Question

Use an integrating factor to solve the following first order linear ODE. xy' + 2y = 3x, y(1) = 3 Find the end behavior of y as x rightarrow infinity.

Answer 1

Solution -->
`y(x) = x + (2)/(x^(2))`

when x --> infinity the y goes to infinity too.

Explanation:

We have the eq.

`x y'+2 y = 3 x`

with y(1) = 3. So a reconfiguration o this last one equation can be like:

`y' + (2)/(x) y = 3`

so is of the form y'+p(x) y = f(x), where the integral factor is given by:

`\mu = \exp[ \int p(x) dx]`

is,

`\mu = \exp[ \int (2)/(x) dx] = x^(2)`

Multiplying the whole equation with this integral factor we can write the expression:

`\int (d)/(dx)[y x^(2)] dx = \int 3x^(2) dx`

and from this we obtain,

`y x^(2) = x^(3) + c,`

So when y(x=1) = 3, c = 2 and the complete solution can be writen as:

`y(x) = x - (2)/(x^(2))`.

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.