# How to solve $y''+by'+y=0$,$b\geq 0$; for what b does $y(t)$ have $\lim\limits_{t\to\infty} y(t)$?

How to solve $$y''+by'+y=0$$ for $b\geq0$ (just need a quick reminder)? For what $b$ does $\lim\limits_{t \to \infty} y(t)$ exist (this is the more important question to me)?

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There is a mild exception that we discuss later, but in general the solutions of the differential equation have the shape $Ae^{\lambda_1 t}+Be^{\lambda_2 t}$ where $\lambda_1$ and $\lambda_2$ are the (possibly non-real) roots of the equation $\lambda^2+b\lambda+1=0$.

By the Quadratic Formula, the roots of this quadratic equation are $\frac{-b \pm\sqrt{b^2-4}}{2}$.

If $b=0$ we are in trouble, the solutions of the DE are linear combinations of $\sin t$ and $\cos t$, and the limit as $t \to \infty$ does not exist, except in the case of the trivial solution $y=0$.

If $0<b<2$, the solutions of the DE are of the shape $e^{-bt/2}$ times a linear combination of sines and cosines, and the $e^{-bt/2}$ term forces the limits to be $0$. And if $b>2$, the two roots $\lambda_1$ and $\lambda_2$ are negative, so again all solutions $\to 0$ as $t\to\infty$.

In the case $b=2$, the equation $\lambda^2+b\lambda+1=0$ has the double root $\lambda=-1$, and the solutions of the differential equation are linear combinations of $e^{-t}$ and $te^{-t}$. Both of these approach $0$ as $t\to \infty$.

You did not ask about $b<0$. In this case, all solutions except the trivial solution $y=0$ blow up as $t\to\infty$.

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How did you get te^(−t)? –  nnb Apr 16 '12 at 13:25
When the characteristic equation has multiple roots, the usual formula for the solutions has to be modified. The modification is part of the general theory of linear differential equations with constant coefficients, takes quite a while to write out. But you can readily check that if $b=2$, then $te^{-t}$ is indeed a solution. –  André Nicolas Apr 16 '12 at 13:35

If you have a equation like a $y''+ b y'+ c = 0$ , you can find the solution using the associated quadratic polynomial to the differential equation : $D^2 + b D + C = 0$. So, resolve it with the quadratic formula $$D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$ You could have complex roots so, in your solutions should appear $\cos x$, $\sin x$ and then, you can propose the solutions of your differential equation, and the form could be: $$y(t)= k e^{D t}$$ so, analyze $\lim\limits_{t\to \infty}y(t)$. Then look how behaves exponential function.

Regards,

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