# Differential equation with inequalities

Given that $y'(x)+p(x)y(x)\geq 0$ and $y(x_0)\geq 0$, how does one go about showing that $y(x)\geq0$ for all $x\in [x_0,\infty)$

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by the way, that expression is usually called a "differential inequality" or a "differential relation". :) – Willie Wong Feb 11 '11 at 16:41

Define $b=y'+py$, so that you have, well, $y'+py=b$ and $b\geq0$. Solve the inhomogeneous equation for $y$, and use now that $b$ is non-negative.

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Thanks, I had tried that before and failed, but tried the same thing again, after your hint, and it worked. – picakhu Feb 11 '11 at 4:18

Can't one reason in the following manner, whitout introducing the function $b$?

Multiplying both sides of the differential inequality $y^\prime (x)+p(x)y(x)\geq 0$ by the positive function $M(x):= \exp \left( \int_{x_0}^x p(t)\ \text{d} t\right)$ yields:

$y^\prime (x) M(x) +y(x)\ p(x)M(x) \geq 0$;

but $M^\prime(x)=p(x) M(x)$, hence the last inequality rewrites:

$\frac{\text{d}}{\text{d} x} \left[ y(x) M(x)\right]\geq 0$,

hence the function $y(x)M(x)$ (which is differentiable in $]x_0,+\infty[$, for it is product of differentiable functions) increases in $[x_0,+\infty[$; this fact implies:

$y(x)M(x)\geq y(x_0)M(x_0)=y(x_0)\geq 0$

and a fortiori $y(x)\geq 0$ for all $x\geq x_0$, which is the claim.

NOTE: The auxiliary funcion $M(x)$ is the reciprocal of the unique solution to the homogeneous ODE:

$\phi^\prime (x)+p(x)\phi(x)=0$

which satisfies $\phi(x_0)=1$. So $M(x)$ is not appeared out of nowhere; in fact the problem you're dealing with can be read as a comparison result between the solution and supersolutions of the problem:

1. $\begin{cases} u^\prime (x)+p(x)u(x)=0 &\text{, in } ]x_0,+\infty[ \\ u(x_0)=y_0\end{cases}$

in the following way: each supersolution of problem 1 is greater than its unique solution in $[x_0,+\infty[$.

P.S.: I didn't check the date before answering. I don't know if this can be considered as necroposting or doesn't matter at all... I'm sorry anyway, next time I'll pay more attenction.

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