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Differential Equations question

2015 · Shift 1 · Q28
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  5. /2015 · Shift 1 · Q28

Differential Equations question

2015 · Shift 1 · Q28

JEE AdvancedMathematicsDifferential EquationsMultiple correct+4 / −2
Let y(x)y(x)y(x) be a solution of the differential equation (1+ex)y′+yex=1.\left( {1 + {e^x}} \right)y' + y{e^x} = 1.(1+ex)y′+yex=1. If y(0)=2y(0)=2y(0)=2, then which of the following statement is (are) true?
  1. A
    y(−4)=0y(-4)=0y(−4)=0
  2. B
    y(−2)=0y(-2)=0y(−2)=0
  3. C
    y(x)y(x)y(x) has a critical point in the interval (−1,0)(-1, 0)(−1,0)
  4. D
    y(x)y(x)y(x) has no critical point in the interval (−1,0)(-1,0)(−1,0)
View written solutionFree

Correct answer: A, C

  1. Given differential equation
(1+ex)y′+yex=1(1+e^x)y' + ye^x = 1(1+ex)y′+yex=1

with initial condition

y(0)=2.y(0)=2.y(0)=2.

We need to solve it and then test the options.

  1. Rewrite in standard linear form

Divide throughout by 1+ex1+e^x1+ex:

y′+ex1+exy=11+ex.y' + \frac{e^x}{1+e^x}y = \frac{1}{1+e^x}.y′+1+exex​y=1+ex1​.

This is a first-order linear differential equation:

y′+P(x)y=Q(x),y' + P(x)y = Q(x),y′+P(x)y=Q(x),

where

P(x)=ex1+ex,Q(x)=11+ex.P(x)=\frac{e^x}{1+e^x}, \qquad Q(x)=\frac{1}{1+e^x}.P(x)=1+exex​,Q(x)=1+ex1​.
  1. Find the integrating factor

The integrating factor is

IF=e∫P(x) dx=e∫ex1+ex dx.IF = e^{\int P(x)\,dx} = e^{\int \frac{e^x}{1+e^x}\,dx}.IF=e∫P(x)dx=e∫1+exex​dx.

Let t=1+ext=1+e^xt=1+ex, so dt=exdxdt=e^x dxdt=exdx. Then

∫ex1+ex dx=∫dtt=ln⁡(1+ex).\int \frac{e^x}{1+e^x}\,dx = \int \frac{dt}{t} = \ln(1+e^x).∫1+exex​dx=∫tdt​=ln(1+ex).

Hence,

IF=eln⁡(1+ex)=1+ex.IF = e^{\ln(1+e^x)} = 1+e^x.IF=eln(1+ex)=1+ex.
  1. Solve the equation

Multiplying the standard form by the integrating factor:

(1+ex)y′+exy=1.(1+e^x)y' + e^x y = 1.(1+ex)y′+exy=1.

The left side is exactly

ddx((1+ex)y)=1.\frac{d}{dx}\big((1+e^x)y\big)=1.dxd​((1+ex)y)=1.

So,

(1+ex)y=x+C.(1+e^x)y = x + C.(1+ex)y=x+C.

Thus,

y(x)=x+C1+ex.y(x)=\frac{x+C}{1+e^x}.y(x)=1+exx+C​.
  1. Use the initial condition

Given y(0)=2y(0)=2y(0)=2:

2=0+C1+e0=C2.2 = \frac{0+C}{1+e^0} = \frac{C}{2}.2=1+e00+C​=2C​.

Therefore,

C=4.C=4.C=4.

So the solution is

y(x)=x+41+ex.\boxed{y(x)=\frac{x+4}{1+e^x}}.y(x)=1+exx+4​​.
  1. Check options A and B

Option A: y(−4)=0y(-4)=0y(−4)=0

y(−4)=−4+41+e−4=0.y(-4)=\frac{-4+4}{1+e^{-4}}=0.y(−4)=1+e−4−4+4​=0.

So A is true.

Option B: y(−2)=0y(-2)=0y(−2)=0

y(−2)=−2+41+e−2=21+e−2≠0.y(-2)=\frac{-2+4}{1+e^{-2}}=\frac{2}{1+e^{-2}} \neq 0.y(−2)=1+e−2−2+4​=1+e−22​=0.

So B is false.

  1. Find critical points in (−1,0)(-1,0)(−1,0)

A critical point occurs where y′(x)=0y'(x)=0y′(x)=0.

From

(1+ex)y′+yex=1,(1+e^x)y' + ye^x = 1,(1+ex)y′+yex=1,

we get

y′=1−yex1+ex.y' = \frac{1-ye^x}{1+e^x}.y′=1+ex1−yex​.

Now substitute

y=x+41+ex.y=\frac{x+4}{1+e^x}.y=1+exx+4​.

Then

yex=(x+4)ex1+ex.ye^x = \frac{(x+4)e^x}{1+e^x}.yex=1+ex(x+4)ex​.

Hence

y′=1−(x+4)ex1+ex1+ex=1+ex−(x+4)ex(1+ex)2=1−(x+3)ex(1+ex)2.y' = \frac{1 - \frac{(x+4)e^x}{1+e^x}}{1+e^x} = \frac{1+e^x-(x+4)e^x}{(1+e^x)^2} = \frac{1-(x+3)e^x}{(1+e^x)^2}.y′=1+ex1−1+ex(x+4)ex​​=(1+ex)21+ex−(x+4)ex​=(1+ex)21−(x+3)ex​.

Therefore critical points satisfy

1−(x+3)ex=01-(x+3)e^x=01−(x+3)ex=0

or

(x+3)ex=1.(x+3)e^x=1.(x+3)ex=1.

Define

f(x)=(x+3)ex−1.f(x)=(x+3)e^x-1.f(x)=(x+3)ex−1.

We check its sign on (−1,0)(-1,0)(−1,0):

f(−1)=(2)e−1−1=2e−1<0f(-1)=(2)e^{-1}-1=\frac{2}{e}-1<0f(−1)=(2)e−1−1=e2​−1<0

(since e>2e>2e>2), and

f(0)=3⋅1−1=2>0.f(0)=3\cdot 1 -1 =2>0.f(0)=3⋅1−1=2>0.

Since fff is continuous, by IVT there exists some c∈(−1,0)c\in(-1,0)c∈(−1,0) such that

f(c)=0.f(c)=0.f(c)=0.

So y′(c)=0y'(c)=0y′(c)=0, i.e. yyy has a critical point in (−1,0)(-1,0)(−1,0).

Thus C is true and D is false.

  1. Final conclusion

Correct statements are:

A, C\boxed{A,\ C}A, C​
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