Question #c4abb

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Question #c4abb

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Answer 1 · 2018-01-03T13:35:23

$y(x) = sinx/x-cosx$

Explanation:

Start by solving the homogeneous equation:

$xdy/dx+y =0$

For $x!=0$ the equation is separable and we have:

$dy/dx =-y/x$

$dy/y = -dx/x$

$lnabsy = -lnabsx+c$

$y=-C/x$

We can now look for a particular solution in the form:

$bary(x) = -C/xv(x)$

$(dbary)/dx = C/x^2v(x) - C/xv'(x)$

Substituting in the original equation:

$x(dbary)/dx +bary = xsinx$

$C/xv(x) -Cv'(x) - C/xv(x) = xsinx$

$v'(x) = - x/Csinx$

$v(x) = -1/C int xsinxdx$

Integrating by parts:

$v(x) = 1/Cxcosx -1/C int cosxdx$

$v(x) = 1/Cxcosx -1/C sinx +C_1$

and so:

$bary(x) = -C/x v(x) = -cosx +sinx/x-(C C_1)/x$

The general solution is then:

$y(x) =-C/x-cosx+sinx/x$

and posing $y(pi) = 1$ we can determine the constant:

$1= -C/pi -cospi + sin pi/pi$

$1= -C/pi +1$

so $C=0$

Finally:

$y(x) = sinx/x-cosx$

Answer 2 · 2018-01-03T14:14:09

y= $-$ cosx+ $sinx/x$

Explanation:

x $dy/dy$ +y=xsinx
$dy/dy$ + $y/x$ = $sinx$

integrating factor,
= $inte^(1/x)$ dx =x

yx= $intsinx.xdx$

integration by parts,
$intsinx.xdx$ =-xcosx+sinx+C

therefore,
xy=-xcosx+sinx+C

y( $pi$ )=1 $rArr$ when x= $pi$ , y=1

$pi$ =- $pi$ cos $pi$ +sin $pi$ +C

c=0

therefore,

xy=-xcosx+sinx
or
y= $-cosx$ + $sinx/x$

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Question #c4abb

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