## anonymous 5 years ago for what values of the (real) constants a and b all solutions of the equations y"+ay'+by=0, are bounded for x between minus infinity and infinity?

1. anonymous

Solve for the solution to the differential equation generally. Use the characteristic equation (e^(rx) substitution results). You should get the quadratic equation.

2. anonymous

helpme, i dont really get it. could u plis explain more in detail?

3. anonymous

That's a differential equation. You have to solve it. There are many different methods to determine the solutions. One of them is by substituting e^(rx) into the solution. You'll realize, that after you plug in the equation, that the e^(rx) divides into the zero. That leaves you with R^2+aR+b=0, which is a simple equation to solve. Just use the quadratic formula, with a and b as your coefficients. Later on, you substitute the solution back into your formula.

4. anonymous

helpmeplease is right, the Characteristic equation is$r^2+ar+b=0$ so the roots are:$r_{1}=\frac{-a+\sqrt{a^2-4b}}{2}, r_{2}=\frac{-a-\sqrt{a^2-4b}}{2}$

5. anonymous

after get the roots, differentiate the solution then substitute them into y",y' and y in the original equation?

6. anonymous

if the roots are real distinct roots use: $y=C_{1}e^{r_{1}x}+C_{1}e^{r_{2}x}$ If the roots are real and repeated where r1=r2 use: $y=C_{1}xe^{r_{1}x}+C_{1}e^{r_{2}x}$ If the roots are Complex then use: $\lambda=\frac{-a}{2}, \mu i=\pm \frac{\sqrt{a^2-4b}}{2}i$ $y=e^{\lambda x}C_{1}\cos(\mu x)+e^{\lambda x}C_{2}\sin(\mu x)$

7. anonymous