Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Start Practicing

Worksheet: Separable Differential Equations

Q1:

Solve the differential equation 𝑦 + π‘₯ 𝑒 = 0 β€² 𝑦 .

  • A 𝑦 = βˆ’ ο€Ή π‘₯ +  l n C 2
  • B 𝑦 = ο€Ύ π‘₯ 2 +  l n C 2
  • C 𝑦 = βˆ’ ο€Ή 2 π‘₯ +  l n C 2
  • D 𝑦 = βˆ’ ο€Ύ π‘₯ 2 +  l n C 2
  • E 𝑦 = ο€Ή 2 π‘₯ +  l n C 2

Q2:

The logistic population model assumes an upper limit 𝐿 , beyond which growth cannot occur. The population 𝑦 ( 𝑑 ) has a rate of change that satisfies for some positive constant π‘˜ . A suitable function 𝑦 ( 𝑑 ) involves a second parameter 𝑏 , determined by how rapid the initial growth is. Without integrating, which of the following could be 𝑦 ( 𝑑 ) ?

  • A 𝑏 𝐿 𝑒 βˆ’ 1   
  • B 𝐿 𝑏 𝑒 βˆ’ 1   
  • C 𝑏 1 + 𝐿 𝑒   
  • D 𝐿 1 + 𝑏 𝑒   
  • E 𝐿 𝑏 𝑒 βˆ’ 1  

Q3:

Solve the differential equation d d 𝑦 π‘₯ = βˆ’ 5 π‘₯ √ 𝑦 .

  • A √ 𝑦 = βˆ’ 5 π‘₯ + 2 C or 𝑦 = 0
  • B √ 𝑦 = βˆ’ 5 π‘₯ 2 + 2 C or 𝑦 = 0
  • C 𝑦 = ο€Ύ βˆ’ 5 π‘₯ 4 +  2 2 C or 𝑦 = 0
  • D √ 𝑦 = βˆ’ 5 π‘₯ 4 + 2 C or 𝑦 = 0
  • E 𝑦 = ο€Ύ βˆ’ 5 π‘₯ 2 +  2 2 C or 𝑦 = 0

Q4:

Suppose a population’s growth is governed by the logistic equation d d 𝑃 𝑑 = 0 . 0 7 𝑃 ο€Ό 1 βˆ’ 𝑃 9 0 0  , where 𝑃 ( 0 ) = 5 0 . Write the formula for 𝑃 ( 𝑑 ) .

  • A 𝑃 ( 𝑑 ) = 9 0 0 1 + 1 7 𝑒 0 . 0 7 𝑑
  • B 𝑃 ( 𝑑 ) = 9 0 0 1 9 βˆ’ 𝑒 βˆ’ 0 . 0 7 𝑑
  • C 𝑃 ( 𝑑 ) = 9 0 0 1 7 + 𝑒 βˆ’ 0 . 0 7 𝑑
  • D 𝑃 ( 𝑑 ) = 9 0 0 1 + 1 7 𝑒 βˆ’ 0 . 0 7 𝑑
  • E 𝑃 ( 𝑑 ) = 9 0 0 1 9 βˆ’ 𝑒 0 . 0 7 𝑑

Q5:

Suppose a population’s growth is governed by the logistic equation d d 𝑃 𝑑 = 0 . 0 6 𝑃 ο€Ό 1 βˆ’ 𝑃 8 0 0  , where 𝑃 ( 0 ) = 8 0 . Write the formula for 𝑃 ( 𝑑 ) .

  • A 𝑃 ( 𝑑 ) = 8 0 0 1 + 9 𝑒 0 . 0 6 𝑑
  • B 𝑃 ( 𝑑 ) = 8 0 0 1 1 βˆ’ 𝑒 βˆ’ 0 . 0 6 𝑑
  • C 𝑃 ( 𝑑 ) = 8 0 0 9 + 𝑒 βˆ’ 0 . 0 6 𝑑
  • D 𝑃 ( 𝑑 ) = 8 0 0 1 + 9 𝑒 βˆ’ 0 . 0 6 𝑑
  • E 𝑃 ( 𝑑 ) = 8 0 0 1 1 βˆ’ 𝑒 0 . 0 6 𝑑

Q6:

Solve the following differential equation by separating it:

