Question

(D^2+4) = sin3x + cos2x​

Answer 1

Answer:

We know that 2sinAcosB = sin(A+B) + sin(A-B)

So, sinAcosB = (1/2){sin(A+B) + sin(A-B)}

=> sin3x cos2x= (1/2){sin 5x + sin x} ………..(1)

Auxiliary Equation is D^2 - 4D +3= 0.

So the roots are 1, 3.

Hence the complementary function

CF is c1 e^x + c2 e^3x……………..(2)

Particular Integral

PI = {1/D^2 - 4D +3} sin3x cos2x

= {1/D^2 - 4D +3}(1/2){sin 5x + sin x} from (1)

= (1/2){1/D^2 - 4D +3} sin 5x

+{1/2){1/D^2 - 4D +3} sin x

= (1/2){1/-1- 4D +3} sin 5x

+(1/2){1/-1- 4D +3} sin x

= (1/2){1/2- 4D} sin 5x

+1/2){1/2- 4D} sin x

= (1/2){(2+4D)/(4–16D^2)} sin 5x

(1/2){(2+4D)/(4–16D^2)} sin x

={(1+2D)/4+400)} sin 5x

+{(1+2D)/4+16)} sin x

= (1/404)(1+2D) sin 5x + (1/20)(1+2D)sin x

= (1/404)(sin 5x+10 cos5x)+(1/20)(sinx+2cos x)

General Solution = CF + PI

Step-by-step explanation:

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