Question

Now shw that if Q(x)->B as x->infintity, then any solution y(x) of equation (9) satisfies y(x)-> B/a as x-> infinity.
equation 9: dy/dx+ay=Q(x)

Answer 1

had a go and thought i'd post as no-one else has had a try, but the integration constant keeps getting in the way

\(y' + a y = Q(x)\)

Using integrating factor

\(y = \dfrac{\int e^{ax} Q(x) dx + C}{e^{ax}}\)


IBP

\(\int e^{ax} Q(x) dx \)

\(= \int \dfrac{d}{dx}(\dfrac{1}{a}e^{ax}) Q(x) dx\)

\(= \dfrac{1}{a}e^{ax} Q(x) - \int \dfrac{1}{a}e^{ax} Q'(x) dx\)

\(\lim\limits_{ x \to \infty} y = \lim\limits_{ x \to \infty} \dfrac{ \dfrac{1}{a}e^{ax} Q(x) - \int \dfrac{1}{a}e^{ax} Q'(x) dx +C}{e^{ax}}\)

\(\lim\limits_{ x \to \infty} Q(x) = B \)

\(=\dfrac{B}{a} - \lim\limits_{ x \to \infty} \dfrac{ \int \dfrac{1}{a}e^{ax} Q'(x) dx +C}{e^{ax}} \qquad \triangle\)


\(\lim\limits_{ x \to \infty} Q(x) = B \implies \lim\limits_{ x \to \infty} Q'(x) = 0\)

So \(\triangle\) becomes

\(=\dfrac{B}{a} - \lim\limits_{ x \to \infty} \dfrac{ \int \dfrac{1}{a}e^{ax} \bullet 0 dx + C}{e^{ax}} \)

\(=\dfrac{B}{a} - \lim\limits_{ x \to \infty} \dfrac{C}{e^{a x}}\)

and that depends on the value of a, so no cigar

looks too complicated an approach too

Answer 2

well you can create counter-examples that show that this only works if a > 0 so that Constant issue goes away

but this is wrong:

\( \lim\limits_{ x \to \infty} \dfrac{ \int \dfrac{1}{a}e^{ax} \color{red}{\bullet 0 }dx + C}{e^{ax}}\)

it really depends on the form of function \(G'(x)\)

Answer 3

Well thank you very much. This helps a lot