Question

Find the particular integral of ( d²+a²) y = cos ax​

Answer 1

Answer:

helo guys its your answer

Step-by-step explanation:

f D is the particular integral of f(D) y = X. 2. 1. X. X dx. D. = ∫ . 3. 1 ax ax. X e. X e dx. D a ... Then to find PI put D – 2 for D in f(D) and take out e–2x. ∴. PI = { }. 2 ... ax f D f D. Put D2 = – a2.

Answer 2

The particular integral of  $(d^2+a^2)y=cos ax$ is $\frac{x}{2a} sinax.$

Step-by-step explanation:

Given:

$(d^2+a^2)y=cos ax$

To find:

The particular integral of $(d^2+a^2)y=cos ax$.

Solution:

$(d^2+a^2)y=cos ax$

Here  $f(0)=0$,

$(d^2+a^2)=0$

$d^2=-a^2$

$d=-ia$

Now the particular integral $(PI),$                              

$PI=\frac{1}{d^2+a^2} cosax$

     $=\frac{1}{(d+ia)(d-ia)} cosax$

Here, $cosax=(\frac{e^{iax}+e^{-ax} }{2i} )$

     $= \frac{1}{(d+ia)(d-ia)} (\frac{e^{iax}+e^{-iax} }{2i} )$

It can be written as,

    $=\frac{1}{2i} (\frac{1}{d+ia} e^{iax} -\frac{1}{d-ia} e^{-iax} )$

Here $d=ia, d=-ia,$

    $=\frac{1}{2i} (\frac{1}{(d-ia)} \frac{1}{2i} )e^{iax} +(\frac{1}{(d+ia)} \frac{-1}{2i} )e^{-iax}$

    $=\frac{1}{2i}(\frac{1}{2ia} (\frac{1}{(d^+ia)}e^{iax} )-\frac{1}{2ia} (\frac{1}{d^-ia} e^{-iax}))$

    $=\frac{1}{2i} (\frac{1}{2ia} xe^{iax} -\frac{1}{2ia} xe^{-iax} )$

    $=\frac{1}{2i}(\frac{x}{ia} (\frac{e^{iax} -e^{-iax} }{2} ))$

    $=\frac{x}{2ai^2} (-sinax)$

Here, $i^{2}=-1$

     $=\frac{x}{2a} sinax$

So the particular integral is  $\frac{x}{2a} sinax.$