Question

Solve: (D² - 4D+3) y=e^x cos2x​

Answer 1

Answer:

The solution is Y = C1 e^3x +C2 e^1x + (-1/8 ×e^x ×(sin2x+cos2x))

Step-by-step explanation:

The given differential equation is,

(D^2-4D+3) y = e^x cos2x

We have to solve this differential equation,

The reduced homogenous equation is,

(D^2-4D+3) y = 0

Let y = e^mx is a solution of the homogenous equation,

Then the auxiliary equation is,

m^2-4m+3 = 0

=> m^2-3m-1m+3 =0

=> (m-3) ×(m-1) = 0

=> m = 3,1

So, the complementary function of the differential equation is,

Yc = C1e^3x + C2^1x

Now we have to find the particular integral,

Yp = 1/((D-1)×(D-3)) (e^x cos2x)

= 1/(D-1) (1/(D-3) (e^xcos2x) )

= 1/(D-1) ( e^x/(D+1-3) (cos2x) )

= 1/(D-1) (e^x/(D-2) (cos2x) )

= 1/(D-1) (e^x (D+2) /(D^2 -4) (cos2x) )

= 1/(D-1) (e^x (D+2) cos2x/(-2^2-4) )

= 1/(D-1) ( e^x(D+2) cos2x/-8)

= 1/(D-1) (e^x/-8 ( -2sin2x +2cos2x))

=1/(D-1) ( e^x/-8 (2(cos2x-sin2x)))

= -1/4 (1/(D-1) (e^x(cos2x-sin2x)))

= -1/4 ( e^x/(D+1-1) (cos2x-sin2x))

= -1/4 ×e^x ( 1/D(cos2x -sin2x))

=-1/4 e^x ((sin2x) /2 +(cos2x) /2)

=-1/8 × e^x ( sin2x+cos2x)

So the solution is,

Y = C1e^3x + C3 e^1x -(1/8×e^x ×(sin2x+cos2x))