Our equation is a linear ordinary differential equation of order 1.
These equations are of the form
(dy)/(dx) + f(x)y(x) = g(x).
We can solve them by multiplying by `e^(int f(x)dx)`
It might be easier to show the method if I solve this problem.
In our case `f(x)=2/x` , `g(x) = sin(x)/x^2` . We got this by dividing both sides by x.
`int f(x) dx = int 2/x dx = 2 ln(x)` ,
We are going to multiply both sides by `e^(2ln(x))=e^(ln(x^2))=x^2`
`x^2(dy)/(dx) + 2xy = sinx`
Now we notice that `(d(x^2y))/(dx) = x^2(dy)/(dx) + 2xy` by the product property.
So our equation is actually
`(d(x^2y))/(dx) = sinx`
Integrating both sides we get
`x^2y = -cosx + C`
Solving for y we get
`y = -(cosx)/(x^2) + C/(x^2)`
Using our intital condition that y(2) = 1
`1 = -(cos(2))/(2^2) + C/(2^2)`
`1+(cos(2))/4 = C/4`
`C = 4 + cos(2)`
So our solution is
`y = -(cos(x))/(x^2) + (4 + cos(2))/x^2`
Hope that helps...