Differential Equations 1 Question 12

12. Let y(x) be a solution of the differential equation (1+ex)y+yex=1. If y(0)=2, then which of the following statement(s) is/are true?

(2015 Adv.)

(a) y(4)=0

(b) y(2)=0

(c) y(x) has a critical point in the interval (1,0)

(d) y(x) has no critical point in the interval (1,0)

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Answer:

Correct Answer: 12. (a, c)

Solution:

  1. Here, (1+ex)y+yex=1

dydx+exdydx+yex=1

dy+exdy+yexdx=dx

dy+d(exy)=dx

On integrating both sides, we get

y+exy=x+C

 Given, y(0)=22+e02=0+CC=4y(1+ex)=x+4y=x+41+ex Now at x=4,y=4+41+e4=0y(4)=0 For critical points, dydx=0 i.e. dydx=(1+ex)1(x+4)ex(1+ex)2=0ex(x+3)1=0 or ex=(x+3)

Clearly, the intersection point lies between (1,0).

y(x) has a critical point in the interval (1,0).



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