Question

Find the
particular integral of
Y" + 2y'+ y = coshx​

Answer 1

SOLUTION

TO DETERMINE

The particular integral of

y" + 2y' + y = cos hx

EVALUATION

Here the given differential equation is

y" + 2y' + y = cos hx

Which can be rewritten as

( D² + 2D + 1 ) y = cos hx

Now the particular integral

$\displaystyle \sf{ = \frac{1}{ {D}^{2} + 2D + 1} \cos hx }$

$\displaystyle \sf{ = \frac{1}{ {(D + 1)}^{2} } \cos hx }$

$\displaystyle \sf{ = \frac{1}{ {(D + 1)}^{2} } \bigg( \: \frac{ {e}^{x} + {e}^{ - x} }{2} \bigg)}$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \frac{1}{ {(D + 1)}^{2} } \: {e}^{x} + \frac{1}{ {(D + 1)}^{2} } \: {e}^{ - x} \: \bigg] }$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \frac{1}{ {(1 + 1)}^{2} } \: {e}^{x} + \frac{ {x}^{2} }{2} \: {e}^{ - x} \: \bigg] }$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \frac{1}{ {(2)}^{2} } \: {e}^{x} + \frac{ {x}^{2} }{2} \: {e}^{ - x} \: \bigg] }$

$\displaystyle \sf{ = \frac{1}{2} \bigg[ \: \: \frac{ {e}^{x} }{4} \: + \frac{ {x}^{2} }{2} \: {e}^{ - x} \: \bigg] }$

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