  • A 𝑦 = ( | π‘₯ | + ) c o s l n C
  • B 𝑦 = ( | π‘₯ | + ) l n c o s C
  • C 𝑦 = ( | π‘₯ | + ) l n s i n C
  • D 𝑦 = ( | π‘₯ | + ) s i n l n C

Q7:

Unlike exponential growth, where a population grows without bound, the logistic model assumes an upper limit 𝐿 , beyond which growth cannot occur. The population 𝑦 ( 𝑑 ) has a rate of change that satisfies for some positive constant π‘˜ . Given a population with 𝑦 ( 0 ) = 𝐿 2 , at what population is the growth zero?

  • A 𝐿 2
  • B0
  • CIt cannot be determined.
  • D 𝐿
  • Enever

Q8:

Find a 1-parameter family of solutions for the following differential equation:

  • A 1 𝑦 + 1 βˆ’ | 𝑦 + 1 | = π‘₯ + 𝑐 l o g
  • B βˆ’ 1 𝑦 + 1 + | 𝑦 + 1 | = π‘₯ + 𝑐 l o g
  • CThere is no solution.
  • D 1 𝑦 + 1 + | 𝑦 + 1 | = π‘₯ + 𝑐 l o g

Q9:

Find the solution of the differential equation given that 𝑦 ( 0 ) = 8 .

  • A 𝑦 = 7 + 𝑒 9 π‘₯
  • B 𝑦 = 7 + 𝑒 βˆ’ π‘₯
  • C 𝑦 = 7 + 𝑒 π‘₯ 9
  • D 𝑦 = 7 + 𝑒 βˆ’ 9 π‘₯

Q10:

Find the implicit solution to the following differential equation:

  • A s i n c o s ( 𝑦 ) + ( π‘₯ ) = 𝐢
  • B c o s s e c ( 𝑦 ) + ( π‘₯ ) = 𝐢
  • C c o s c s c ( 𝑦 ) + ( π‘₯ ) = 𝐢
  • D c o s s i n ( 𝑦 ) + ( π‘₯ ) = 𝐢

Q11:

Which of the following is a solution of π‘₯ + 𝑦 𝑦 β€² = 0 defined for all βˆ’ 4 < π‘₯ < 4 ?

  • A 𝑦 = √ 1 6 + π‘₯ 2
  • B 𝑦 = √ 4 βˆ’ π‘₯ 2
  • C 𝑦 = √ 4 + π‘₯ 2
  • D 𝑦 = √ 1 6 βˆ’ π‘₯ 2

Q12:

Find a relation between 𝑒 and 𝑑 given that d d 𝑒 𝑑 = 1 + 𝑑 𝑒 𝑑 + 𝑒 𝑑 4 2 4 2 .

  • A 𝑒 5 + 𝑒 2 = βˆ’ 1 𝑑 + 𝑑 + 5 2 3 C
  • B 𝑒 5 + 𝑒 2 = 1 𝑑 + 𝑑 3 + 5 2 3 C
  • C 𝑒 + 𝑒 = βˆ’ 1 𝑑 + 𝑑 3 + 5 2 3 C
  • D 𝑒 5 + 𝑒 2 = βˆ’ 1 𝑑 + 𝑑 3 + 5 2 3 C
  • E 𝑒 + 𝑒 = 1 𝑑 + 𝑑 3 + 5 2 3 C

Q13:

Find the solution for the following differential equation for 𝑦 ( 0 ) = 2 :

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

Q14:

Solve the differential equation d d 𝑧 𝑑 + 𝑒 = 0 2 𝑑 + 2 𝑧 .

  • A 𝑧 = βˆ’ 1 2 ο€Ύ 𝑒 2 +  l n C 2 𝑑
  • B 𝑧 = βˆ’ 1 2 ο€Ή 2 𝑒 +  l n C 2 𝑑
  • C 𝑧 = 1 2 ο€Ή 𝑒 +  l n C 2 𝑑
  • D 𝑧 = βˆ’ 1 2 ο€Ή 𝑒 +  l n C 2 𝑑

Q15:

Suppose a population grows according to a logistic model with an initial population of 1 000 and a carrying capacity of 10 000. If the population grows to 2 500 after one year, what will the population be after another three years?

Q16:

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