Complex number and product of roots

Question

For a second order ODE

y''+10y'+ 21y=0

which was reduced to this quadratic expression x^2+10x+21=0

• is there any way to tell whether the expression is bounded that is y(x) is either periodic or has a limit 0 as x tends to infinity?

*Does periodic means having only complex solutions?

Answer 1

You only need to calculate the discriminant, that is $\Delta=b^2-4ac$, if $\Delta >0$ the equation has two dinstinct real solutions, if $\Delta=0$ then it has a repeated real solution and if $\Delta<0$ then it has no real solutions. Note that a quadratic with real coefficents cannot have a complex and a real solution.

Answer 2

Suppose that you want to solve the quadratic equation, $ax^2 + bx + c = 0$. The quadratic formula gives the two roots as

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Now, take a look at the surd, $\sqrt{b^2 - 4ac}$. If $b^2 - 4ac$ was negative, then the roots would be complex, because the square root of a negative real number is imaginary.

Intuitively, if $b^2 - 4ac$ is positive, then there will be two distinct real roots. If $b^2 - 4ac$ is zero, then both roots will be the same.

This $b^2 - 4ac$ is what we call the discriminant of the quadratic equation.

Answer 3

To answer the second part, in the differential equation $$ ay''(x)+by'(x)+cy(x)=0 $$ you would need $b=0$ and $ac>0$ to get periodic solutions. With complex, but not purely imaginary eigenvalues you get oscillating solutions where the amplitude changes with an exponential function.