What will be the solution of the mentioned problem????
Ify=cos3x
then, y_n=?
If
then,
1 Answer
y_n = (d^n)/(dx^n) cos3x = { ((-1)^(n/2 ) \ 3^n \ sin 3x,n " even"), ((-1)^((n+1)/(2)) \ 3^n \ cos 3x,n " odd") :}
Explanation:
We have:
y = cos3x
Using the notation
Differentiating once wrt
y_1 = (-sin3x)(3) = -3sin3x
Differentiating further times we get:
y_2 = (-3)(cos3x)(3) \ \ \ \ \ \ = -3^2cos3x
y_3 = (-3^2)(-sin3x)(3) = +3^3sin3x
y_4 = (3^3)(cos3x)(3) \ \ \ \ \ \ \ \ \ \= +3^4cos3x
y_5 = (3^4)(-sin3x)(3) \ \ \ \ \ \= -3^5sin3x
vdots
And a clear pattern is now forming, and the
y_n = (d^n)/(dx^n) cos3x = { ((-1)^(n/2 ) \ 3^n \ sin 3x,n " even"), ((-1)^((n+1)/(2)) \ 3^n \ cos 3x,n " odd") :}