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solve the following using LT
i) d2x/dt2 + dx/dt +x = 1; x'(0)=x(0)=0
ii) d2x/dt2 + 2dx/dt +x = 1; x'(0)=x(0)=0
iii) d4x/dt24 + d3x/dt3 = cost ; x'''(0)=x'(0)=0, x''(0)=1
To solve these differential equations using Laplace Transform (LT), follow these steps:
1. Apply the Laplace Transform to both sides of the equation.
2. Use initial conditions to simplify the transformed equation.
3. Solve for X(s) in the Laplace domain.
4. Apply the Inverse Laplace Transform to find x(t).
d2 x
i) dt2
+ dx
dt
+ x = 1, x′ (0) = x(0) = 0
Step 1: Apply Laplace Transform
Using L{x′ (t)} = sX(s) − x(0) and L{x′′ (t)} = s2 X(s) − sx(0) − x′ (0), the equation
becomes:
1
s2 X(s) + sX(s) + X(s) =
.
s
Step 2: Simplify
1
X(s)(s2 + s + 1) = .
s
1
X(s) = .
s(s2 + s + 1)
Step 3: Partial Fraction Decomposition
Decompose:
1 A Bs + C
= + .
s(s2 + s + 1) s2 + s + 1
Solving for A, B, C :
A = 1, B = −1, C = 1.
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Thus:
1 s−1
X(s) = − 2 .
s s +s+1
Step 4: Inverse Laplace Transform
1 s−1
x(t) = L−1 { } − L−1 { 2 }.
s +s+1
L−1 { 1s } = 1,
L−1 { s2 +s+1
s
} = e−t cos (
2
3
t ),
1
L−1 { s2 +s+1 }=
2 −t
3
e sin ( 2
3
t ).
Thus:
x(t) = 1 − e−t [cos ( t) − sin ( t)] .
3 2 3
2 3 2
2
ii) ddtx2
+ 2 dx
dt
+ x = 1, x′ (0) = x(0) = 0
Step 1: Apply Laplace Transform
1
s2 X(s) + 2sX(s) + X(s) = .
s
Step 2: Simplify
1
X(s)(s2 + 2s + 1) = .
s
1
X(s) = .
s(s + 1)2
Step 3: Partial Fraction Decomposition
1 A B C
= + + .
s(s + 1)2 s + 1 (s + 1)2
Solving for A, B, C :
A = 1, B = −1, C = 1.
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Thus:
1 1 1
X(s) = − + .
s s + 1 (s + 1)2
Step 4: Inverse Laplace Transform
1 1 1
x(t) = L−1 { } − L−1 { } + L−1 { }.
s+1 (s + 1)2
L−1 { 1s } = 1,
L−1 { s+1
1
} = e−t ,
L−1 { (s+1)
1
2 } = te
−t
.
Thus:
x(t) = 1 − e−t + te−t .
4
d3 x
iii) ddtx4 + dt3 = cos(t), x′′′ (0) = x′ (0) = 0, x′′ (0) = 1
Step 1: Apply Laplace Transform
s
s4 X(s) − s3 x′′ (0) + s3 X(s) = .
s2 +1
Using x′′ (0) = 1, x′′′ (0) = 0:
s
s4 X(s) − s3 + s3 X(s) = .
s2 + 1
Step 2: Simplify
s
X(s)(s4 + s3 ) = + s3 .
s +1
2
s
s2 +1
+ s3
X(s) = .
s4 + s3
Step 3: Decompose and Simplify
Decompose and take inverse Laplace transforms for each term (algebraic steps omitted for
brevity).
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x(t)= ?
Let’s determine the full solution for x(t) explicitly. The given differential equation is:
d4 x d3 x
+ 3 = cos(t),
dt4
dt
with initial conditions:
x(0) = 0, x′ (0) = 0, x′′ (0) = 1, x′′′ (0) = 0.
Step 1: Laplace Transform of the Equation
The Laplace transform gives:
d4 x d3 x
L { 4 } + L { 3 } = L{cos(t)}.
dt dt
Using L{x(n) (t)} = sn X(s) − sn−1 x(0) − sn−2 x′ (0) − ⋯ − x(n−1) (0), we get:
s
s4 X(s) − s3 x′′ (0) + s3 X(s) − s2 x′′ (0) = .
s2 + 1
Substitute the initial conditions:
x(0) = 0, x′ (0) = 0, x′′ (0) = 1, x′′′ (0) = 0.
This gives:
s
s4 X(s) − s3 + s3 X(s) − s2 = .
s2 + 1
Simplify:
s
X(s)(s4 + s3 ) = + s3 − s2 .
s +1
2
Step 2: Solve for X(s)
Factor X(s):
3 2
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s 3 2
s2 +1 + s − s
X(s) = .
s3 (s + 1)
Simplify the numerator:
s s3 s2
X(s) = 3 + − .
s (s + 1)(s2 + 1) s3 (s + 1) s3 (s + 1)
Split into terms:
s 1 1
X(s) = + − .
s3 (s + 1)(s2 + 1) s + 1 s(s + 1)
Step 3: Inverse Laplace Transform
s
Term 1: s3 (s+1)(s 2 +1)
s
Decompose s3 (s+1)(s 2 +1) into partial fractions:
s A B C D Es + F
= + + + + .
s3 (s + 1)(s2 + 1) s2 s3 s+1 s2 + 1
This requires solving for coefficients A, B, C, D, E, F . After solving:
A = 0, B = 0, C = 1, D = 0, E = 0, F = 1.
Thus:
s 1 1
= + .
s3 (s + 1)(s2 + 1) s3 s2 + 1
Take the inverse Laplace transform term-by-term:
t2
L−1 { s13 } =
2
,
L−1 { s21+1 } = sin(t).
So the inverse transform of this term is:
t2
{ 3 } = + sin(t).
−1 s
L
s (s + 1)(s2 + 1) 2
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1
Term 2: s+1
The inverse Laplace transform is:
1
L−1 { } = e−t .
s+1
1
Term 3: s(s+1)
1
Decompose s(s+1) :
1 1 1
= − .
s(s + 1) s s+1
Take the inverse Laplace transform:
L−1 { 1s } = 1,
1
L−1 { s+1 } = e−t .
Thus:
1
L−1 { } = 1 − e−t .
s(s + 1)
Step 4: Combine All Terms
Now, combine the inverse transforms of all terms:
t2
x(t) = ( + sin(t)) + e−t − (1 − e−t ).
2
Simplify:
t2
x(t) = + sin(t) + 2e−t − 1.
2
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Final Answer:
t2
x(t) = + sin(t) + 2e−t − 1.
2
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