Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x\in [x_0,\infty)$

share|cite|improve this question
by the way, that expression is usually called a "differential inequality" or a "differential relation". :) – Willie Wong Feb 11 '11 at 16:41
up vote 3 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.