y ''−4y'+3y = 6t−8, y(0) = 0, y'(0) = 0

Question

Solve the initial value problem below using

the Laplace transform

Please help

Answer 1

Laplace Transforms change differential equations into more manageable algebraic equations.

Let's define L{y(t)} = F(s), so we are going to use the definition int_{lb=0}_{ub=∞} (y(t)*exp(-st)dt to convert from the t variable to the s variable

Now using integration by parts we can determine what L{y'(t)} is as well as L{y''(t)} is with respect to the definition L{y(t)} or F(s). In doing so we will need to utilize the initial conditions.

L{y'(t)} = s*F(s) - y(0)

L{y"(t)} = s²F(s) - s*y(0) - y'(0)

L{6t-8} = 6 * 1/s² - 8/s

So, we have...

s²F(s) - s*y(0) - y'(0) - 4*(s*F(s) - y(0)) + 3F(s) = (6 - 8s)/s²

s²F(s) - s*0 - 0 - 4*(s*F(s) - 0) + 3F(s) = (6 - 8s)/s²

s²F(s) - 4sF(s) + 3F(s) = (6 - 8s)/s²

F(s) = (6 - 8s)/s² * 1/(s² - 4s + 3)

F(s) = -2(4s - 3)/s² * 1/(s² - 4s + 3)

F(s) = -2(4s - 3)/s² * 1/(s - 3)(s - 1)

F(s) = -2(4s - 3) / s²(s - 3)(s - 1)

Now we need to break into partial fractions to use the inverse Laplace Transform to solve the problem...

A/s² + B/(s-1) + C/(s-3) = (4s - 3) / s²(s - 3)(s - 1)

I don't need to worry about the overall constant factor of -2

Finding the common denominator, we get three (four but with same answer) equations in three unknowns...

B = -C. ( to cancel the s³ terms)

-B + A -3C = 0 (to cancel the s² term)

-4A = 4 to account for the s term

-3A = 3 to account for the constant term

A = -1

Plugging back into the second equation with help fo first equations gives...

-B + 1 + 3B = 0 ---> B = 1/2

Therefore C = -1/2

F(s) = -2 [ -1/s² + 1/(2*(s-3)) - 1/(2*(s-1))

Using the inverse transform now is trivial...

y(t) = L^-1{F(s)} = -2 * (-t + 1/2*e^(3t) - 1/2*e^(t))

y(t) = 2t + e^3t - e^t