## Monday, August 19, 2013

### Differential Equations #7: Laplace Transforms and Systems of Differential Equations

In this blog we will look at how Laplace transforms assist us in solving systems of differential equations.

In the examples tonight, I have design the y(x) and z(x) as the equations to solve for, and designate:

L(y) = F(s) (which I will shorten to F)
L(y') = s*F - y(0)

L(z) = G(s) (which I will shorten to G)
L(z') = s*G - z(0)

As a reference:
L( f(t) ) → F(s)
1 → 1/s
t^n → n!/(s^(n+1))
e^(a*t) → 1/(s-a)
sin(a*t) → a/(s^2 + a^2)
t*sin(a*t) → (2*a*s)/((s^2 + a^2)^2)
cos(a*t) → s/(s^2 + a^2)
t*cos(a*t) → (s^2 - a^2)/((s^2 + a^2)^2)
sinh(a*t) → a/(s^2-a^2)
cosh(a*t) → s/(s^2-a^2)
e^(a*t)*sin(b*t) → b/((s-a)^2 + b^2)
e^(a*t)*cos(b*t) → (s-a)/((s-a)^2 + b^2)
e^(a*t)*sinh(b*t) → b/((s-a)^2 - b^2)
e^(a*t)*cosh(b*t) → (s-a)/((s-a)^2 - b^2)

For this, it would be best to show how to solve systems by example.

Examples:

1. y' = 2*y + z
z' = -2y
Initial Conditions: y(0)=1, z(0)=-1

Using the transforms above, the result is:
s*F - 1 = 2*F + G
s*G + 1 = -2*F

-1 = (2-s)*F + G
1 = -2*F - s*G

Solving for F and G:

F = (s-1)/(s^2 - 2*s + 2)
G = (-s)/(s^2 - 2*s + 2)

Taking inverse Laplace Transforms for both F and G:

F = (s-1)/(s^2 - 2*s + 2)
F = (s-1)/(s^2 - 2*s + 1 + 1)
F = (s-1)/((s-1)^2 + 1)
L⁻¹(F) = y = e^x * cos x

G = (-s)/(s^2 - 2*s + 2)
G = (-s)/((s - 1)^2 + 1)
G = (-s - 1 + 1)/((s-1)^2 + 1)
G = -(s - 1)/((s-1)^2 + 1) - 1/((s-1)^2 + 1)
L⁻¹(G) = z = -e^-x * cos x - e^x * sin x

y = e^x * cos x
z = -e^-x * cos x - e^x * sin x

2. y' + z = x
z' + 4*y = 0
Initial Conditions: y(0)=1, z(0)=-1

L(x) = 1/s^2

Then:
s*F - 1 + G = 1/s^2
s*G + 1 + 4*F = 0

s*F + G = 1/s^2 + 1
4 * F + s*G = -1

F = (s^2 + s + 1)/(s^3 - 4*s)
G = (-s^3 - 4*s^2 - 4)/(s^4 - 4*s^2)

Finding y and z:

F = (s^2 + s + 1)/(s^3 - 4*s)
F = -1/4 * 1/s + 3/8 * 1/(s+2) + 7/8 * 1/(s-2)
L⁻¹(F) = y = -1/4 + 3/8 * e^(-2x) + 7/8 * e^(2x)

G = (-s^3 - 4*s^2 - 4)/(s^4 - 4*s^2)
G = 1/s^2 + 3/4 * 1/(s + 2) - 7/4 * 1/(s-2)
L⁻¹(G) = z = x + 3/4 * e^(-2*x) - 7/4 * e^(2*x)

y = -1/4 + 3/8 * e^(-2x) + 7/8 * e^(2x)
z = x + 3/4 * e^(-2*x) - 7/4 * e^(2*x)

3. y' + z' = 3
y - z = x
Initial Conditions: y(0) = 1, z(0) = 0

s*F - 1 + s*G = 3/s
F - G = 1/s^2

F = (s + 4)/(2*s^2)
F = s/(2*s^2) + 4/(2*s^2)
F = 1/(2*s) + 2/(s^2)
L⁻¹(F) = y = 1/2 + 2*x

G = (s + 2)/(2*s^2)
G = s/(2*s^2) + 2/(2*s^2)
G = 1/(2*s) + 1/(s^2)
L⁻¹(G) = z = 1/2 + x

y = 1/2 + 2*x
z = 1/2 + x

This concludes our section for Differential Equations. Hope this is helpful and thanks as always! Eddie

Update: There will be more of the Differential Equations series! Look for additional posts during September 2013. Eddie - 8/30/13

This blog is property of Edward Shore. 2013

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