In the last post, Differential Equations #5: An intro to Laplace Transforms, we introduced the concept of taking Laplace transforms and inverse Laplace transforms.

Two Laplace transforms we need to take note of:

L( f'(t) ) = s * F(s) - f(0)

L( f"(t) ) = s^2 * F(s) - s * f(0) - f'(0)

The presence of f(0) and f'(0) indicates that can use appropriate conditions to solve the differential equation.

To solve the differential equation using Laplace transformation:

1. Perform a Laplace transform on each term. Remember that

L(y(x)) = F(s)

L(y'(x)) = s*F(s) - y(0)

L(y"(x)) = s^2*F(s) - s*y(0) - y'(0)

2. Solve for F(s). If necessary, use algebraic manipulation to get F(s) in a working form.

3. Find the inverse Laplace transform for F(s). The final solution is:

y(x) = L⁻¹(F(s))

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)

**Examples:**

In our examples, I am going to shorten F(s) to F.

1. y' + 2*y = e^x, y(0) = 3

L(y') = s*F - y(0) = s*F - 3

L(2*y) = 2*F

L(e^x) = 1/(s-1)

Then:

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

Solving for F yields:

F = 1/((s-1)*(s+3)) + 3/(s+3)

= (3*s - 2)/((s-1)*(s+3))

= (1/3)/(s-1) + (8/3)/(s+3) (by partial decomposition of fractions)

Then y = L⁻¹(F)

= L⁻¹(1/3)/(s-1) + (8/3)/(s+3))

= 1/3*L⁻¹(1/(s-1)) + 8/3*L⁻¹(1/(s+3))

= 1/3 * e^-x + 8/3 * e^(2*x)

Final answer y(x) = 1/3 * e^-x + 8/3 * e^(2*x)

2. y" - y' = e^x, y(0)=1, y'(0) = 0

L(y") = s^2*F - s*y(0) - y'(0) = s^2*F - s

L(-y) = -L(y) = -(s*F - y(0)) = -s*F + 1

L(e^x) = 1/(s-1)

Then:

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

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

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

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

Since s^3 - 2*s^2 + s = s * (s^2 - 2*s + 1) = s * (s - 1)^2,

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

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

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

By partial decomposition of fractions:

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

By inverse Laplace transform:

L⁻¹(F) = y = 2 - e^x + x*e^x

3. y' + y = sin x, y(0) = π/2

L(y') = s*F - π/2

L(y) = s*F

L(sin x) = 1/(s^2 + 1)

(s*F - π/2) + F = 1/(s^2 + 1)

(s + 1) * F = 1/(s^2 + 1) + π/2

F = 1/((s^2 + 1)*(s + 1)) + π/(2*(s+1))

Simplification and partial decomposition of fractions:

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

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

L⁻¹(F) =

(1/2 + π/2)*e^-x - 1/2 * cos x + 1/2 * sin x

4. y' + 2*y = x^2

L(y') = s*F - 2

L(2*y) = 2*F

L(x^2) = 2/s^3

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

(s + 2) * F = 2/s^3 + 2

F = (2 + 2*s^3)/(s^3 * (s + 2))

F = 1/s^3 - 1/2 * 1/s^2 + 1/4 * 1/s + 7/4 * 1/(s+2)

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

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

Thanks as always, comments and questions are welcomed!

This blog is property of Edward Shore. 2013

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