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.

Solution -->

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

Step-by-step explanation:

We have the eq.

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

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

is,

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

and from this we obtain,

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

.

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.

10 * 563.67 = 5636.7

13.46

step-by-step explanation:

16.80 x .20 = 13.46