56. The differential equation	dy	+ 4y = 5
    	dx

    
    is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of the differential equation is
    

1. 1. y = e^4x + 5
      

   
   
   

   2. y = e^–4x + 1.25
      

   
   
   

   3. y = e^–4x + 5
      

   
   
   

   4. y = e^4x + 1.25

   
   
   

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1. View Hint View Answer Discuss in Forum

   	dy	+ 4y = 5
   	dx

   
   I.F. = e^{∫4 dx} = e^{4 x}
   ⇒ y × I.F. = ∫ 5 × (I.F.)dx + C
   

   ⇒ y . e4 x = 5.	e4 x	+ C
   	4

   
   

   ⇒ 2.25 × 1 = 5.	e0	+ C
   	4

   
   ⇒ C = 1
   

   y . e4 x = 5.	e4 x	+ 1
   	4

   
   

   ⇒ y =	5	+ e-4 x
   	4

   
   ⇒ y = e^{-4 x} + 1.25
   

   Correct Option: B

   	dy	+ 4y = 5
   	dx

   
   I.F. = e^{∫4 dx} = e^{4 x}
   ⇒ y × I.F. = ∫ 5 × (I.F.)dx + C
   

   ⇒ y . e4 x = 5.	e4 x	+ C
   	4

   
   

   ⇒ 2.25 × 1 = 5.	e0	+ C
   	4

   
   ⇒ C = 1
   

   y . e4 x = 5.	e4 x	+ 1
   	4

   
   

   ⇒ y =	5	+ e-4 x
   	4

   
   ⇒ y = e^{-4 x} + 1.25
   

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57. For the equation	dy	+ 7x2y = 0 , if y(0) = 3 / 7 ,
    	dx

    
    then the value of y(1) is
    

1. 1. 3	e-7 / 3
      7

      
      

   
   
   

   2. 3	e-3 / 7
      7

      
      

   
   
   

   3. 7	e-3 / 7
      3

      
      

   
   
   

   4. 7	e-7 / 3
      3

   
   
   

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1. View Hint View Answer Discuss in Forum

   From question, we have
   

   	dy	+ 7x2y = 0 , y(0) =	3	, y(1) = ?
   	dx		7

   
   

   	dy	= -7x2y
   	dx

   
   

   	dy	= -7x2dx
   	y

   
   

   ln y = -	7x3	+ C
   	3

   
   y = e^{{ (-7x3 / 3) + C}}
   

   y(0) = eC =	3
   	7

   
   

   C = ln		3
   		7

   
   

   ln y = -	7x3	+ ln		3
   	3			7

   
   

   ln y - ln		3		= -	7x3
   		7			3

   
   

   ln		7y		= -	7x3
   		3			3

   
   

   y =	3	e (-7x3/ 3)
   	7

   
   

   ∴ y(1) =	3	e (-7/ 3)
   	7

   Correct Option: A

   From question, we have
   

   	dy	+ 7x2y = 0 , y(0) =	3	, y(1) = ?
   	dx		7

   
   

   	dy	= -7x2y
   	dx

   
   

   	dy	= -7x2dx
   	y

   
   

   ln y = -	7x3	+ C
   	3

   
   y = e^{{ (-7x3 / 3) + C}}
   

   y(0) = eC =	3
   	7

   
   

   C = ln		3
   		7

   
   

   ln y = -	7x3	+ ln		3
   	3			7

   
   

   ln y - ln		3		= -	7x3
   		7			3

   
   

   ln		7y		= -	7x3
   		3			3

   
   

   y =	3	e (-7x3/ 3)
   	7

   
   

   ∴ y(1) =	3	e (-7/ 3)
   	7

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58. If y is the solution of the differential equation y3	dy	+ x3 = 0 ,
    	dx

    
    y(0) = 1 the value of y(–1) is
    

1. 1. –2
      

   
   
   

   2. –1
      

   
   
   

   3. 0
      

   
   
   

   4. 1
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   y3	dy	= -x3
   	dx

   
   y³ dy = -x³ dx
   ∫ y³ dy = -∫ x³ dx
   

   y4	=	− x4	+ c
   4		4

   
   

   	x4 + y4	= c
   	4

   
   y(0) = 1 ,
   

   	0 + 1	= c
   	4

   
   

   c =	1
   	4

   
   x⁴ + y⁴ = 1
   y = 4√1 - x⁴
   when x = -1
   y = 0

   Correct Option: C

   y3	dy	= -x3
   	dx

   
   y³ dy = -x³ dx
   ∫ y³ dy = -∫ x³ dx
   

   y4	=	− x4	+ c
   4		4

   
   

   	x4 + y4	= c
   	4

   
   y(0) = 1 ,
   

   	0 + 1	= c
   	4

   
   

   c =	1
   	4

   
   x⁴ + y⁴ = 1
   y = 4√1 - x⁴
   when x = -1
   y = 0

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59. Consider the following differential equation :
    

    	dy	= -5y ; initial condition; y = 2 at t = 0 The value of y at t = 3 is
    	dt

    
    

1. 1. -5e^-10
      

   
   
   

   2. 2e^-10
      

   
   
   

   3. 2e^-15
      

   
   
   

   4. -15e²
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   dy	= -5y
   dt

   
   

   dy	= -5dt
   y

   
   By Integrating above equation, we get
   ln y = 5t + c
   at t = 0 , y = 2
   ln 2 = c
   So, ln y = – 5t + ln 2
   

   ln	y	= -5t
   	2

   
   

   	y	= e-5t
   	2

   
   y = 2e^-5t
   at t = 3
   y = 2 e^-15

   Correct Option: C

   dy	= -5y
   dt

   
   

   dy	= -5dt
   y

   
   By Integrating above equation, we get
   ln y = 5t + c
   at t = 0 , y = 2
   ln 2 = c
   So, ln y = – 5t + ln 2
   

   ln	y	= -5t
   	2

   
   

   	y	= e-5t
   	2

   
   y = 2e^-5t
   at t = 3
   y = 2 e^-15

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60. The distance between the origin and the point nearest it on the surface z² = 1 + xy is
    

1. 1. 1
      

   
   
   

   2. √3
      2

      
      

   
   
   

   3. √3
      

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Take a point on the curve z = 1, x = 0, y = 0
   Length between origin and this point
   l = √( 1- 0) + (0 - 0) + (0 - 0 ) = 1
   This is minimum length because all options have length greater than l.

   Correct Option: A

   Take a point on the curve z = 1, x = 0, y = 0
   Length between origin and this point
   l = √( 1- 0) + (0 - 0) + (0 - 0 ) = 1
   This is minimum length because all options have length greater than l.

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