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Differential Equations question

2021 · 26 Aug · Shift 1 · Q24
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  5. /2021 · 26 Aug · Shift 1 · Q24

Differential Equations question

2021 · 26 Aug · Shift 1 · Q24

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be a solution curve of the differential equation (y+1)tan⁡2x dx+tan⁡x dy+y dx=0(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0(y+1)tan2xdx+tanxdy+ydx=0, x∈(0,π2)x \in \left( {0,{\pi \over 2}} \right)x∈(0,2π​). If lim⁡x→0+xy(x)=1\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1x→0+lim​xy(x)=1, then the value of y(π4)y\left( {{\pi \over 4}} \right)y(4π​) is :
  1. A
    −π4- {\pi \over 4}−4π​
  2. B
    π4−1{\pi \over 4} - 14π​−1
  3. C
    π4+1{\pi \over 4} + 14π​+1
  4. D
    π4{\pi \over 4}4π​
View written solutionFree

Correct answer: D

  1. Rewrite the differential equation

Given

(y+1)tan⁡2x dx+tan⁡x dy+y dx=0.(y+1)\tan^2 x\,dx+\tan x\,dy+y\,dx=0.(y+1)tan2xdx+tanxdy+ydx=0.

Combine the dxdxdx terms:

tan⁡x dy+((y+1)tan⁡2x+y)dx=0.\tan x\,dy+\big((y+1)\tan^2 x+y\big)dx=0.tanxdy+((y+1)tan2x+y)dx=0.

So,

tan⁡xdydx+(y+1)tan⁡2x+y=0.\tan x\frac{dy}{dx}+ (y+1)\tan^2 x+y=0.tanxdxdy​+(y+1)tan2x+y=0.

Divide by tan⁡x\tan xtanx (valid since x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2), so tan⁡x>0\tan x>0tanx>0):

dydx+(y+1)tan⁡x+ycot⁡x=0.\frac{dy}{dx}+ (y+1)\tan x+y\cot x=0.dxdy​+(y+1)tanx+ycotx=0.

Expand:

dydx+y(tan⁡x+cot⁡x)+tan⁡x=0.\frac{dy}{dx}+y(\tan x+\cot x)+\tan x=0.dxdy​+y(tanx+cotx)+tanx=0.

Hence,

dydx+y(tan⁡x+cot⁡x)=−tan⁡x.\frac{dy}{dx}+y(\tan x+\cot x)=-\tan x.dxdy​+y(tanx+cotx)=−tanx.
  1. Simplify the coefficient

Use

tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x.\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} =\frac{\sin^2 x+\cos^2 x}{\sin x\cos x} =\frac{1}{\sin x\cos x}.tanx+cotx=cosxsinx​+sinxcosx​=sinxcosxsin2x+cos2x​=sinxcosx1​.

Thus the linear ODE is

dydx+1sin⁡xcos⁡xy=−tan⁡x.\frac{dy}{dx}+\frac{1}{\sin x\cos x}y=-\tan x.dxdy​+sinxcosx1​y=−tanx.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫1sin⁡xcos⁡x dx.I.F.=e^{\int \frac{1}{\sin x\cos x}\,dx}.I.F.=e∫sinxcosx1​dx.

Now,

1sin⁡xcos⁡x=tan⁡x+cot⁡x,\frac{1}{\sin x\cos x}=\tan x+\cot x,sinxcosx1​=tanx+cotx,

so

∫1sin⁡xcos⁡x dx=∫(tan⁡x+cot⁡x)dx=−ln⁡(cos⁡x)+ln⁡(sin⁡x)=ln⁡(tan⁡x).\int \frac{1}{\sin x\cos x}\,dx= \int (\tan x+\cot x)dx = -\ln(\cos x)+\ln(\sin x) =\ln(\tan x).∫sinxcosx1​dx=∫(tanx+cotx)dx=−ln(cosx)+ln(sinx)=ln(tanx).

