An equation of the form
dy/dx = f(x)
that has derivative in it is called differential equation.
What is general solution ?
A general solution to a differential equation is one that has a constant in it. It is really a family of functions that solves the differential equation.
What is particular solution ?
A particular solution to a differential equation is one that satisfies an initial value.
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
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM