Question

Question #c4abb

Answer 1

`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

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`