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Differential Equations question

2022 · 27 Jun · Shift 1 · Q28
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  5. /2022 · 27 Jun · Shift 1 · Q28

Differential Equations question

2022 · 27 Jun · Shift 1 · Q28

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If dydx+2x−y(2y−1)2x−1=0{{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0dxdy​+2x−12x−y(2y−1)​=0, x, y > 0, y(1) = 1, then y(2) is equal to :
  1. A
    2+log⁡232 + {\log _2}32+log2​3
  2. B
    2+log⁡322 + {\log _3}22+log3​2
  3. C
    2−log⁡322 - {\log _3}22−log3​2
  4. D
    2−log⁡232 - {\log _2}32−log2​3
View written solutionFree

Correct answer: D

  1. Given differential equation
dydx+2x−y(2y−1)2x−1=0,x>0, y>0,y(1)=1\frac{dy}{dx}+\frac{2^{x-y}(2^y-1)}{2^x-1}=0, \qquad x>0,\ y>0, \qquad y(1)=1dxdy​+2x−12x−y(2y−1)​=0,x>0, y>0,y(1)=1

We need to find y(2)y(2)y(2).

  1. Simplify the differential equation

Notice that

2x−y(2y−1)=2x−2x−y.2^{x-y}(2^y-1)=2^x-2^{x-y}.2x−y(2y−1)=2x−2x−y.

A better simplification is:

2x−y(2y−1)=2x−y⋅2y−2x−y=2x−2x−y.2^{x-y}(2^y-1)=2^{x-y}\cdot 2^y-2^{x-y}=2^x-2^{x-y}.2x−y(2y−1)=2x−y⋅2y−2x−y=2x−2x−y.

But the cleanest form comes by writing

2x−y(2y−1)=2x−y2y−2x−y=2x−2x−y.2^{x-y}(2^y-1)=2^{x-y}2^y-2^{x-y}=2^x-2^{x-y}.2x−y(2y−1)=2x−y2y−2x−y=2x−2x−y.

So the DE becomes

dydx=−2x−y(2y−1)2x−1.\frac{dy}{dx}=-\frac{2^{x-y}(2^y-1)}{2^x-1}.dxdy​=−2x−12x−y(2y−1)​.

Now divide numerator and denominator carefully:

2x−y(2y−1)=2x(1−2−y).2^{x-y}(2^y-1)=2^x\left(1-2^{-y}\right).2x−y(2y−1)=2x(1−2−y).

Hence

dydx=−2x(1−2−y)2x−1.\frac{dy}{dx}=-\frac{2^x(1-2^{-y})}{2^x-1}.dxdy​=−2x−12x(1−2−y)​.

This is separable.

  1. Separate variables

Rearrange:

dy1−2−y=−2x2x−1 dx.\frac{dy}{1-2^{-y}}=-\frac{2^x}{2^x-1}\,dx.1−2−ydy​=−2x−12x​dx.

Now simplify the left side:

1−2−y=2y−12y1-2^{-y}=\frac{2^y-1}{2^y}1−2−y=2y2y−1​

so

11−2−y=2y2y−1.\frac{1}{1-2^{-y}}=\frac{2^y}{2^y-1}.1−2−y1​=2y−12y​.

Thus

2y2y−1 dy=−2x2x−1 dx.\frac{2^y}{2^y-1}\,dy=-\frac{2^x}{2^x-1}\,dx.2y−12y​dy=−2x−12x​dx.
  1. Integrate both sides

We use the substitution:

  • On the left, let u=2y−1u=2^y-1u=2y−1, then du=2yln⁡2 dydu=2^y\ln 2\,dydu=2yln2dy.
  • On the right, let v=2x−1v=2^x-1v=2x−1, then dv=2xln⁡2 dxdv=2^x\ln 2\,dxdv=2xln2dx.

Therefore,

∫2y2y−1 dyn=1ln⁡2∫duu=1ln⁡2ln⁡∣2y−1∣,\int \frac{2^y}{2^y-1}\,dy n=\frac{1}{\ln 2}\int \frac{du}{u} =\frac{1}{\ln 2}\ln|2^y-1|,∫2y−12y​dyn=ln21​∫udu​=ln21​ln∣2y−1∣,

and

∫2x2x−1 dx=1ln⁡2ln⁡∣2x−1∣.\int \frac{2^x}{2^x-1}\,dx =\frac{1}{\ln 2}\ln|2^x-1|.∫2x−12x​dx=ln21​ln∣2x−1∣.

So after integration,

1ln⁡2ln⁡(2y−1)=−1ln⁡2ln⁡(2x−1)+C.\frac{1}{\ln 2}\ln(2^y-1)=-\frac{1}{\ln 2}\ln(2^x-1)+C.ln21​ln(2y−1)=−ln21​ln(2x−1)+C.

Multiplying by ln⁡2\ln 2ln2,

ln⁡(2y−1)=−ln⁡(2x−1)+C1.\ln(2^y-1)=-\ln(2^x-1)+C_1.ln(2y−1)=−ln(2x−1)+C1​.

Hence

ln⁡((2y−1)(2x−1))=C1.\ln\big((2^y-1)(2^x-1)\big)=C_1.ln((2y−1)(2x−1))=C1​.

So

(2y−1)(2x−1)=C.(2^y-1)(2^x-1)=C.(2y−1)(2x−1)=C.
  1. Use the initial condition y(1)=1y(1)=1y(1)=1

Substitute x=1x=1x=1, y=1y=1y=1:

(21−1)(21−1)=1⋅1=1.(2^1-1)(2^1-1)=1\cdot 1=1.(21−1)(21−1)=1⋅1=1.

Thus

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

Therefore the solution satisfies

(2y−1)(2x−1)=1.(2^y-1)(2^x-1)=1.(2y−1)(2x−1)=1.
  1. Find y(2)y(2)y(2)

Put x=2x=2x=2:

(2y(2)−1)(22−1)=1.(2^{y(2)}-1)(2^2-1)=1.(2y(2)−1)(22−1)=1.

That is,

(2y(2)−1)⋅3=1(2^{y(2)}-1)\cdot 3=1(2y(2)−1)⋅3=1

so

2y(2)−1=132^{y(2)}-1=\frac{1}{3}2y(2)−1=31​ 2y(2)=43.2^{y(2)}=\frac{4}{3}.2y(2)=34​.

Taking log⁡2\log_2log2​:

y(2)=log⁡2(43)=log⁡24−log⁡23=2−log⁡23.y(2)=\log_2\left(\frac{4}{3}\right)=\log_2 4-\log_2 3=2-\log_2 3.y(2)=log2​(34​)=log2​4−log2​3=2−log2​3.
  1. Match with options
y(2)=2−log⁡23y(2)=2-\log_2 3y(2)=2−log2​3

This is Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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