Question

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

dy/dt  − y = 1,    y(0) = 0

y(t) =

Answer 1

$\bigstar$ Using laplace transform

Theorem - L{f'(t)=sf(s)-f(0)

We have

L($dy\over dt$ ) - L (y) = L(1)

sY(s) - y(0)- Y(s)= $1\over s$

(s-1)Y(s)=$1\over s$

Y(s)=$1\over s(s-1)$

Where Y(s) = L{y(t)}

$\bigstar$ In order to find y(t) , we find inverse laplace transform of Y(s) , since

$1\over s(s-1)$ = $1\over s-1$ - $1\over s$

Now , we have using theorem a= 1 and a=0

y(t) = L^-1$1\over s(s-1)$ =

=L^-1$1\over s-1$ - L^-1$1\over s$

= e^t-1

$\boxed{y(t)=e^t-1}$