# Thread: second order differential equation

1. ## second order differential equation

solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

general solution (alpha=2, beta=3)
y=Ae^[(2+3i)x] + Be^[(2-3i)x]
y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

how to get from the first step to the second step? someone please explain to me thanks =)

2. Originally Posted by yen yen
solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

general solution (alpha=2, beta=3)
y=Ae^[(2+3i)x] + Be^[(2-3i)x]
y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

how to get from the first step to the second step? someone please explain to me thanks =)

$\displaystyle y=Ae^{(2+3i)x} + Be^{(2-3i)x}$

which may be written:

$\displaystyle y=Ae^{2x}e^{3ix} + Be^{2x}e^{-3ix}$

Now expand the two expontials with imaginary powers into $\displaystyle \sin$'s and $\displaystyle \cos$'s and rearrange and re-lable the constanants.

(that is use: $\displaystyle e^{3ix}=\cos(3x)+i \sin(3x)$ and $\displaystyle e^{-3ix}=\cos(3x)-i \sin(3x)$ )

CB

3. Originally Posted by CaptainBlack

Now expand the two expontials with imaginary powers into $\displaystyle \sin$'s and $\displaystyle \cos$'s and rearrange and re-lable the constanants.

(that is use: $\displaystyle e^{3ix}=\cos(3x)+i \sin(3x)$ and $\displaystyle e^{-3ix}=\cos(3x)-i \sin(3x)$ )

CB
i get that but how do you get to
y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

wouldnt u get
y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???

4. Originally Posted by yen yen
i get that but how do you get to
y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

wouldnt u get
y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???
Now collect the terms with cos and sign together to get:

$\displaystyle y=(A+B)e^{2x}\cos(3x)+(A-B)e^{2x} i \sin(3x)$

and put $\displaystyle C_1=(A+B)$ and $\displaystyle C_2=i(A-B)$

CB