To solve the initial-value problem y' - y = 2 cos(6t), y(0) = 0 using the Laplace transform, we first take the Laplace transform of both sides of the equation.
L[y'] - L[y] = L[2 cos(6t)]
Using the property of the Laplace transform that L[ y' ] = sY(s) - y(0) and L[ cos(6t) ] = s/( s^2 + 36 ), this becomes:
sY(s) - y(0) - Y(s) = 2 * s / ( s^2 + 36 )
Substituting y(0) = 0, we get:
sY(s) - Y(s) = 2 * s / ( s^2 + 36 )
Factoring out Y(s), we get:
( s - 1 ) * Y(s) = 2 * s / ( s^2 + 36 )
Solving for Y(s), we get:
Y(s) = 2 * s / ( ( s - 1 ) * ( s^2 + 36 ) )
Using partial fractions, we can write Y(s) as:
Y(s) = A / ( s - 1 ) + B * s / ( s^2 + 36 )
Multiplying both sides by the denominator on the right-hand side and substituting s = 1, we get:
2 = A / ( 1 - 1 ) + B * 1 / ( 1^2 + 36 )
2 = B / 37
Thus, B = 74.
Substituting B in the previous equation and simplifying, we get:
Y(s) = 2 / ( s - 1 ) + 2s / ( s^2 + 36 )
Taking the inverse Laplace transform of Y(s) using a table or a software, we get:
y(t) = 2 * e^t + sin(6t) / 3
Therefore, the solution to the initial-value problem y' - y = 2 cos(6t), y(0) = 0 using the Laplace transform is y(t) = 2 * e^t + sin(6t) / 3.