# Second-order Linear Equations

## Homework Statement

Suppose y(t) is a complex-valued solution of y'' +py' + qy=0 where p and q are real numbers. Show that if y(t)=yre(t) + iyim(t), where yre(t) and yim(t) are real valued, then both yre(t) and yim(t) are solutions of the second-order equation.

## Homework Equations

We can use the idea that a complex numbers is zero as long as it's real and imaginary parts are equal to zero.

## The Attempt at a Solution

I do not have an attempt at this solution. I do not know where to begin.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Suppose y(t) is a complex-valued solution of y'' +py' + qy=0 where p and q are real numbers. Show that if y(t)=yre(t) + iyim(t), where yre(t) and yim(t) are real valued, then both yre(t) and yim(t) are solutions of the second-order equation.

## Homework Equations

We can use the idea that a complex numbers is zero as long as it's real and imaginary parts are equal to zero.

## The Attempt at a Solution

I do not have an attempt at this solution. I do not know where to begin.

Begin by plugging y(t)=yre(t) + iyim(t) into your DE and see what happens.

then I would get y''re(t) + py're(t) +qyre(t) +y''im(t) +py'im(t) +qyim(t)=0

so the real part would add up too zero and the imaginary part would add up to zero.

would this be correct?

vela
Staff Emeritus
Homework Helper