Worksheet: Initial Value Problems

In this worksheet, we will practice finding a specific solution to a separable differential equation given an initial value.

Q1:

Find the equation of the curve that passes through the point ( βˆ’ 3 , 2 ) given that the gradient of the tangent at any point is βˆ’ 4 π‘₯ 7 𝑦 .

  • A 7 𝑦 = βˆ’ 4 π‘₯ + C
  • B 7 𝑦 = βˆ’ 4 π‘₯ + 6 4  
  • C 7 𝑦 = βˆ’ 4 π‘₯ + 7 9  
  • D 7 𝑦 = βˆ’ 4 π‘₯ + 7 9 2  

Q2:

Find the equation of the curve that passes through the point ( 0 , βˆ’ 1 ) given d d 𝑦 π‘₯ = βˆ’ 6 π‘₯ βˆ’ 4 4 𝑦 + 1 3 .

  • A 4 𝑦 + 1 3 𝑦 = βˆ’ 3 π‘₯ βˆ’ 4 π‘₯ βˆ’ 1 1  
  • B 2 𝑦 + 1 3 𝑦 = βˆ’ 3 π‘₯ βˆ’ 4 π‘₯ βˆ’ 1 1  
  • C 2 𝑦 + 1 3 𝑦 = βˆ’ 6 π‘₯ βˆ’ 4 π‘₯ βˆ’ 9  
  • D 4 𝑦 + 1 3 𝑦 = βˆ’ 6 π‘₯ βˆ’ 4 π‘₯ βˆ’ 9  
  • E 2 𝑦 + 1 3 𝑦 = βˆ’ 3 π‘₯ βˆ’ 4 π‘₯ βˆ’ 9  

Q3:

Find the solution of the differential equation d d 𝑦 π‘₯ = π‘₯ 𝑒  that satisfies the initial condition 𝑦 ( 0 ) = 0 .

  • A 𝑦 = βˆ’ ο€Ό βˆ’ 1 2 π‘₯  l n 
  • B 𝑦 = βˆ’ ο€Ή βˆ’ π‘₯ + 1  l n 
  • C 𝑦 = βˆ’ ο€Ό 1 2 π‘₯ + 1  l n 
  • D 𝑦 = βˆ’ ο€Ή βˆ’ π‘₯  l n 
  • E 𝑦 = βˆ’ ο€Ό βˆ’ 1 2 π‘₯ + 1  l n 

Q4:

Find the solution of the differential equation d d 𝑃 𝑑 = √ 𝑃 𝑑 that satisfies the initial condition 𝑃 ( 1 ) = 2 .

  • A √ 𝑃 = 1 3 𝑑 + 1 3 + √ 2  
  • B √ 𝑃 = 1 3 𝑑 βˆ’ 1 3 + √ 2  
  • C √ 𝑃 = 1 3 𝑑 βˆ’ √ 2 βˆ’ 1 3  
  • D √ 𝑃 = 1 3 𝑑 βˆ’ 3 4 + √ 2  
  • E √ 𝑃 = βˆ’ 1 3 𝑑 βˆ’ √ 2 + 1 3  

Q5:

Suppose that d d c o s s i n 𝑦 π‘₯ = 4 π‘₯ βˆ’ 4 2 π‘₯ 4 𝑦 + 9 and 𝑦 = 0 when π‘₯ = 0 . Find 𝑦 in terms of π‘₯ .

  • A 9 𝑦 βˆ’ 4 𝑦 = 2 π‘₯ βˆ’ 4 2 π‘₯ βˆ’ 4 c o s s i n 
  • B 9 𝑦 βˆ’ 4 𝑦 = 4 π‘₯ βˆ’ 2 2 π‘₯ βˆ’ 4 c o s s i n 
  • C 9 𝑦 + 4 𝑦 = 2 π‘₯ + 2 2 π‘₯ + 4 c o s s i n 
  • D 9 𝑦 βˆ’ 4 𝑦 = 2 π‘₯ βˆ’ 2 2 π‘₯ βˆ’ 4 c o s s i n 
  • E 9 𝑦 + 4 𝑦 = 4 π‘₯ + 2 2 π‘₯ + 4 c o s s i n 

Q6:

A relation 𝑓 ( π‘₯ , 𝑦 ) = 0 is implicitly differentiated to obtain d d 𝑦 π‘₯ = 2 π‘₯ + 5 2 𝑦 + 5 . Find the relation given that when 𝑦 = 3 , π‘₯ = 3 .

  • A π‘₯ + 5 π‘₯ + 5 𝑦 βˆ’ 9 = 0 
  • B π‘₯ βˆ’ 5 𝑦 βˆ’ 3 = 0 
  • C π‘₯ + 5 π‘₯ βˆ’ 2 𝑦 βˆ’ 5 𝑦 = 0  
  • D π‘₯ + 5 π‘₯ βˆ’ 𝑦 βˆ’ 5 𝑦 = 0  

Q7:

Find the equation of the curve that passes through the point ( βˆ’ 8 , 1 ) given that the gradient of the tangent at any point is equal to 2 times the square of the 𝑦 coordinate.

