# 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.

thanks for your help :)

LCKurtz
Science Advisor
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.

thanks for your help :)

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
Science Advisor
Homework Helper
Education Advisor
Where'd the factor of ##i## go?

oops so it would be y''re(t) + py're(t) +qyre(t) +iy''im(t) +ipy'im(t) +iqyim(t)=0

lets take yre as a solution, y''re + py're+qyre=0
then take yim as a solution, y''im+py'im +qyim=0
the multiply with i both sides of the 2nd eqn,
i(y''im+py'im +qyim)=0

linear combination of particular solutions isalso a solution,

general solution y(t)=yre(t) + i yim(t)