Worksheet: Initial Value Problems of Differential Equations

In this worksheet, we will practice solving initial value problems of differential equations.

Q1:

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

  • A 𝑦 = 4 𝑥 + 3 𝑥 2
  • B 𝑦 = 8 𝑥 + 3 𝑥 + 1 2 2
  • C 𝑦 = 8 𝑥 + 3 𝑥 2
  • D 𝑦 = 4 𝑥 + 3 𝑥 + 1 2 2

Q2:

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

  • A 𝑒 + 𝑒 = 2 𝑥 𝑦
  • B 𝑒 + 𝑒 = 2 𝑥 𝑦
  • C 𝑒 + 𝑒 = 2 𝑥 𝑦
  • D 𝑒 + 𝑒 = 2 𝑥 𝑦

Q3:

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

  • A 𝑦 = 𝑒 1 + 𝑥
  • B 𝑦 = 𝑒 𝑥 1
  • C 𝑦 = 𝑒 1 𝑥
  • D 𝑦 = 𝑒 1 𝑥

Q4:

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

  • A 𝑦 = | 2 𝑥 + 2 | l n
  • B 𝑦 = | 𝑥 + 2 | l n
  • C 𝑦 = | | 𝑥 2 + 2 | | l n
  • D 𝑦 = | | 𝑥 + 2 | | l n 2

Q5:

Find the particular solution for the following separable differential equation: c o s d d ( 𝑦 ) 𝑦 𝑥 𝑥 = 0 , 𝑦 ( 0 ) = 0 .

  • A 𝑦 = 𝑥 2 s i n 2
  • B 𝑦 = 𝑥 2 c o s 2
  • C 𝑦 = 𝑥 2 2
  • D 𝑦 = 𝑥 2 s i n 1 2

Q6:

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

  • A 𝑦 = 5 + 1 1 𝑒 4 𝑥
  • B 𝑦 = 5 + 1 1 𝑒 𝑥
  • C 𝑦 = 5 + 1 1 𝑒 𝑥 4
  • D 𝑦 = 5 + 1 1 𝑒 4 𝑥

Q7:

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

  • A 𝑦 = ( 2 𝑒 + 2 ) 𝑥 1 2
  • B 𝑦 = ( 2 𝑒 + 2 ) 𝑥 1 2
  • C 𝑦 = ( 2 𝑒 + 2 ) 𝑥 1 2
  • D 𝑦 = 1 + 𝑒 𝑥

Q8:

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

  • A 𝑦 = 𝑥 2 𝑥 3 + 1 2 3
  • B 𝑦 = 2 𝑥 + 3 𝑥 + 1 2 3
  • C 𝑦 = 2 𝑥 3 𝑥 + 1 2 3
  • D 𝑦 = 𝑥 2 + 𝑥 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 𝑡 = 0 s e c o n d s , find the general solution. 𝐶 is the capacitance of the capacitor. 𝑅 is the resistance of the resistor.

  • A 𝑄 = 𝑄 𝑒 0 𝑡 𝑅 𝐶
  • B 𝑄 = 𝑄 𝑒 0 𝑅 𝐶 𝑡
  • C 𝑄 = 𝑄 𝑒 0 𝑅 𝐶 𝑡
  • D 𝑄 = 𝑄 𝑒 0 𝑡 𝑅 𝐶

Q10:

Suppose that d d s i n c o s 𝑦 𝑥 = 5 2 𝑥 𝑦 3 2 and 𝑦 = 3 𝜋 4 when 𝑥 = 𝜋 6 . Find 𝑦 in terms of 𝑥 .

  • A 1 3 𝑦 3 = 2 𝑥 7 4 t a n c o s
  • B t a n c o s 𝑦 3 = 2 𝑥 + 1 7 4
  • C 1 3 𝑦 3 = 2 𝑥 7 4 t a n s i n
  • D 3 𝑦 3 = 5 2 2 𝑥 + 1 7 4 t a n c o s
  • E 1 3 𝑦 3 = 5 2 2 𝑥 1 7 4 t a n c o s

Q11:

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

  • A 𝑢 = 𝑡 + 𝑡 + 9 2 2 t a n
  • B 𝑢 = 𝑡 2 𝑡 + 9 2 2 t a n
  • C 𝑢 = 𝑡 + 2 𝑡 2 2 t a n
  • D 𝑢 = 𝑡 + 2 𝑡 + 9 2 2 t a n
  • E 𝑢 = 𝑡 + 2 𝑡 9 2 2 t a n

Q12:

Find the solution of the differential equation 𝑥 𝑥 = 𝑦 1 + 1 + 3 𝑦 𝑦 l n 2 that satisfies the initial condition 𝑦 ( 1 ) = 1 .

  • A 1 2 𝑥 𝑥 1 4 𝑥 = 1 9 3 𝑦 + 1 + 1 2 𝑦 2 2 2 2 l n 3 2
  • B 1 2 𝑥 𝑥 + 1 4 𝑥 + 4 1 3 6 = 1 9 3 𝑦 + 1 + 1 2 𝑦 2 2 2 2 l n 3 2
  • C 1 2 𝑥 𝑥 + 1 4 𝑥 = 1 9 3 𝑦 + 1 + 1 2 𝑦 2 2 2 2 l n 3 2
  • D 1 2 𝑥 𝑥 1 4 𝑥 + 5 9 3 6 = 1 9 3 𝑦 + 1 + 1 2 𝑦 2 2 2 2 l n 3 2
  • E 1 2 𝑥 𝑥 1 4 𝑥 + 5 9 3 6 = 2 3 3 𝑦 + 1 + 1 2 𝑦 2 2 2 2 l n 3 2

Q13:

Solve the differential equation d d 𝑦 𝑥 𝑥 + 4 = 3 for 𝑦 given that 𝑦 ( 2 ) = 0 .

