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Lesson: Initial Value Problems of Differential Equations

In this lesson, we will learn how to solve initial value problems of differential equations.

Worksheet • 22 Questions

Q1:

Find a particular solution for the following differential equation for which 𝑦 ( 0 ) = 1 2 :

  • A 𝑦 = 4 π‘₯ + 3 π‘₯ + 1 2 2
  • B 𝑦 = 8 π‘₯ + 3 π‘₯ + 1 2 2
  • C 𝑦 = 8 π‘₯ + 3 π‘₯ 2
  • D 𝑦 = 4 π‘₯ + 3 π‘₯ 2

Q2:

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

  • A 𝑒 + 𝑒 = 2 π‘₯ βˆ’ 𝑦
  • B 𝑒 + 𝑒 = 2 π‘₯ 𝑦
  • C 𝑒 + 𝑒 = 2 βˆ’ π‘₯ βˆ’ 𝑦
  • D 𝑒 + 𝑒 = 2 βˆ’ π‘₯ 𝑦

Q3:

Find the solution for the following differential equation for 𝑦 ( 1 ) = 1 :

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

Q4:

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

  • A 𝑦 = | | π‘₯ + 2 | | l n 2
  • B 𝑦 = | π‘₯ + 2 | l n
  • C 𝑦 = | | π‘₯ 2 + 2 | | l n
  • D 𝑦 = | 2 π‘₯ + 2 | l n

Q5:

Find the particular solution for the following separable differential equation: c o s d d ( 𝑦 ) 𝑦 π‘₯ βˆ’ π‘₯ = 0 , 𝑦 ( 0 ) = 0 .

  • A 𝑦 = ο€Ύ π‘₯ 2  s i n βˆ’ 1 2
  • B 𝑦 = ο€Ύ π‘₯ 2  c o s 2
  • C 𝑦 = π‘₯ 2 2
  • D 𝑦 = ο€Ύ π‘₯ 2  s i n 2

Q6:

Find the solution for the following differential equation for 𝑦 ( 0 ) = 1 6 :

  • A 𝑦 = 5 + 1 1 𝑒 4 π‘₯
  • B 𝑦 = 5 + 1 1 𝑒 π‘₯
  • C 𝑦 = 5 + 1 1 𝑒 π‘₯ 4
  • D 𝑦 = 5 + 1 1 𝑒 βˆ’ 4 π‘₯

Q7:

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

  • A 𝑦 = 1 + 𝑒 π‘₯
  • B 𝑦 = ( 2 𝑒 + 2 ) π‘₯ 1 2
  • C 𝑦 = ( βˆ’ 2 𝑒 + 2 ) π‘₯ 1 2
  • D 𝑦 = ( βˆ’ 2 𝑒 + 2 ) βˆ’ π‘₯ 1 2

Q8:

Find the solution for the following differential equation for 𝑦 ( 0 ) = 1 :

  • A 𝑦 = π‘₯ 2 + π‘₯ 3 + 1 2 3
  • B 𝑦 = 2 π‘₯ + 3 π‘₯ + 1 2 3
  • C 𝑦 = βˆ’ 2 π‘₯ βˆ’ 3 π‘₯ + 1 2 3
  • D 𝑦 = βˆ’ π‘₯ 2 βˆ’ π‘₯ 3 + 1 2 3

Q9:

For a circuit containing a capacitor and a resistor, the first-order differential equation that describes a discharging capacitor is If 𝑄 0 represents the charge within the capacitor at 𝑑 = 0 s e c o n d s , find the general solution. 𝐢 is the capacitance of the capacitor. 𝑅 is the resistance of the resistor.

  • A 𝑄 = 𝑄 𝑒 0 βˆ’ 𝑑 𝑅 𝐢
  • B 𝑄 = 𝑄 𝑒 0 βˆ’ 𝑅 𝐢 𝑑
  • C 𝑄 = 𝑄 𝑒 0 𝑅 𝐢 𝑑
  • D 𝑄 = 𝑄 𝑒 0 𝑑 𝑅 𝐢

Q10:

Suppose that d d s i n c o s 𝑦 π‘₯ = 5 2 π‘₯ ο€» 𝑦 3  2 and 𝑦 = 3 πœ‹ 4 when π‘₯ = πœ‹ 6 . Find 𝑦 in terms of π‘₯ .

