CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let y=f(x) be a non-negative function defined on [0,2] satisfying the differential equation y3y′′+1=0. If f(1)=0 and f(1)=1, then 
  1. the maximum value of f(x) is 1.
  2. the minimum value of f(x) is 0. 
  3. the solution y=f(x) of the given differential equation represents a semi-circle with centre (1,0).
  4. the area bounded by the curve y=f(x) and xaxis is π4 sq. units.


Solution

The correct options are
A the maximum value of f(x) is 1.
B the minimum value of f(x) is 0. 
C the solution y=f(x) of the given differential equation represents a semi-circle with centre (1,0).
y3y′′=1
yy′′=yy3
Integrating both the sides,
(y)22=12y2+C
f(1)=0, f(1)=1C=12
(y)22=12y212
y=±1y2y
y1y2y=±1
Integrating both the sides,
y1y2y dx=±dx

Put 1y2=z22yydx=2zdz
z dzz=±x+c
z=±x+c
1y2=±x+c

y(1)=1c=1
1y2=±x1
1y2=x1 or x+1
Squaring both the sides, we get
1y2=x22x+1 
y2=2xx2     ...(1)
y=2xx2=f(x)

From eqn (1),
x2+y22x=0 
(x1)2+y2=1,  x[0,2]


Clearly, x=1 is the point of maxima.
Maximum value of f(x)=21=1 

Minimum value of f(x)=0 at x=0,2

Area bounded by the curve y=f(x) and xaxis is :
π×122=π2 sq. units.  


  
 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...



footer-image