  • A 𝑦 = 3 2 ( 𝑥 ) 3 𝜋 8 t a n
  • B 𝑦 = 3 𝑥 2 3 𝜋 8 t a n
  • C 𝑦 = 3 2 𝑥 2 + 3 𝜋 8 t a n
  • D 𝑦 = 3 2 𝑥 2 3 𝜋 8 t a n
  • E 𝑦 = 3 𝑥 2 + 3 𝜋 8 t a n

Q14:

The gradient of the tangent to a curve is 𝑦 + 3 4 𝑥 4 . Find the equation of the curve given that the curve passes through the point ( 5 , 3 ) .

  • A 2 𝑦 + 3 = 8 4 𝑥 4 2
  • B 2 𝑦 + 3 = 2 4 𝑥 4 2
  • C 𝑦 + 3 = 1 4 4 𝑥 4 2
  • D 2 𝑦 + 3 = 1 2 4 𝑥 4 2

Q15:

Find the solution of the differential equation 𝑦 𝑥 = 𝑎 + 𝑦 t a n , where 0 < 𝑥 < 𝜋 2 , that satisfies the initial condition 𝑦 𝜋 3 = 𝑎 .

  • A 𝑦 = 𝑎 4 𝑎 3 𝑥 s i n
  • B 𝑦 = 4 𝑎 𝑥 𝑎 c o s
  • C 𝑦 = 𝑎 4 𝑎 𝑥 c o s
  • D 𝑦 = 4 𝑎 3 𝑥 𝑎 s i n
  • E 𝑦 = 4 𝑎 𝑥 𝑎 s i n

Q16:

Solve the differential equation d d 𝑦 𝑥 𝑥 9 = 1 for 𝑦 given that 𝑦 ( 5 ) = 3 l n .

  • A 𝑦 = 1 3 𝑥 9 𝑥 + 1 3 4 5 l n l n
  • B 𝑦 = 𝑥 + 𝑥 9 + 3 l n l n
  • C 𝑦 = 𝑥 + 𝑥 9 2 3 l n l n
  • D 𝑦 = 𝑥 + 𝑥 9 3 l n l n
  • E 𝑦 = 1 3 𝑥 9 𝑥 + 1 3 4 5 l n l n

Q17:

Find the solution of the differential equation d d l n 𝐿 𝑡 = 𝑘 𝐿 𝑡 2 that satisfies the initial condition 𝐿 ( 1 ) = 1 .

  • A 𝐿 = 1 𝑘 𝑡 𝑡 + 𝑘 𝑡 + 1 l n
  • B 𝐿 = 1 𝑘 𝑡 𝑡 + 𝑘 𝑡 + 1 𝑘 l n
  • C 𝐿 = 1 𝑘 𝑡 𝑡 𝑘 𝑡 + 1 + 𝑘 l n
  • D 𝐿 = 1 𝑘 𝑡 𝑡 𝑘 𝑡 + 1 + 𝑘 l n
  • E 𝐿 = 1 𝑘 𝑡 𝑡 + 𝑘 𝑡 + 1 l n

Q18:

Solve the differential equation 𝑥 𝑦 𝑥 = 𝑥 4 d d for 𝑦 given that 𝑦 ( 2 ) = 0 .

  • A 𝑦 = 𝑥 4 + 2 𝑥 2 s e c
  • B 𝑦 = 𝑥 4 ( 𝑥 ) s e c
  • C 𝑦 = 2 𝑥 + 1
  • D 𝑦 = 𝑥 4 2 𝑥 2 s e c
  • E 𝑦 = 2 𝑥 1

Q19:

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

  • A 𝑦 = 1 + 2 𝑒 3 3 𝑥 1 2
  • B 𝑦 = 1 + 2 𝑒 3 3 𝑥 1 2
  • C 𝑦 = 1 + 2 𝑒 3 3 𝑥 1 2 2
  • D 𝑦 = 1 + 2 𝑒 3 3 𝑥 1 2 2

Q20:

Find the solution of the differential equation 𝑥 + 𝑦 𝑥 + 3 𝑦 𝑥 = 0 2 2 d d that satisfies the initial condition 𝑦 ( 1 ) = 2 .

  • A 𝑦 = 3 𝑥 + 3 + 1 4 3 2
  • B 𝑦 = 𝑥 + 3 + 6 3 2
  • C 𝑦 = 3 𝑥 + 3 3 2
  • D 𝑦 = 3 𝑥 + 3 + 2 3 2
  • E 𝑦 = 𝑥 + 3 6 2

Q21:

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

  • A ( 𝑦 + 1 ) ( 1 𝑥 ) = 𝑒 𝑥
  • B ( 𝑦 1 ) ( 1 + 𝑥 ) = 𝑒 𝑥
  • C ( 𝑦 1 ) ( 1 + 𝑥 ) = 𝑒 𝑥
  • D ( 𝑦 + 1 ) ( 1 𝑥 ) = 𝑒 𝑥

Q22:

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

  • A 𝑦 2 𝑥 5 = 0 , 𝑦 + 2 𝑥 + 7 = 0
  • B 2 𝑦 𝑥 4 = 0 , 2 𝑦 + 𝑥 + 8 = 0
  • C 𝑦 + 2 𝑥 + 3 = 0 , 𝑦 2 𝑥 1 = 0
  • D 2 𝑦 + 𝑥 = 0 , 2 𝑦 𝑥 + 4 = 0

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