# Thread: [SOLVED] non-homogeneous differential eqn. - method of undeterminded coefficients

1. ## [SOLVED] non-homogeneous differential eqn. - method of undeterminded coefficients

Here is the equation:
$y''+3y'+2y=2cos(2x)-6sin(2x)$

I am supposed to solve this differential equation using 2 different methods. Method 1, I understand fine, but method 2 is what I need help understanding.

METHOD 1: Find complete solution using real trial functions. By using $Asin(2x)+Bcos(2x)$, differentiating it and putting it back into the origional equation I got the answer of $y=C_{1} e^{-x}+C_{2} e^{-2}+\frac{3}{5}sin(2x)+\frac{4}{5}cos(2x)$ which IS correct. I got this method down.

METHOD 2: Here are my given directions for "using a complex valued approach"
a) Solve the equation $y''+3y'+2y=2cos(2x)-6sin(2x)=e^{2ix}$ for a patricular solution $z_{p}=c e^{2ix}$, where $c$ is a complex constant.
b) Expanding $z_{p}=c e^{2ix}=z_{1}(x)+i z_{2}(x)$;
c)Noting that: $z_{1}(x)=Re[ce^{2ix}] and z_{2}(x)=Im[ce^{2ix}]$ are solutions to both solutions to $cos(2x)$ and $sin(2x)$ respectively.
d)Now by taking the right combination of $z_{1}(x), z_{2}(x)$ to obtain a particular solution to the original differential equation.

Hopefully I made what I need to do in method 2 clear, unfortuneatly, I am not clear on how to do it. If anyone would be willing to run throught the solution procedure using this "complex valued approach" I would be thankful.

2. Start with the equation $y''+3y'+2y=e^{2ix}$ and we know a particular solution to the homogeneous equation $y''+3y+2y=0$ is $y_1=e^{-x}$. Now let $y=e^{-x}v$ and $w=v'$ and run through that routine in the book to get down to the expression $e^{-x}w'+e^{-x}w=e^{x(2+2i)}$. Integrate it once to get $w=\frac{1}{2i+2}e^{x(2i+1)}+c_1 e^{-x}$. Integrate it again to get $v=-c_1 e^{-x}+\frac{1}{(2i+2)(2i+1)}e^{x(2i+1)}+c2$. Back-substitute to obtain the solution

$y=c_1 e^{-x}-c_2 e^{-2x}+\frac{1}{(2i+2)(2i+1)}e^{2ix}$

to the non-homogeneous equation.

Now the part with c1 and c2 represent the solution to the homogeneous case so that a particular solution is just the other part which I can write is as:

$y_p(x)=\frac{1}{(2i+2)(2i+1)}\left(\cos(2x)+i\sin( 2x)\right)$

Now, if I expand $y_p(x)$ into it's real and imaginary parts, then the real part, when substituted into the ODE, will give $\cos(2x)$ and the imaginary part will give $\sin(2x)$ and therefore, the function $y(x)=2\text{Re}\,y_p(x)-6\text{Im}\,y_p(x)$ is a solution to:

$y''+3y'+2y=2\cos(2x)-6\sin(2x)$

3. Thanks for the help. I just need a little clarification.

Originally Posted by shawsend
Now let $y=e^{-x}v$ and $w=v'$ and run through that routine in the book to get down to the expression $e^{-x}w'+e^{-x}w=e^{x(2+2i)}$.
I understand almost everything put the part quoted above. I am unsure of how you got: $e^{-x}w'+e^{-x}w=e^{x(2+2i)}$. Unfortuneately, I do not have a book because my professor thinks that his handouts in class are better than a textbook when his handouts do not have a single example of this method (or any examples ) or any more explanation that what I gave you in the problem...

We can choose any of the homogeneous solutions: $e^{-x}, e^{-2x}$ right?
These are not particular solutions (because they only satisfy the homogenoues equation) or not?

4. Originally Posted by snaes
Thanks for the help. I just need a little clarification.

I understand almost everything put the part quoted above. I am unsure of how you got: $e^{-x}w'+e^{-x}w=e^{x(2+2i)}$. Unfortuneately, I do not have a book because my professor thinks that his handouts in class are better than a textbook when his handouts do not have a single example of this method (or any examples ) or any more explanation that what I gave you in the problem...

We can choose any of the homogeneous solutions: $e^{-x}, e^{-2x}$ right?
These are not particular solutions (because they only satisfy the homogenoues equation) or not?
Just substitute $ye^{-x}$ into the original DE to arrive at the DE in w. Check out any DE text book on the subject of "Reduction of Order" and work through the example to better understand it. You can use any of the two general solutions and $e^{-x}$ is a particular solution while $c_1 e^{-x}$ is a general solution.