Question

Problem 2: Use the Laplace Transforms to solve:

Answer 1

Using the given table of transforms,

$L\left\{ y'' - y' - 2y \right\} = L\left\{ 1-x \right\}$

$(s^2 Y(s) - sy(0) - y'(0)) - (s Y(s) - y(0)) - 2 Y(s) = \dfrac1s - \dfrac1{s^2}$

(where Y(s) is the Laplace transform of y(x))

Solve for Y(s) :

$(s^2 - s - 2) Y(s) - s = \dfrac1s - \dfrac1{s^2}$

$(s^2 - s - 2) Y(s) - s = \dfrac{s-1}{s^2}$

$(s^2 - s - 2) Y(s) = \dfrac{s-1}{s^2} + s$

$(s^2 - s - 2) Y(s) = \dfrac{s^3 + s - 1}{s^2}$

$Y(s) = \dfrac{s^3 + s - 1}{s^2 (s^2 - s - 2)}$

$Y(s) = \dfrac{s^3 + s - 1}{s^2 (s + 1) (s - 2)}$

Decompose the right side into partial fractions:

$Y(s) = \dfrac as + \dfrac b{s^2} + \dfrac c{s+1} + \dfrac d{s-2}$

Solve for the coefficients.

$\implies s^3 + s - 1 = as(s+1)(s-2) + b(s+1)(s-2) + cs^2(s-2) + ds^2(s+1)$

$\implies s^3 + s - 1 = -2 b + (-2 a - b) s + (-a + b - 2 c + d) s^2 + (a + c + d) s^3$

$\implies \begin{cases}-2b = -1 \\ -2a-b = 1 \\ -a+b-2c+d = 0 \\ a+c+d = 1 \end{cases} \implies a=-\dfrac34, b=\dfrac12, c=1, d=\dfrac34$

Then

$Y(s) = -\dfrac34 \times \dfrac1s + \dfrac12 \times \dfrac1{s^2} + \dfrac1{s+1} + \dfrac34 \times\dfrac1{s-2}$

Take the inverse transform and solve for y(x) :

$F(s) = \dfrac1s \implies f(x) = 1$

$F(s) = \dfrac1{s^2} \implies f(x) = x$

Using the frequency-shifting property,

$F(s) = \dfrac1{s+1} \implies f(x) = e^{-x}$

$F(s) = \dfrac1{s-2} \implies f(x) = e^{2x}$

So, the particular solution to the ODE is

$\boxed{y(x) = -\dfrac34 + \dfrac x2 + e^{-x} + \dfrac{3e^{2x}}4}$