Question

Particular integral for (D^2-4D+D)y=cos2x is ?

Answer 1

Answer:

Step-by-step explanation:

Hi

Answer 2

P.I = 1/52 (-6sin2x-4cos2x)

Explanation:

Given:

(D²-4D+D)y=cos2x

To find:

Particular Integral

Formula:

$P.I= \frac{1}{f(D)}cosax$  

To find P.I

$P.I= \frac{1}{D^{2-4D+D} }cosax$  

$P.I= \frac{1}{D^{2}-3D } cos2x$  

Replace D² by -(2)²

==> $\frac{1}{-(2)^{2} -3D } cos2x$

==>   $\frac{1}{-4 -3D } cos2x$

==> $\frac{-1}{4 +3D }cos2x$

==> Taking Conjugate

==> $\frac{-1}{3D+4 }cos2x$

==> $\frac{-1}{3D+4 }\frac{3D-4}{3D-4} cos2x$

==> (a+b)(a-b)=a²-b²

==> $\frac{-1(3D-4)}{(3D)^{2} -4^{2} }cos2x$

==> $\frac{-1(3D-4)}{9D^{2} -16 }cos2x$

==> Replace D² by -(2)²

==> $\frac{-1(3D-4)}{9(-(2^{2})) -16 }cos2x$

==> $\frac{-1(3D-4)}{9(-(4)) -16 }cos2x$

==> $\frac{-1(3D-4)}{-36 -16 }cos2x$

==> $\frac{-1(3D-4)}{-(36+16 ) }cos2x$

==> $\frac{-1(3D-4)}{-(52) }cos2x$

==> $\frac{(3D-4)}{52 }cos2x$

==> 1/52 (3D-4)cos2x

==> 1/52 (3Dcos2x-4cos2x)

==> Dcos2x = -2sin2x

==> 1/52 (3(-2)sin2x-4cos2x)

==>1/52 (-6sin2x-4cos2x)

P.I = 1/52 (-6sin2x-4cos2x)