  • A 𝑦 = βˆ’ 1 2 π‘₯ + 1 5
  • B 𝑦 = βˆ’ 1 2 π‘₯ βˆ’ 1 7
  • C 𝑦 = βˆ’ 1 2 π‘₯ + 1 7
  • D 𝑦 = 1 2 π‘₯ βˆ’ 1 5

Q8:

Suppose that d d s i n c o s 𝑦 π‘₯ = 3 π‘₯ 4 2 𝑦   and 𝑦 = πœ‹ 4 when π‘₯ = πœ‹ 2 . Find 𝑦 in terms of π‘₯ .

  • A 8 𝑦 + 8 4 𝑦 = 6 π‘₯ βˆ’ 6 2 π‘₯ βˆ’ πœ‹ s i n s i n
  • B 8 𝑦 βˆ’ 8 4 𝑦 = 6 π‘₯ + 6 2 π‘₯ βˆ’ πœ‹ s i n s i n
  • C 8 𝑦 + 2 4 𝑦 = 6 π‘₯ βˆ’ 3 2 π‘₯ βˆ’ πœ‹ s i n s i n
  • D 8 𝑦 βˆ’ 2 4 𝑦 = 6 π‘₯ + 3 2 π‘₯ βˆ’ πœ‹ s i n s i n

Q9:

Find the solution of the differential equation d d 𝑦 π‘₯ + 9 𝑦 = 6 3 given that 𝑦 ( 0 ) = 8 .

  • A 𝑦 = 7 + 𝑒   
  • B 𝑦 = 7 + 𝑒  
  • C 𝑦 = 7 + 𝑒  
  • D 𝑦 = 7 + 𝑒  

Q10:

Find the solution for the following differential equation for 𝑦 ( 0 ) = 2 : d d 𝑦 π‘₯ = π‘₯ 𝑦 . οŠͺ 

  • A 𝑦 = 1 5 π‘₯ + 8 
  • B 𝑦 = ο€Ό 3 5 π‘₯ + 8    
  • C 𝑦 = ο€Ή 1 5 π‘₯ + 8    
  • D 𝑦 = 3 5 π‘₯ + 8 

Q11:

Find the solution of the differential equation d d s i n 𝑦 π‘₯ = π‘₯ π‘₯ 𝑦 that satisfies the initial condition 𝑦 ( 0 ) = βˆ’ 6 .

  • A 𝑦 = 2 ( π‘₯ π‘₯ βˆ’ π‘₯ ) + 3 6  s i n c o s
  • B 𝑦 = ( π‘₯ βˆ’ π‘₯ π‘₯ ) + 3 6  s i n c o s
  • C 𝑦 = 2 ( π‘₯ βˆ’ π‘₯ π‘₯ )  s i n c o s
  • D 𝑦 = 2 ( π‘₯ βˆ’ π‘₯ π‘₯ ) βˆ’ 3 6  s i n c o s
  • E 𝑦 = 2 ( π‘₯ βˆ’ π‘₯ π‘₯ ) + 3 6  s i n c o s

Q12:

Find the solution of the differential equation π‘₯ π‘₯ = 𝑦 ο€» 1 + √ 1 + 3 𝑦  𝑦 β€² l n  that satisfies the initial condition 𝑦 ( 1 ) = βˆ’ 1 .

  • A 1 2 π‘₯ π‘₯ + 1 4 π‘₯ + 4 1 3 6 = 1 9 ο€Ή 3 𝑦 + 1  + 1 2 𝑦     l n  
  • B 1 2 π‘₯ π‘₯ βˆ’ 1 4 π‘₯ = 1 9 ο€Ή 3 𝑦 + 1  + 1 2 𝑦     l n  
  • C 1 2 π‘₯ π‘₯ βˆ’ 1 4 π‘₯ + 5 9 3 6 = 1 9 ο€Ή 3 𝑦 + 1  + 1 2 𝑦     l n  
  • D 1 2 π‘₯ π‘₯ βˆ’ 1 4 π‘₯ + 5 9 3 6 = 2 3 ο€Ή 3 𝑦 + 1  + 1 2 𝑦     l n  
  • E 1 2 π‘₯ π‘₯ + 1 4 π‘₯ = 1 9 ο€Ή 3 𝑦 + 1  + 1 2 𝑦     l n  

Q13:

Find the solution for the following differential equation for 𝑦 ( 2 ) = 1 : 𝑦 𝑦 + π‘₯ π‘₯ = 3 π‘₯ 𝑦 π‘₯ . d d d 

  • A 𝑦 = ο„ž 1 + 2 𝑒 3     
  • B 𝑦 = ο„ž 1 + 2 𝑒 3      
  • C 𝑦 = ο„ž 1 + 2 𝑒 3      
  • D 𝑦 = ο„ž 1 + 2 𝑒 3       

Q14:

Find the solution of the differential equation π‘₯ + 𝑦 √ π‘₯ + 3 𝑦 π‘₯ = 0   d d that satisfies the initial condition 𝑦 ( βˆ’ 1 ) = βˆ’ 2 .

  • A 𝑦 = βˆ’  3 √ π‘₯ + 3 + 1 4  
  • B 𝑦 = βˆ’  3 √ π‘₯ + 3 + 2  
  • C 𝑦 = βˆ’ √ π‘₯ + 3 οŽ₯ 
  • D 𝑦 = βˆ’  √ π‘₯ + 3 + 6  
  • E 𝑦 = βˆ’  3 √ π‘₯ + 3  

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