Question

If` y= e^2x sin  x/2 cos   x/2 sin3x. "find"  y_n`

Answer 1

`y=e^(2x) sin  x/2 cos  x/2 sin 3x xx2/2`

=`1/2e^2x[sin2(x/2)]sin3    `{∵ 2 sinAcosA=sin2A}`

=`1/2 e^(2x) sinxsin3x xx2/2`

=`1/4e^(2x)[cos(3x-x)-cos(3x+x)]`  ` {∵2sinAsinB=cos(A-B)-cos(A+B)}`

∴`y=1/4[e^(2x)cos2x-e^(2x)cos4x]` 

Taking `n^(th)` order derivative,`y_n=1/4{d_n/dx^n(e^(2x)cos2x)-d^n/dx^n(e^(2x) cos4x)}`                    …(1) 

We know, If y =`e^(ax)cos(bx+c),y_n=r^n e^(ax)cos(bx+c+n∅)`                        …(2) 

Here `a = 2, c = 0, b_1 = 2 and b_2 = 4` 
∴ `r_1=sqrt(a^2+b_1^2)=sqrt(2^2+2^2)=sqrt8=8^1/2 and r2 =sqrt(a^2+b_2^2)=sqrt(2^2+4^2)=sqrt20=20 1/2`                                  …(3) 

And, `∅_1=tan (-1^b1)/a=tan^-1 2/2=tan^-1 1=pi/4   & ∅=tan ^-1b^2/a=tan (-1b_2)/a=tan^-1 2/4=tan -1 1/2 ` …(4)

∴ From (1),(2),(3) and (4),

`Y_n=1/4{(8 ^(1/2))^n e^2x cos (2x+0+0 n∅_1)+(20^(1/2))^n e^2x cos(4x+0+n∅_2)}`

∴`y=1/4 e^(2x) [8^(n/2) cos (2x+(npi)/4)+20^(n/2)cos(4x+n∅_2)]`Where` ∅_=tan^-1  1/2`