Solve:
`(dy)/(dx) +(3y)/(x) = g(x)`, where g(x) = `{{:(1,if0lexle1),(1/x,if xgt1):}`
`y(1/2)=1/8` and y(x) is continous on `[0,infty]`.

Explanation

For `0lexlt1`
`(dy)/(dx)+(3y)/(x)=1`
I.F. `e^(int(3/xdx))=e^(3log_(e)x)=x^(3)`
`therefore` Solution is `yx^(3) = intx^(3)dx=1/4x^(4)+C`
`rArr y(1/2) = 1/8`, we get C=0
So, `y=x/4, 0lexlt1`
For `xgt1`
`(dy)/(dx) +3/xy=1/x`
I.F. `=x^(3)`
`therefore` Solution is, `yx^(3)=intx^(2)dx=1/3x^(3)+C`
So, `y=1/3+C/x^(3),xgt1`
Thus, `y={{:(x/4, 0lexlt1),(1/3+C/x^(3),xgt1):}`
Since `y(x)` is continous, we have
`lim_(hto0)x/4=lim_(xto1^(+))(1/3+C/x^(3))`
`rArr C=-1/12`
So, `y={{:(x/4, 0lexle1),(1/3-1/(12x^(3)),xgt1):}`