Answer:
Solution --> [tex] y(x) = x + \frac{2}{x^{2}} [/tex]
when x --> infinity the y goes to infinity too.
Step-by-step explanation:
We have the eq.
[tex] x y'+2 y = 3 x [/tex]
with y(1) = 3. So a reconfiguration o this last one equation can be like:
[tex] y' + \frac{2}{x} y = 3 [/tex]
so is of the form y'+p(x) y = f(x), where the integral factor is given by:
[tex] \mu = \exp[ \int p(x) dx] [/tex]
is,
[tex] \mu = \exp[ \int \frac{2}{x} dx] = x^{2} [/tex]
Multiplying the whole equation with this integral factor we can write the expression:
[tex] \int \frac{d}{dx}[y x^{2}] dx = \int 3x^{2} dx [/tex]
and from this we obtain,
[tex] y x^{2} = x^{3} + c,[/tex]
So when y(x=1) = 3, c = 2 and the complete solution can be writen as:
[tex] y(x) = x - \frac{2}{x^{2}} [/tex].
And we can see that when x --> infinity the second term of the solution is practically zero and the first is infinity so y also goes to infinity when x does.
