Question

(D^2+1) y=xcosx to find particular integral in mathe matics in engineering

Answer 1

Step-by-step explanation:

(D^2+1) y=xcosx to find particular integral in mathe matics in engineering answer

Answer 2

The requried answer is $yp=$ $\frac{1}{P(D)} xcosx = \frac{x}{8}cosx +\frac{1}{32}sinx$.

Integration : An integral assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. The process of finding integrals is called integration.

Particular integral : Particular integral is the function obtained by operation 1/∅(x) on f(x). This involves integration and can get laborious at times. As f(x)  changes, particular integral changes; but as long as ∅(D) does not changes; the complimentary function will remain same.

We need to find the particular integral of (D^2+1) y=xcosx.

The particular integral can be obtained by using the operator $\frac{1}{D}$. One has the

P(D) = $D^{2} +9, then \frac{1}{P(D)} = \frac{1}{D^{2}+9 }$ . The particular integral is obtained.

by using the relation $\frac{1}{P(D)}xf(x) = x\frac{1}{P(D)} f(x) - (\frac{P'(D)}{(P(D))^{2} } f(x).$ Then

$yp=$ $\frac{1}{P(D)} xcosx = \frac{x}{8}cosx +\frac{1}{32}sinx$

Henced the requried answer is $yp=$ $\frac{1}{P(D)} xcosx = \frac{x}{8}cosx +\frac{1}{32}sinx$.

To learn more about particular integral follow the given link

https://brainly.in/question/4630073?

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