ODE involving expoentials, L'Holital's rule, Help

[creative22]

The question is: Consider the ODE y'(x)+ay(x)=g(x), where a denotes a positive constant, and g contained in C(0, +infinity), Suppose futher that g->b > 0 as x->+infinity. Show that any solution, say phi, of the ODE, has the property that phi->b/a as x-> +infinity.

Hint: Start by finding the general solution of the given equation. Note L'Hopital's rule can be used to simplify the reasoning.

We were giving the hint to use the equation
phi(x)=e^(integral from x0 to x (P(t)dt)) * {const + integral (x0 to x) e^ (-integral (x to t) P(s) ds) * q(t) dt } Assuming p,q are constant in (0, +infinity) and phi(x) is constant.

I tried changing the variables of the equation in the hint to make it work for the one in the problem. However, I haven't progressed very far in finding the general solution to the differential equation. Honestly, I am not sure where to begin with this problem. Any advice and guidance on starting the problem would be great. Thanks.

 

[Deleted member 4993]

The question is: Consider the ODE y'(x)+ay(x)=g(x), where a denotes a positive constant, and g contained in C(0, +infinity), Suppose futher that g->b > 0 as x->+infinity. Show that any solution, say phi, of the ODE, has the property that phi->b/a as x-> +infinity.

Hint: Start by finding the general solution of the given equation. Note L'Hopital's rule can be used to simplify the reasoning.

We were giving the hint to use the equation
phi(x)=e^(integral from x0 to x (P(t)dt)) * {const + integral (x0 to x) e^ (-integral (x to t) P(s) ds) * q(t) dt } Assuming p,q are constant in (0, +infinity) and phi(x) is constant. <<< Are you sure this is what the problem states?

I tried changing the variables of the equation in the hint to make it work for the one in the problem. However, I haven't progressed very far in finding the general solution to the differential equation. Honestly, I am not sure where to begin with this problem. Any advice and guidance on starting the problem would be great. Thanks.

The general solution for y is:

\(\displaystyle y \ = \ e^{-ax}\cdot \int e^{ax}\cdot g(x) dx\)

 

[Deleted member 4993]

I was wondering if you could explain or show me how you determined what the solution was to
viewtopic.php?f=15&t=43407&p=168792&e=168792

This is a "standard" solution of first-order-linear ODE - look-up in almost any standard textbook.