The discriminant of a quadratic equation is calculated through the equation,
D = b² - 4ac
In the given above, a = 5, b = -4, and c = 3. Substituting these values to the equation,
D = (-4)² - 4(5)(3) = -44
Since the discriminant is negative, the roots are non-real.
L[ y(t) ] = Y(s)
L[ y' ] = sY-y(0) = sY-0 = sY
L[ y'' ] = s^2*Y - sy(0) - y'(0) = s^2*Y - 1
laplace trasform both sides
L[ y'' - 6y' + 9y ] = L[ t ]
= L[ y'' ] - 6 L[ y' ] + 9 L[ y ] = 1/s
= [ s^2*Y - 1 ] - 6[ sY ] + 9Y = 1/s
( s^2 - 6s + 9 ) Y - 1 = 1/s
⇒ ( s^2 - 6s + 9 ) Y = (1/s) + 1
⇒Y = [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ]
Let inverse laplace trasform , find y(t) :
y(t) = L^(-1)[ Y(s) ] = L^(-1) { [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ] }
= [ (1/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9) ] + [ t*e^(3t) ]
= (4/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9)
Answer: 190
Step-by-step explanation:
Answer:
What grade math is this??
Step-by-step explanation:
80% because its a huge amount the bank account goes to
