Solving 2nd Order DEs w/ Undermined Coefficients & e^x, cosx, sinx

[jaydnul]

1. For a non-homogeneous, linear, 2nd order DE, using the method of undermined coefficients, what do you use for the particular solution for $e^x$ and $cosx$ or $sinx$. For example, if it is $x^2$, you use $Ax^2+Bx+C$.

2.For $y''-2y'+2y=0$ i put $y(x)=c_1e^{(1+i)x}+c_2e^{(1-i)x}$ but got it wrong. The right answer was written as $y(x)=e^{ix}(c_1cosx+c_2sinx)$. I always assumed that an answer using $e^x$ was right. How do I know when to use $e^x$ and when to use $sinx$ or $cosx$?

Thanks a bunch!

 

[Simon Bridge]

1.
You use the solution that fits with the rest of the DE.

How do you know in advance? By experience. Painful experience. Builds character.
That's why you are given these exercises - to gain the experience.

Have you seen:
http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx

i.e. if the inhomogeniety is e^x, then try Ae^x as a particular solution.

2.
For y''-2y'+2y = 0, you guess $y=e^{\lambda x}$ - which ends up with your solution right?
- expand the "right" answer in terms of powers of e and see if it matches yours.

 

[jaydnul]

Thanks Simon. I'm still a little confused. What do you mean by expanding it in terms of powers of e?

 

[Simon Bridge]

You can express sine and cosine as a sum of complex exponentials.
Look up Euler's formula.

general solutions of the form $y=Ae^{\lambda_1x}+Be^{\lambda_2 x}$
and of form $y=e^{ax}C\sin(x)+D\cos(x)$ should be equivalent in this case.

 

[jaydnul]

Oh ok i see. Thanks