrac{dt}{dy}=-\tan y\dfrac{dx}{dt}dydx​=dtdx​dydt​=−tanydtdx​:

−tan⁡ydxdt−3tan⁡ytx=tan⁡yt2.-\tan y\frac{dx}{dt}-\frac{3\tan y}{t}x=\frac{\tan y}{t^2}.−tanydtdx​−t3tany​x=t2tany​.

Since 0<y<π20<y<\frac{\pi}{2}0<y<2π​, we have tan⁡y≠0\tan y\neq 0tany=0, so divide by tan⁡y\tan ytany:

−dxdt−3tx=1t2.-\frac{dx}{dt}-\frac{3}{t}x=\frac{1}{t^2}.−dtdx​−t3​x=t21​.

Thus

dxdt+3tx=−1t2.\frac{dx}{dt}+\frac{3}{t}x=-\frac{1}{t^2}.dtdx​+t3​x=−t21​.
  1. Solve the linear ODE

We have

dxdt+3tx=−1t2.\frac{dx}{dt}+\frac{3}{t}x=-\frac{1}{t^2}.dtdx​+t3​x=−t21​.

Integrating factor:

I.F.=e∫3tdt=t3.\text{I.F.}=e^{\int \frac{3}{t}dt}=t^3.I.F.=e∫t3​dt=t3.

Multiply throughout by t3t^3t3:

t3dxdt+3t2x=−t.t^3\frac{dx}{dt}+3t^2x=-t.t3dtdx​+3t2x=−t.

So

ddt(t3x)=−t.\frac{d}{dt}(t^3x)=-t.dtd​(t3x)=−t.

Integrate:

t3x=−t22+C.t^3x=-\frac{t^2}{2}+C.t3x=−2t2​+C.

Therefore

x=−12t+Ct3.x=-\frac{1}{2t}+\frac{C}{t^3}.x=−2t1​+t3C​.

Now substitute back t=ln⁡(cos⁡y)t=\ln(\cos y)t=ln(cosy):

x(y)=−12ln⁡(cos⁡y)+C(ln⁡(cos⁡y))3.x(y)=-\frac{1}{2\ln(\cos y)}+\frac{C}{(\ln(\cos y))^3}.x(y)=−2ln(cosy)1​+(ln(cosy))3C​.
  1. Use the condition x(π3)=12ln⁡2x\left(\frac{\pi}{3}\right)=\dfrac{1}{2\ln 2}x(3π​)=2ln21​

At y=π3y=\frac{\pi}{3}y=3π​,

cos⁡π3=12,\cos\frac{\pi}{3}=\frac12,cos3π​=21​,

so

ln⁡(cos⁡π3)=ln⁡(12)=−ln⁡2.\ln(\cos \tfrac{\pi}{3})=\ln\left(\frac12\right)=-\ln 2.ln(cos3π​)=ln(21​)=−ln2.

Thus

x(π3)=−12(−ln⁡2)+C(−ln⁡2)3=12ln⁡2−C(ln⁡2)3.x\left(\frac{\pi}{3}\right)=-\frac{1}{2(-\ln2)}+\frac{C}{(-\ln2)^3} =\frac{1}{2\ln2}-\frac{C}{(\ln2)^3}.x(3π​)=−2(−ln2)1​+(−ln2)3C​=2ln21​−(ln2)3C​.

Given this equals 12ln⁡2\dfrac{1}{2\ln2}2ln21​, we get

12ln⁡2−C(ln⁡2)3=12ln⁡2  ⟹  C=0.\frac{1}{2\ln2}-\frac{C}{(\ln2)^3}=\frac{1}{2\ln2} \implies C=0.2ln21​−(ln2)3C​=2ln21​⟹C=0.

Hence the solution is

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

Since

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

we get

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

Now

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

x(π6)=−1ln⁡3−ln⁡4=1ln⁡4−ln⁡3.x\left(\frac{\pi}{6}\right)=-\frac{1}{\ln3-\ln4} =\frac{1}{\ln4-\ln3}.x(6π​)=−ln3−ln41​=ln4−ln31​.

This matches

1ln⁡m−ln⁡n\frac{1}{\ln m-\ln n}lnm−lnn1​

with

m=4,n=3.m=4,\quad n=3.m=4,n=3.

These are coprime, so

mn=12.mn=12.mn=12.
  1. Final answer
12\boxed{12}12​

The derived answer matches the stored correct answer.

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