Maths-
General
Easy

Question

For the primitive integral equation y d x plus y to the power of 2 end exponent d y equals x d y comma x element of R comma y greater than 0 comma y equals y open parentheses x close parentheses comma y open parentheses 1 close parentheses equals 1 comma then y open parentheses negative 3 close parentheses is

  1. 3    
  2. 2    
  3. 1    
  4. 5    

hintHint:

We are given the primitive intergral equation. We are given the conditions of it. We are asked to find the value of equation when x = - 3. We will rearrange and integrate the equation. Then we will substitute the equation.

The correct answer is: 3


    The given equation is  ydx + y2dy = xdx
    The conditions are as follows:
    y > 0
    y = y(x)
    y(1) = 1
    We have to find y(-3)
    y d x space plus space y squared d y space equals space x d y
y d x space minus space x d y space equals space minus y squared d y
D i v i d i n g space b o t h space t h e space s i d e s space b y space y squared

fraction numerator y d x space minus space x d y over denominator y squared end fraction equals space minus d y
I
f space w e space s e e space t h e space l e f t space s i d e space i s space d o n e space b y space u space b y space v space m e t h o d. space
I t space i s space t h e space d i f f e r e n t i a t i o n space o f space x over y space u sin g space t h e space m e t h o d.
S o comma space w e space c a n space w r i t e space i t space a s

d open parentheses fraction numerator x over denominator y space end fraction close parentheses space equals space minus d y

N o w comma space w e space w i l l space i n t e g r a t e space b o t h space t h e space s i d e s.
    integral d open parentheses x over y close parentheses space equals space minus integral d y
space space space space space x over y space equals space minus y space plus space c
W e space h a v e space t h e space e q u a t i o n.
    We will find the value of c now. We will use the conditions given to us.
    y left parenthesis 1 right parenthesis space equals space 1
rightwards double arrow x space equals space 1 space a n d space y space equals space 1
S u b s t i t u t i n g space i n space t h e space e q u a t i o n space w e space g e t comma
x over y equals space minus y space plus space c
1 over 1 equals space minus 1 space plus space c
1 space plus space 1 space equals space c
R e a r r a n g i n g space w e space g e t space
c space equals space 2
    The equation becomes
    x over y equals space minus space y space plus space 2
N o w comma space w e space w i l l space f i n d space t h e space v a l u e space o f space y left parenthesis negative 3 right parenthesis
H e r e comma space x space equals space minus 3
S u b s t i t u t i n g space t h e space v a l u e space i n space a b o v e space e q u a t i o n space w e space g e t comma
fraction numerator negative 3 over denominator y space end fraction equals space minus y space plus space 2
M u l t i p l y i n g space y space o n space b o t h space t h e space s i d e
minus 3 space space equals space minus y squared space plus space 2 y
R e a r r a g i n g space t h e space e q u a t i o n
minus y squared space plus space 2 y space plus space 3 space equals space 0
space W e space w i l l space u s e space f a c t o r i z i a t i o n space m e t h o d
minus y squared space minus y space plus space 3 y space plus space 3 space equals space 0
minus y left parenthesis y space plus space 1 right parenthesis space plus space 3 left parenthesis y space plus space 1 right parenthesis space equals space 0
left parenthesis 3 minus y right parenthesis space left parenthesis y space plus space 1 right parenthesis space equals space 0
    It means (3 - y) = 0 or (y + 1) = 0
    3 - y = 0
    y = 3
    y + 1 = 0
    y = - 1
    But, we have condition that y > 0
    So, y ≠ 1.
    Therefore, y(-3) = 3
    The right option is 3.

    For such questions, we should know different method of differentiation and integration.

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