Therefore,

I.F.=eln⁡(tan⁡x)=tan⁡x.I.F.=e^{\ln(\tan x)}=\tan x.I.F.=eln(tanx)=tanx.
  1. Solve the equation

Multiplying the ODE by tan⁡x\tan xtanx:

tan⁡xdydx+ytan⁡x(tan⁡x+cot⁡x)=−tan⁡2x.\tan x\frac{dy}{dx}+y\tan x(\tan x+\cot x)=-\tan^2 x.tanxdxdy​+ytanx(tanx+cotx)=−tan2x.

But

ddx(ytan⁡x)=tan⁡xdydx+ysec⁡2x.\frac{d}{dx}(y\tan x)=\tan x\frac{dy}{dx}+y\sec^2 x.dxd​(ytanx)=tanxdxdy​+ysec2x.

Also,

tan⁡x(tan⁡x+cot⁡x)=tan⁡2x+1=sec⁡2x.\tan x(\tan x+\cot x)=\tan^2 x+1=\sec^2 x.tanx(tanx+cotx)=tan2x+1=sec2x.

So the left side becomes

ddx(ytan⁡x)=−tan⁡2x.\frac{d}{dx}(y\tan x)=-\tan^2 x.dxd​(ytanx)=−tan2x.

Integrate:

ytan⁡x=∫−tan⁡2x dx+C.y\tan x=\int -\tan^2 x\,dx +C.ytanx=∫−tan2xdx+C.

Using

tan⁡2x=sec⁡2x−1,\tan^2 x=\sec^2 x-1,tan2x=sec2x−1,

we get

∫−tan⁡2x dx=∫(1−sec⁡2x)dx=x−tan⁡x.\int -\tan^2 x\,dx=\int (1-\sec^2 x)dx=x-\tan x.∫−tan2xdx=∫(1−sec2x)dx=x−tanx.

Hence,

ytan⁡x=x−tan⁡x+C.y\tan x=x-\tan x+C.ytanx=x−tanx+C.

Therefore,

y=x−tan⁡x+Ctan⁡x.y=\frac{x-\tan x+C}{\tan x}.y=tanxx−tanx+C​.
  1. Use the condition lim⁡x→0+xy(x)=1\lim_{x\to 0^+}x y(x)=1limx→0+​xy(x)=1

We have

xy=x⋅x−tan⁡x+Ctan⁡x.xy=x\cdot \frac{x-\tan x+C}{\tan x}.xy=x⋅tanxx−tanx+C​.

As x→0+x\to 0^+x→0+, tan⁡x∼x\tan x\sim xtanx∼x. So:

  • if C≠0C\neq 0C=0, then xy∼x⋅Cx=C,xy\sim x\cdot \frac{C}{x}=C,xy∼x⋅xC​=C, more precisely the limit becomes CCC.

Let us compute directly:

lim⁡x→0+xy=lim⁡x→0+x(x−tan⁡x+C)tan⁡x.\lim_{x\to 0^+}xy= \lim_{x\to 0^+}\frac{x(x-\tan x+C)}{\tan x}.x→0+lim​xy=x→0+lim​tanxx(x−tanx+C)​.

Since x−tan⁡x→0x-\tan x\to 0x−tanx→0 and xtan⁡x→1\frac{x}{\tan x}\to 1tanxx​→1,

lim⁡x→0+xy=C.\lim_{x\to 0^+}xy=C.x→0+lim​xy=C.

Given this limit is 111, we get

C=1.C=1.C=1.

So the solution is

y=x−tan⁡x+1tan⁡x.y=\frac{x-\tan x+1}{\tan x}.y=tanxx−tanx+1​.
  1. Compute y(π4)y\left(\frac{\pi}{4}\right)y(4π​)

Since

tan⁡π4=1,\tan\frac{\pi}{4}=1,tan4π​=1,

we have

y(π4)=π4−1+11=π4.y\left(\frac{\pi}{4}\right)=\frac{\frac{\pi}{4}-1+1}{1}=\frac{\pi}{4}.y(4π​)=14π​−1+1​=4π​.
  1. Check options

Thus,

y(π4)=π4.y\left(\frac{\pi}{4}\right)=\frac{\pi}{4}.y(4π​)=4π​.

This matches Option D.

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