Answer to Question #177430 in Differential Equations for Promise Omiponle

Question

Use the Laplace transform to solve the given initial-value problem.

y'' + 5y' + 4y = 0,    y(0) = 1,  y'(0) = 0

y(t) =

Answer 1

L{y′′}+5L{y′}+4L{y}=0

$s^2L\{y\}−s+5sL\{y\}−5+4L\{y\}=0$

Since, according to the table,

$L\{y′′\}=s^2y(s)−sy(0)−y′(0),$

$L\{y′\}=sy(s)−y(0),$

and we have the initial conditions that y(0)=1,y′(0)=0. Gather appropriately:

$L\{y\}(s^2+5s+4)=s+5$

⟹$y=L^{−1}\{\frac{s+5}{(s+4)(s+1)}\}$

Now we have to use partial fraction decomposition,

$\frac{A}{s+4}+\frac{B}{s+1}=\frac{s+5}{(s+4)(s+1)}$

Multiplying through by the denominator on the right side of the equation to solve for A and B, we arrive at:

A(s+1)+B(s+4)=s+5

In order to solve for A we set s=−4, so that B is multiplied by zero, then we have that −3A=1⟹A=−13, similarly if s=−1⟹B=43. Substituting those values in:

$y=L^{−1}\{\frac{\frac{−1}{3}}{s+4}\}+L^{−1}\{\frac{\frac{4}{3}}{s+1}\}$

Looking at the Laplace transform table, we see that after factoring out the numerator of each fraction, we are left taking the inverse Laplace transform of an expression of the form $\frac{1}{s−(−a)}$ which has inverse Laplace transform of eat. Thus we arrive at:

$y=\frac{−1}{3}e^{−4t}+\frac{4}{3}e^{−t}$