Question

Consider the initial value problem y′+4y=48t,y(0)=9. y′+4y=48t,y(0)=9. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other

Answer 1

Answer:

$sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2}$

Step-by-step explanation:

given is the Differential equation in I order linear as

$y′+4y=48t,y(0)=9.$

Take Laplace on both sides

$L(y') +4L(y) = 48L(t)\\sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2} \\Y(s) [s+4]=\frac{48}{s^2}+9\\Y(s) = \frac{1}{s^2(s+4)}+\frac{9}{s+4}$

Now if we take inverse we get y(t) the solution

Thus the algebraic equation would be$sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2}$