# Proving solution to ODE is decreasing

If I have an equation of the form $$y’=A(x)y+B(x)$$ with an initial condition $$y(0)=0$$ where $$A(x),B(x)$$ are negative and decreasing for all $$x>0$$ with the following limits being finite and negative: $$\lim_{t\rightarrow \infty} A(x),B(x)=L,M .$$, how can I show that the function solution $$y(x)$$ is decreasing? The functions $$A(x),B(x)$$ I cannot express them in an explicit way unfortunately.

So my attempt so far is: Solving the equation generally yields easily that $$y(x)$$ is a negative solution. I also know that $$e^{\int A(x) dx}y(x)$$ is decreasing too. Because $$B(x)$$ is negative then for some “initial” interval $$[0,x_0]$$ I have that $$y’(x)<0%$$ so I was thinking to prove by contradiction that there’s an interval $$[x_0,x_1]$$ where $$y’(x)>0$$ by comparing maybe to the equation $$y’=A(x)y(x)$$ but not sure how to approach.

Help is appreciated

• But I wrote strictly that A(x) and B(x) are negative! Mar 8 at 5:48

I don't think that is true? Consider the ODE $$y'(x)=-e^xy(x)-x$$ with solution $$y(x)=e^{-e^x}\int_0^x-e^{e^t}t\mathrm{d}t$$ where $$y'(x)=e^{x-e^x}\int_0^xe^{e^t}t\mathrm{d}t-x$$ computing numerically, is positive at $$x=2$$. And finite limits can be achieved by connecting $$-e^x$$ and $$x$$ with functions that do converge at infinity.