  • A 3 ο€» 𝑦 3  = βˆ’ 5 2 2 π‘₯ + 1 7 4 t a n c o s
  • B 1 3 ο€» 𝑦 3  = βˆ’ 5 2 2 π‘₯ βˆ’ 1 7 4 t a n c o s
  • C t a n c o s ο€» 𝑦 3  = 2 π‘₯ + 1 7 4
  • D 1 3 ο€» 𝑦 3  = 2 π‘₯ βˆ’ 7 4 t a n s i n
  • E 1 3 ο€» 𝑦 3  = 2 π‘₯ βˆ’ 7 4 t a n c o s

Q11:

Find the solution of the differential equation d d s e c 𝑒 𝑑 = 𝑑 + 𝑑 𝑒 2 that satisfies the initial condition 𝑒 ( 0 ) = βˆ’ 3 .

  • A 𝑒 = 𝑑 + 2 𝑑 + 9 2 2 t a n
  • B 𝑒 = 𝑑 + 2 𝑑 βˆ’ 9 2 2 t a n
  • C 𝑒 = 𝑑 βˆ’ 2 𝑑 + 9 2 2 t a n
  • D 𝑒 = 𝑑 + 2 𝑑 2 2 t a n
  • E 𝑒 = 𝑑 + 𝑑 + 9 2 2 t a n

Q12:

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

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

Q13:

Solve the differential equation d d 𝑦 π‘₯ ο€Ή π‘₯ + 4  = 3  for 𝑦 given that 𝑦 ( 2 ) = 0 .

  • A 𝑦 = 3 2 ο€» π‘₯ 2  βˆ’ 3 πœ‹ 8 t a n  
  • B 𝑦 = 3 ο€» π‘₯ 2  + 3 πœ‹ 8 t a n  
  • C 𝑦 = 3 ο€» π‘₯ 2  βˆ’ 3 πœ‹ 8 t a n  
  • D 𝑦 = 3 2 ο€» π‘₯ 2  + 3 πœ‹ 8 t a n  
  • E 𝑦 = 3 2 ( π‘₯ ) βˆ’ 3 πœ‹ 8 t a n  

Q14:

The gradient of the tangent to a curve is ο„ž βˆ’ 𝑦 + 3 4 π‘₯ βˆ’ 4 . Find the equation of the curve given that the curve passes through the point ( 5 , 3 ) .

  • A βˆ’ 2 √ βˆ’ 𝑦 + 3 = 1 2 √ 4 π‘₯ βˆ’ 4 βˆ’ 2
  • B 2 √ βˆ’ 𝑦 + 3 = 2 √ 4 π‘₯ βˆ’ 4 βˆ’ 2
  • C βˆ’ √ βˆ’ 𝑦 + 3 = 1 4 √ 4 π‘₯ βˆ’ 4 βˆ’ 2
  • D βˆ’ 2 √ βˆ’ 𝑦 + 3 = 8 √ 4 π‘₯ βˆ’ 4 βˆ’ 2

Q15:

Find the solution of the differential equation 𝑦 β€² π‘₯ = π‘Ž + 𝑦 t a n , where 0 < π‘₯ < πœ‹ 2 , that satisfies the initial condition 𝑦 ο€» πœ‹ 3  = π‘Ž .

  • A 𝑦 = 4 π‘Ž √ 3 π‘₯ βˆ’ π‘Ž s i n
  • B 𝑦 = 4 π‘Ž π‘₯ βˆ’ π‘Ž s i n
  • C 𝑦 = 4 π‘Ž π‘₯ βˆ’ π‘Ž c o s
  • D 𝑦 = π‘Ž βˆ’ 4 π‘Ž π‘₯ c o s
  • E 𝑦 = π‘Ž βˆ’ 4 π‘Ž √ 3 π‘₯ s i n

Q16:

Solve the differential equation d d 𝑦 π‘₯ √ π‘₯ βˆ’ 9 = 1  for 𝑦 given that 𝑦 ( 5 ) = 3 l n .

