Find the general solution of y"-4y'-5y=sin4t

y = - 21/697 sin (4t) + 16/697 cos (4t) + C₁e^ (-t) + C₂e^ (5t)

Step-by-step explanation:

y" - 4y' - 5y = sin (4t)

This is a second-order, non-homogeneous equation with constant coefficients. The general solution is the sum of the complementary and particular solutions.

First, find the complementary solution:

y" - 4y' - 5y = 0

s² - 4s - 5 = 0

(s - 5) (s + 1) = 0

s = - 1 or 5

y = C₁e^ (-t) + C₂e^ (5t)

Next, find the particular solution. Since g (t) = sin (4t):

y = A sin (4t) + B cos (4t)

y' = 4A cos (4t) - 4B sin (4t)

y" = - 16A sin (4t) - 16B cos (4t)

Substituting:

y" - 4y' - 5y = sin (4t)

-16A sin (4t) - 16B cos (4t) - 4 (4A cos (4t) - 4B sin (4t)) - 5 (A sin (4t) + B cos (4t)) = sin (4t)

-16A sin (4t) - 16B cos (4t) - 16A cos (4t) + 16B sin (4t) - 5A sin (4t) - 5B cos (4t) = sin (4t)

(16B - 21A) sin (4t) - (16A + 21B) cos (4t) = sin (4t)

Match the coefficients:

16B - 21A = 1, 16A + 21B = 0

Solve the system of equations:

A = - 21/16 B

16B - 21 (-21/16 B) = 1

16B + 441/16 B = 1

697B/16 = 1

B = 16/697

A = - 21/697

Therefore, the general solution is:

y = - 21/697 sin (4t) + 16/697 cos (4t) + C₁e^ (-t) + C₂e^ (5t)

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