Use Laplace Transform To Solve The Differential Equation

Use Laplace Transform to solve Differential Equation HELP!!?

Use Laplace Transform and another method to solve the following DE and
compare the results
x'' - x' -12x = sin(3t) + e^(4t)
x(0) = 2 ; x'(0) = 1
--------------------------------------...
I am very confused as to how to tackle to problem. I am also asked to use another method to solve the problem and I don't know how to do that. Any help is appreciated.
This is what I tried:
L{x''} - L{x'} - 12L{x} = L{sin(3t)} + L{e^(4t)}
Let L{x} = X(s). .. thus
s^2 X(s) -1 - sX(s) - 12X(s) = 3/(s^2 - 9) + 1/(s-4)
I then solved for X(s) and now I'm stuck

Solve the following differential equation using laplace transforms?

s f(s) - T(0) = k [ f(s) - 50 / s + 20 e^(- h s ) / s ]--->

f(s) [ s - k ] = T(0) + k [ 20 e^(-hs) / s - 50 / s ]--->

f(s) = T(0) / [ s - k ] + k [ 20 e^(-hs) - 50 ] / { s [ s - k ] }---->

T(t) = T(0) e(kt) + 50 - 50 e^(kt) + 20 U(t-h) [ - 1 + e^(kt - kh ) ]

Solving systems of Differential Equations using Laplace transform?

Applying the Laplace Transform to each equation, we obtain
s X(s) - 7 = 5 X(s) - 2 Y(s)
s Y(s) + 2 = -3 X(s) + Y(s).

Simplifying, we obtain
(s - 5) X(s) + 2 Y(s) = 7
3 X(s) + (s - 1) Y(s) = -2.

Solving this by Cramer's Rule or otherwise, we obtain
X(s) = (7s - 3) / (s^2 - 6s - 1) and Y(s) = (-2s - 11) / (s^2 - 6s - 1).

Completing the square, we have
X(s) = [7(s - 3) + 18] / [(s - 3)^2 - 10] and Y(s) = [-2(s - 3) - 17] / [(s - 3)^2 - 10].

Inverting term by term:
x(t) = e^(3t) [7 cosh(t/√10) + (18/√10) sinh(t/√10)]
y(t) = e^(3t) [-2 cosh(t/√10) - (17/√10) sinh(t/√10)]

I hope this helps!

Differential Equations: Finding y(t) using Laplace transforms?

i assumed that
y'' + 2y' + 10y = g(t)

when g(t)=10
y'' + 2y '+ 10y =10
Step 1: We use Laplace transform on the differential equation,

L[ y'' + 2y' + 10y]=L[10]
L[y'' ]+L[2y] +L[10y] = L[10]
[ s² L(y) - sy(0) - y'(0)] +2[sL(y)-y(0)] +10L(y) = L(10)

let Y= L[y] and we know L[10]= 10/s
thus,
s²Y +1 + 2sY+2 +10Y =10/s

s²Y + 2sY +10Y+3 =10/s

Y(s²+2s+10)=10/s -3 ----(1)

Step 2 : Let Q(s)= 1 /(s²+2s+10) ,then multply both side of (1) by Q(s)

Y= (10/s -3)(1/(s²+2s+10))

=[(10-3s)/s][(1/(s²+2s+10)]
=[(10-3s)]/[s(s²+2s+10)]


Step 3: y(t)=L^(-1)[Y]


then use partial fraction or any other method

Differential Equations - Nontrivial Solution/Laplace Transform?

Let X(s) be the Laplace transform of x(t). The Laplace transforms of the terms in the equation are:
L[5 x] = 5 X(s)
L[-6 x'] = -6 (s X(s) - x(0))
L[5 t x'] = -5 d/ds L[x'] = -5 (s X'(s) + X(s))
L[t x''] = - d/ds L[x''] = -2 s X(s) - s^2 X'(s) + x(0)

Putting everything together and using x(0) = 0:
(s + 5) X' + 8 X = 0
The solution of this equation is
X(s) = C/(s + 5)^8
with a constant C.

The inverse Laplace transform of this X(s) is
x(t) = C e^(-5 t) t^7 / 5040
with x(1) = e^(-5) we get C = 5040 and thus
x(t) = t^7 e^(-5t)

Use laplace transform to solve the given system of differential equations?

Rewrite as
(s - 1) L[x] + 2 L[y] = -1
5 L[x] + (-s - 1) L[y] = -2

Multiply top equation by 5, the bottom equation by (s - 1), and subtract:
5(s - 1) L[x] + 10 L[y] = -5
- [5(s - 1) L[x] + (-s - 1)(s - 1) L[y] = -2(s - 1)]
--------------------------------------...
(10 + (s^2 - 1)) L[y] = -5 + 2(s - 1)
==> L[y] = (2s - 7)/(s^2 + 9).

Multiply top equation by (s+1), the bottom equation by 2, and add:
(s - 1)(s+1) L[x] + 2(s+1) L[y] = -(s+1)
+ [10 L[x] + 2(-s - 1) L[y] = -4]
--------------------------------------...
(s^2 + 9) L[x] = -s - 5
==> L[x] = (-s - 5)/(s^2 + 9)

Inverting yields
x(t) = -cos(3t) - (5/3) sin(3t)
y(t) = 2 cos(3t) - (7/3) sin(3t).

I hope this helps!

Differential Equations: Finding w using the Laplace Transform?

