Using complex numbers for 2nd order DE

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• Mar 25th 2010, 06:04 PM
snaes
Using complex numbers for 2nd order DE
I can change this equation, using Eulers method, and get the answers using $\displaystyle sin$ and $\displaystyle cos$, but using complex numbers is apparently faster and we have to do it this way for my professor...so if you could help me figure this out I would appreciate it.

$\displaystyle y''-8y'+25y=3 e^{4x} cos(3x)$

HERE IS WHAT I DID:
First, I'll solve the homogeneous case to find the complementary solution and the roots.
$\displaystyle r^{2}-8r+25=0$
$\displaystyle r=4+3i$ or $\displaystyle r=4-3i$
$\displaystyle y_{c}=c_{1} e^{4x}cos(3x)+c_{2}e^{4x}sin(3x)$

Now, to find the solution $\displaystyle y=y_{c}+y_{p}$, where $\displaystyle y_{p}$ is the particular solution to the non-homogeneous equation, I need help. I guessed a few functions but none seemed to work out.

Here is what Ive tried:
Since $\displaystyle 3e^{4x}cos(3x)=3e^{(4+3i)x}$...well the real part I think
and
$\displaystyle 4+3i$ is a root of the homogenous equatoin, so I need multiply my guess for $\displaystyle y_{p}$ by $\displaystyle x$

From here I cannot figure out what function I need to use...
• Mar 25th 2010, 07:38 PM
Prove It
Quote:

Originally Posted by snaes
I can change this equation, using Eulers method, and get the answers using $\displaystyle sin$ and $\displaystyle cos$, but using complex numbers is apparently faster and we have to do it this way for my professor...so if you could help me figure this out I would appreciate it.

$\displaystyle y''-8y'+25y=3 e^{4x} cos(3x)$

HERE IS WHAT I DID:
First, I'll solve the homogeneous case to find the complementary solution and the roots.
$\displaystyle r^{2}-8r+25=0$
$\displaystyle r=4+3i$ or $\displaystyle r=4-3i$
$\displaystyle y_{c}=c_{1} e^{4x}cos(3x)+c_{2}e^{4x}sin(3x)$

Now, to find the solution $\displaystyle y=y_{c}+y_{p}$, where $\displaystyle y_{p}$ is the particular solution to the non-homogeneous equation, I need help. I guessed a few functions but none seemed to work out.

Here is what Ive tried:
Since $\displaystyle 3e^{4x}cos(3x)=3e^{(4+3i)x}$...well the real part I think
and
$\displaystyle 4+3i$ is a root of the homogenous equatoin, so I need multiply my guess for $\displaystyle y_{p}$ by $\displaystyle x$

From here I cannot figure out what function I need to use...

Since the complementary solution involves the RHS, you will need to guess a particular solution of the form

$\displaystyle y_p = A\color{red}x\color{black}e^{4x}\cos{3x} + B\color{red}x\color{black}e^{4x}\sin{3x}$.
• Mar 26th 2010, 05:48 AM
snaes
Thanks, but is there any way to convert this into a "complex" valued expression. I need to use something like: $\displaystyle a x e^{(4+3i)x}$.

 for myhomework we are supposed to do each problem with real valued (sine and cosine) which i can do AND using some $\displaystyle e^{a+bi}$ kind of form, which i cannot do. Any help would be great.