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Differential Equations question

2021 · 31 Aug · Shift 2 · Q28
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  5. /2021 · 31 Aug · Shift 2 · Q28

Differential Equations question

2021 · 31 Aug · Shift 2 · Q28

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If dydx=2xy+2y.2x2x+2x+ylog⁡e2{{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}dxdy​=2x+2x+yloge​22xy+2y.2x​, y(0) = 0, then for y = 1, the value of x lies in the interval :
  1. A
    (1, 2)
  2. B
    (12,1]\left( {{1 \over 2},1} \right](21​,1]
  3. C
    (2, 3)
  4. D
    (0,12]\left( {0,{1 \over 2}} \right](0,21​]
View written solutionFree

Correct answer: A

  1. Given differential equation

We are given

dydx=2xy+2y⋅2x2x+2x+yln⁡2,y(0)=0.\frac{dy}{dx}=\frac{2^x y+2^y\cdot 2^x}{2^x+2^{x+y}\ln 2}, \qquad y(0)=0.dxdy​=2x+2x+yln22xy+2y⋅2x​,y(0)=0.

First simplify the expression by factoring 2x2^x2x from numerator and denominator:

dydx=2x(y+2y)2x(1+2yln⁡2)=y+2y1+2yln⁡2.\frac{dy}{dx}=\frac{2^x(y+2^y)}{2^x(1+2^y\ln 2)}=\frac{y+2^y}{1+2^y\ln 2}.dxdy​=2x(1+2yln2)2x(y+2y)​=1+2yln2y+2y​.

So the differential equation becomes

dydx=y+2y1+2yln⁡2.\frac{dy}{dx}=\frac{y+2^y}{1+2^y\ln 2}.dxdy​=1+2yln2y+2y​.

This is separable.

  1. Separate variables

Invert both sides:

dxdy=1+2yln⁡2y+2y.\frac{dx}{dy}=\frac{1+2^y\ln 2}{y+2^y}.dydx​=y+2y1+2yln2​.

Notice that

ddy(y+2y)=1+2yln⁡2.\frac{d}{dy}(y+2^y)=1+2^y\ln 2.dyd​(y+2y)=1+2yln2.

Hence,

dxdy=ddy(y+2y)y+2y.\frac{dx}{dy}=\frac{\frac{d}{dy}(y+2^y)}{y+2^y}.dydx​=y+2ydyd​(y+2y)​.

Therefore,

dxdy=ddy(ln⁡(y+2y)).\frac{dx}{dy}=\frac{d}{dy}\bigl(\ln(y+2^y)\bigr).dydx​=dyd​(ln(y+2y)).

Integrating with respect to yyy,

x=ln⁡(y+2y)+C.x=\ln(y+2^y)+C.x=ln(y+2y)+C.
  1. Use the initial condition

Given y(0)=0y(0)=0y(0)=0, so when x=0x=0x=0, y=0y=0y=0.

Substitute into the solution:

0=ln⁡(0+20)+C=ln⁡(1)+C=0+C.0=\ln(0+2^0)+C=\ln(1)+C=0+C.0=ln(0+20)+C=ln(1)+C=0+C.

So,

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

Thus the solution is

x=ln⁡(y+2y).x=\ln(y+2^y).x=ln(y+2y).
  1. Find xxx when y=1y=1y=1

Substitute y=1y=1y=1:

x=ln⁡(1+21)=ln⁡3.x=\ln(1+2^1)=\ln 3.x=ln(1+21)=ln3.

Now,

ln⁡3≈1.0986.\ln 3\approx 1.0986.ln3≈1.0986.

So,

1<ln⁡3<2.1<\ln 3<2.1<ln3<2.

Hence xxx lies in the interval

(1,2).(1,2).(1,2).
  1. Check options
  • A: (1,2)(1,2)(1,2) ✅
  • B: (12,1]\left(\frac12,1\right](21​,1] ❌
  • C: (2,3)(2,3)(2,3) ❌
  • D: (0,12]\left(0,\frac12\right](0,21​] ❌

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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