  • A 𝑦 = ο€» π‘₯ + √ π‘₯ βˆ’ 9  βˆ’ 3 l n l n 
  • B 𝑦 = βˆ’ 1 3 √ π‘₯ βˆ’ 9 π‘₯ + 1 3 4 5 l n l n 
  • C 𝑦 = ο€» π‘₯ + √ π‘₯ βˆ’ 9  + 3 l n l n 
  • D 𝑦 = ο€» π‘₯ + √ π‘₯ βˆ’ 9  βˆ’ 2 3 l n l n 
  • E 𝑦 = 1 3 √ π‘₯ βˆ’ 9 π‘₯ + 1 3 4 5 l n l n 

Q17:

Find the solution of the differential equation d d l n 𝐿 𝑑 = π‘˜ 𝐿 𝑑 2 that satisfies the initial condition 𝐿 ( 1 ) = βˆ’ 1 .

  • A 𝐿 = βˆ’ 1 π‘˜ 𝑑 𝑑 βˆ’ π‘˜ 𝑑 + 1 + π‘˜ l n
  • B 𝐿 = βˆ’ 1 π‘˜ 𝑑 𝑑 + π‘˜ 𝑑 + 1 l n
  • C 𝐿 = 1 π‘˜ 𝑑 𝑑 + π‘˜ 𝑑 + 1 βˆ’ π‘˜ l n
  • D 𝐿 = 1 π‘˜ 𝑑 𝑑 βˆ’ π‘˜ 𝑑 + 1 + π‘˜ l n
  • E 𝐿 = 1 π‘˜ 𝑑 𝑑 + π‘˜ 𝑑 + 1 l n

Q18:

Solve the differential equation π‘₯ 𝑦 π‘₯ = √ π‘₯ βˆ’ 4 d d  for 𝑦 given that 𝑦 ( 2 ) = 0 .

  • A 𝑦 = √ π‘₯ βˆ’ 4 βˆ’ 2 ο€» π‘₯ 2     s e c
  • B 𝑦 = 2 π‘₯ βˆ’ 1
  • C 𝑦 = √ π‘₯ βˆ’ 4 βˆ’ ( π‘₯ )    s e c
  • D 𝑦 = βˆ’ 2 π‘₯ + 1
  • E 𝑦 = √ π‘₯ βˆ’ 4 + 2 ο€» π‘₯ 2     s e c

Q19:

Find the solution for the following differential equation for 𝑦 ( 2 ) = 1 :

  • A 𝑦 = ο„ž 1 + 2 𝑒 3 3 π‘₯ βˆ’ 1 2 2
  • B 𝑦 = ο„ž 1 + 2 𝑒 3 3 π‘₯ βˆ’ 1 2
  • C 𝑦 = ο„ž 1 + 2 𝑒 3 βˆ’ 3 π‘₯ βˆ’ 1 2 2
  • D 𝑦 = ο„ž 1 + 2 𝑒 3 βˆ’ 3 π‘₯ βˆ’ 1 2

Q20:

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

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

Q21:

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

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

Q22:

The gradient of the tangent of a curve is βˆ’ 4 π‘₯ + 4 3 𝑦 + 3 and the curve passes through the point ( βˆ’ 2 , βˆ’ 3 ) . Find the equation of the normal to the curve at the point where the π‘₯ -coordinate is βˆ’ 2 .

  • A 2 𝑦 + π‘₯ = 0 , 2 𝑦 βˆ’ π‘₯ + 4 = 0
  • B 2 𝑦 βˆ’ π‘₯ βˆ’ 4 = 0 , 2 𝑦 + π‘₯ + 8 = 0
  • C 𝑦 + 2 π‘₯ + 3 = 0 , 𝑦 βˆ’ 2 π‘₯ βˆ’ 1 = 0
  • D 𝑦 βˆ’ 2 π‘₯ βˆ’ 5 = 0 , 𝑦 + 2 π‘₯ + 7 = 0
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