To use Laplace transform you need an a "true" initial value problem. that means initial conditions like:
w(0)=...
w'(0=...
etc.

To achieve this, substitute
τ = t + 1 <=> t = τ-1

Because
dτ/dt = 1
this won't change the differentials of w, i.e.
w' = dw/dt = dw/dτ ∙ dτ/dt = dw/dτ
w'' = d²w/dt² = d(dw/dt)/dt = d(dw/dτ)/dτ ∙ dτ/dt = d²w/dτ²

Hence:
w'' - 2∙w' + w = 6∙t(τ-1) - 2
<=>
d²w/dτ² - 2∙dw/dτ + w = 6∙(τ-1) - 2
<=>
d²w/dτ² - 2∙dw/dτ + w = 6∙τ - 8
with
w(0) = 3
w'(0) = 7

use Laplace transform tor transform from w(τ) to W(s)
ℒ{ d²w/dτ² - 2∙dw/dτ + w } = ℒ{ 6∙τ - 8 }
<=>
ℒ{d²w/dτ²} - 2∙ℒ{dw/dτ} + ℒ{w} = ℒ{6∙τ} - ℒ{8}
=>
s²∙W - s∙w(0) - w'(0) - 2∙(s∙W - w(0)) + W = 6/s² - 8/s
<=>
s²∙W - s∙3 - 7 - 2∙s∙W + 2∙3 + W = 6/s² - 8/s
<=>
(s² - 2∙s + 1)∙W = 6/s² - 8/s + 1 + 3∙s
<=>
(s - 1)²∙W = (6 - 8∙s + s² + 3∙s³)/s²
<=>
W = (3∙s³ + s² - 8∙s + 6) / [s²∙(s-1)²]

decompose RHS to partial fractions
(3∙s³ + s² - 8∙s + 6) / [s∙(s-1)²] = A/s + B/s² + C/(s-1) + D/(s-1)²
<=>
3∙s³ + s² - 8∙s + 6 = A∙s∙(s-1)² + B∙(s-1)² + C∙s²∙(s-1) + D∙s²
<=>
3∙s³ + s² - 8∙s + 6 = s³∙(A + C) + s²∙(-2A + B - C + D) + s∙(A - 2B) + B
comparing coefficients of same power of s leads to
A + C = 3
-2A + B - C + D = 1
A - 2B = 8
B = 6
=>
A = 20
B = 6
C = -17
D = 18

Hence:
W = 20/s + 6/s² - 17/(s-1) + 18/(s-1)²

take inverse transform
w = ℒ⁻¹{20/s + 6/s² - 17/(s-1) + 18/(s-1)²}
= 20∙ℒ⁻¹{1/s} + 6∙ℒ⁻¹{1/s²} - 17∙ℒ⁻¹{1/(s-1)} + 18∙ℒ⁻¹{1/(s-1)²}
= 20∙ (1) + 6∙(τ) - 17∙(e^(τ)) + 18∙(τ∙e^(τ))
= 20 + 6∙τ + (18∙τ - 17)∙e^(τ)
back substitution
= 20 + 6∙(t+1) + (18∙(t+1) - 17)∙e^(t+1)
= 26 + 6∙t + (18∙t + 1)∙e^(t+1)

How do you solve the differential equation y'' - 3y' - 4y = 3; initial conditions: y(0) = 1, y'(0) = 2 using Laplace transforms?

y'' - 3y' - 4y = 3

Laplace transform:
[ s² Y - s y(0) - y'(0) ] - 3 [ s Y - y(0) ] - 4 Y = 3/s

Simplifying:
[ s² Y - s - 2 ] - 3 [ s Y - 1 ] - 4 Y = 3/s
s² Y - s - 2 - 3 s Y + 3 - 4 Y = 3/s
(s² - 3s - 4) Y - s + 1 = 3/s
(s - 4)(s + 1) Y = 3/s + s - 1
(s - 4)(s + 1) Y = 3/s + (s² - s)/s
(s - 4)(s + 1) Y = (s² - s + 3)/s
Y = (s² - s + 3) / (s(s - 4)(s + 1))

Use partial fraction decomposition (here's where you said you needed help).
In the denominator, we have three factors: s, s-4, and s+1. Write the fraction as the sum of three fractions, such that each fraction has one of the factors as a denominator:
Y = A/s + B/(s-4) + C/(s+1)

Use least common denominator to combine back into one fraction (I won't bother writing the denominator out, but it's still there):
Y = A(s-4)(s+1)/... + Bs(s+1)/... + Cs(s-4)/...
Y = ( A(s-4)(s+1) + Bs(s+1) + Cs(s-4) ) / ...

Since this fraction is the same as the fraction before PFD, the numerators must be equal:
A(s-4)(s+1) + Bs(s+1) + Cs(s-4) = s² - s + 3
A(s² - 3s - 4) + B(s² + s) + C(s² - 4s) = s² - s + 3
(A + B + C)s² + (-3A + B - 4C)s + (-4A) = s² - s + 3

Matching coefficients:
A + B + C = 1
-3A + B - 4C = -1
-4A = 3

Solving the system:
A = -3/4
B = 3/4
C = 1

Y = -3/4 (1/s) + 3/4 (1/(s-4)) + (1/(s+1))

Taking inverse Laplace transform:
y = -3/4 + 3/4 e⁴ˣ + e⁻ˣ