# 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.

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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
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)