Use Laplace Transforms to solve the given initial-value problems:

Question

Use Laplace Transforms to solve the given initial-value problems:

y'' + 5y' + 6y = 0

y(a) = 0, y'(a) = 1, (a > 0)

Answer 1

Given:

y'' + 5y' + 6y = 0

y(a) = 0, y'(a) = 1, (a > 0)

Find:

y(t) = ?

Solution:

Initial Value Problems ( IVPs ) don't "go well" when its initial value is at something other than 0 when using Laplace transforms, so, I recommend redefining some parts of the problem to mold it to the usual Laplace framework

I recommend defining some function, call it 'u' (doesn't have to be u but most other substitutions in "impossible" integrals are u-substitutions so if you can't beat them, join them) and have it dependent upon some independent variable, let's call 'b' (it can be whatever you like, 'b' was just easy to type)

u(b) will be equated to being similar to y(t) in this way:

b = t - a

So when t = a, a - a = b = 0

Also: t = b + a (thus dt/db = 1, this will become helpful in just a second)

We also need the function 'u' to be able to fit to the given Second Order Differential Equation:

y( t ) = u( b ) ⇒ d/db ( y( t ) ) = u'(b) = dy/dt * dt/db¹ (here's where it is helpful)

u'(b) = y'(t)

The same logic can be used to find: u''(b) = y''(t) (chain-rule sure is nice)

Thus:

u'' + 5u' + 6u = 0

u(0) = 0, u'(0) = 1

Then, the usual Laplace transform

s² U - s u(0)⁰ - u'(0)¹ + 5s U - 5 u(0)⁰ + 6 U = 0

U ( s² + 5s + 6 ) = 1

U(s) = 1 / [(s+2)(s+3)] = A / (s+2) + B / (s+3)

There are many methods for partial fractions, whatever way you take, A = 1, B = -1

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

u(b) = e^-2b - e^-3b

Note: Because the exponentials on the right side of the equal sign are still functions of 'b', it's best to represent the function 'y' like it is a function of 'b' to eliminate as much confusion as possible

y( b + a ) = e^-2b - e^-3b

Replace every b with t - a

Replace every b + a with t

y(t) = e^{-2 ( t - a)} - e^{-3 ( t - a )}

∴ y(t) = e^{-2t + 2a} - e^{-3t + 3a}

I hope this helps! Message me in the comments with any questions, comments, or concerns! I'll be honest, this is my first attempt at an IVP where the initial value is not 0 so it may be a little weird, I will admit. It was weird teaching it to myself!

Answer 2

ηLet new variable: x = η + a; and y(x) = y(η + a);

Then we get equation: y''(η + a) + 6y'(η + 3) + y(η + 3) = 0; u(η) = y(η + 3);

u''(η) +6u' + 5 = 0, y(η + a) = u(η) u(0) = 0; u'(0) = 1

Laplace Transform: s²F(s) - 1 + 5sF(s) + 6F(s) = 0; F(s) = 1/[(s + 2)(s + 3)] = 1/(s + 2) - 1/(s + 3)

Inverse Laplace: u(η) = e^-2η - e^-3η_; y(x) = e^{-2(x - a)} - e^{-3(x - a)}