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Worksheet: Separable Differential Equations

In this worksheet, we will practice solving separable differential equations.

Q1:

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

Q2:

Solve the differential equation d d 𝑦 π‘₯ = βˆ’ 5 π‘₯ 𝑦 2 2 .

  • A 𝑦 = 1 5 π‘₯ + 3 C or 𝑦 = 0
  • B 𝑦 = βˆ’ 3 5 π‘₯ + 3 C or 𝑦 = 0
  • C 𝑦 = 1 1 5 π‘₯ + 3 C or 𝑦 = 0
  • D 𝑦 = 3 5 π‘₯ + 3 C or 𝑦 = 0
  • E 𝑦 = βˆ’ 1 1 5 π‘₯ + 3 C or 𝑦 = 0

Q3:

Solve the differential equation d d l n 𝐻 𝑅 = 𝑅 𝐻 √ 1 + 𝑅 𝐻 2 2 .

  • A βˆ’ 𝐻 𝐻 βˆ’ 1 𝐻 = 1 2 ο€Ή 1 + 𝑅  + l n C 2 3 2
  • B βˆ’ 𝐻 𝐻 βˆ’ 1 𝐻 = 2 3 ο€Ή 1 + 𝑅  + l n C 2 3 2
  • C l n C 𝐻 𝐻 + 1 𝐻 = 1 3 ο€Ή 1 + 𝑅  + 2 3 2
  • D βˆ’ 𝐻 𝐻 βˆ’ 1 𝐻 = 1 3 ο€Ή 1 + 𝑅  + l n C 2 3 2
  • E βˆ’ 𝐻 𝐻 + 1 𝐻 = 1 3 ο€Ή 1 + 𝑅  + l n C 2 3 2

Q4:

Find a relation between 𝑦 and π‘₯ , given that π‘₯ 𝑦 𝑦 β€² = π‘₯ βˆ’ 5 2 .

  • A 𝑦 = 2 π‘₯ βˆ’ 1 0 | π‘₯ | + 2 2 l n C
  • B 𝑦 = π‘₯ βˆ’ 5 | π‘₯ | + 2 2 l n C
  • C 𝑦 = π‘₯ 2 βˆ’ 5 | π‘₯ | + 2 2 l n C
  • D 𝑦 = π‘₯ βˆ’ 1 0 | π‘₯ | + 2 2 l n C
  • E 𝑦 = 2 π‘₯ βˆ’ 1 0 π‘₯ + 2 2 l n C

Q5:

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

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

Q7:

Solve the differential equation d d 𝑦 π‘₯ + 𝑦 = 1 .

  • A 𝑦 = π‘₯ + 𝑒 C βˆ’ π‘₯
  • B 𝑦 = π‘₯ 𝑒 + 𝑒 βˆ’ π‘₯ βˆ’ π‘₯ C
  • C 𝑦 = 1 + 𝑒 C π‘₯
  • D 𝑦 = 1 + 𝑒 C βˆ’ π‘₯
  • E 𝑦 = π‘₯ + 𝑒 C π‘₯

Q8:

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

Q9:

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

Q10:

Solve the following differential equation: ( 𝑒 βˆ’ 5 ) 𝑦 β€² = 2 + π‘₯ 𝑦 c o s .

  • A 𝑒 βˆ’ 5 𝑦 = π‘₯ + π‘₯ + 𝑦 s i n C
  • B 𝑒 βˆ’ 5 𝑦 = 2 π‘₯ βˆ’ π‘₯ + 𝑦 s i n C
  • C 𝑒 βˆ’ 5 = 2 π‘₯ + π‘₯ + 𝑦 s i n C
  • D 𝑒 βˆ’ 5 𝑦 = 2 π‘₯ + π‘₯ + 𝑦 s i n C
  • E 𝑒 βˆ’ 5 = 2 π‘₯ βˆ’ π‘₯ + 𝑦 s i n C

Q11:

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

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

Q12:

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

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

Q13:

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

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

Q14:

Find the solution of the differential equation d d 𝑦 π‘₯ = π‘₯ 𝑒 𝑦 that satisfies the initial condition 𝑦 ( 2 ) = 0 .

  • A 𝑦 = βˆ’ ο€Ό 1 2 π‘₯ βˆ’ 1  l n 2
  • B 𝑦 = βˆ’ ο€Ή βˆ’ π‘₯ + 5  l n 2
  • C 𝑦 = βˆ’ ο€Ό βˆ’ 1 2 π‘₯ + 2  l n 2
  • D 𝑦 = βˆ’ ο€Ό βˆ’ 1 2 π‘₯ + 3  l n 2
  • E 𝑦 = βˆ’ ο€Ή βˆ’ π‘₯ + 4  l n 2

Q15:

Solve the following differential equation: d d 𝑝 𝑑 = 𝑑 𝑝 βˆ’ 5 𝑝 + 𝑑 βˆ’ 5 2 2 .

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

Q16:

Solve the differential equation d d s e c πœƒ 𝑑 = 𝑑 πœƒ πœƒ 𝑒 𝑑 2 .

  • A πœƒ πœƒ = βˆ’ 𝑒 2 + s i n C βˆ’ 𝑑 2
  • B πœƒ πœƒ + πœƒ = 𝑒 + s i n c o s C βˆ’ 𝑑 2
  • C βˆ’ πœƒ πœƒ βˆ’ πœƒ = βˆ’ 𝑒 2 + s i n c o s C βˆ’ 𝑑 2
  • D πœƒ πœƒ + πœƒ = βˆ’ 𝑒 2 + s i n c o s C βˆ’ 𝑑 2
  • E βˆ’ πœƒ πœƒ βˆ’ πœƒ = 𝑒 + s i n c o s C βˆ’ 𝑑 2

Q17:

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

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