FINDING PARTICULAR SOLUTION OF DIFFERENTIAL EQUATIONS GIVEN CONDITIONS

Problem 1 :

Solve the initial value problem

Solution :

Problem 2 :

Solve the initial value problem

Solution :

Problem 3 :

Solution :

Problem 4 :

Solution :

Problem 5 :

Given the velocity 

V = ds/dt = 32t - 2

and the initial position of the body as s(1/2) = 4. Find the body's position at the time t.

Solution :

ds/dt = 32t - 2

ds =  ∫(32t - 2) dt

s = 32t²/2 - 2t + C

s = 16t² - 2t + C ----(1)

When t = 1/2, s = 4

4 = 16(1/2)² - 2(1/2) + C

4 = 4 - 1 + C

C = 1

Applying the value of C in (1), we get

s = 16t² - 2t + 1

Problem 6 :

f'(x) = 4x, f(0) = 6

Solution :

f'(x) = 4x

dy/dx = 4x

dy = 4x dx

Finding antiderivative on both sides.

∫ dy =  ∫ 4x dx

y = 4x²/2 + C

y = 2x² + C ----(1)

6 = 2(0)² + C

C = 6

Applying the value of C in (1), we get

y = 2x² + 6

Problem 7 :

h'(t) = 8t³ + 5, h(1) = -4

Solution :

h'(t) = 8t³ + 5

dh/dt = 8t³ + 5 

dh = (8t³ + 5) dt

Finding antiderivative on both sides.

∫ dh =  ∫ (8t³ + 5) dt

h = 8t⁴/4 + 5t + C

h = 2t⁴ + 5t + C -------(1)

When t = 1, h = -4

-4 = 2(1)⁴ + 5(1) + C

C = -4 - 2 - 5

C  = -11

By applying the value of C in (1), we get

h = 2t⁴ + 5t - 11

Problem 8 :

f''(x) = 2, f'(2) = 5, f(2) = 10

Solution :

f''(x) = 2

Integrating on both sides.

f'(x) = 2x + C₁

f'(2) = 2(2) + C₁

5 = 4 + C₁

C₁ = 5 - 4

C₁ = 1

f'(x) = 2x + 1

Integrating on both sides again.

f(x) = 2x²/2 + x + C₂

f(x) = x² + x + C₂ -----(2)

Applying the condition f(2) = 10

10 = 2² + 2 + C₂ 

C₂ = 10 - 4 - 2

C₂ = 4

Applying the value of C₂ in (2), we get

f(x) = x² + x + 4

Problem 9 :

Given the acceleration

a = d² s/dt² = -4sin 2t

initial velocity v (0) = 2, and the initial position of the body as s (0) = -3. Find the body's position at time t.

Solution :

a = d² s/dt² = -4sin 2t

Integrating on both sides.

ds/dt = -4(-cos 2t)/2 + C₁

ds/dt = 2cos 2t + C₁ -----(1)

v = ds/dt

2 = 2 cos 2(0) + C₁

2  = 2(1) + C₁

 C₁ = 0

Applying the value of C₁ in (1), we get

ds/dt = 2cos 2t + 0

Integrating again, we get

s = (2 sin 2t)/2 + C₂

s = sin 2t + C₂ -----(2)

Given condition s(0) = -3

-3 = sin 2(0) + C₂

C₂ = -3 - 0

C₂ = -3

Applying the value of C₂ in (2), we get

s = sin 2t - 3