Find the complete solution of y 5y 6y e4t use method of u

Find the complete solution of: y\" + 5y\' + 6y = e^-4t. (use method of undetermined coefficients.)

Solution

Consider the differential equation

y\'\' + 5y\' + 6y = e^-4t    ----------------- ( 1 )

The characteristic equation is r² + 5r +6 = 0 => ( r + 2 )( r + 3 ) = 0

The roots of the characteristic equation are r = -2 and r = -3.

The complementary solution is y_c(t) = C₁ e^-2t + C₂ e^-3t

Now we find the particular solution y_p(t) using the method of undetermined coefficients.

we have g(t) = e^-4t . Let the the particular solution be y_p(t) = Ae^-4t

Then y\'_p(t) = -4Ae^-4t    and   y\'\'_p(t) = 16Ae^-4t

Substitute these results in ( 1 ) , we get

16Ae^-4t + 5( -4Ae-4t   ) + 6( Ae^-4t ) = e^-4t

( 16A - 20A + 6A )e^-4t = e^-4t

2Ae^-4t = e^-4t

^{Equating on both sides , we get}

2A = 1 => A = 1/2

Therefore, the particular solution y_p(t) = (1/2)e^-4t

The complete solution of ( 1 ) is y(t) = y_c(t) + y_p(t)

Therefore, y(t) = C₁ e^-2t + C₂ e^-3t + (1/2)e^-4t