# Homework Help: Second-order Linear Equations

1. Apr 17, 2016

### Dusty912

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

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

2. Apr 17, 2016

### LCKurtz

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

3. Apr 17, 2016

### Dusty912

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.

4. Apr 17, 2016

### Dusty912

would this be correct?

5. Apr 17, 2016

### vela

Staff Emeritus
Where'd the factor of $i$ go?

6. Apr 17, 2016

### Dusty912

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

7. Apr 17, 2016

### Hiranya Pasan

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)