Undetermined Coefficients Problem

[Lancelot59]

I'm given:
y''-y'+\frac{y}{4}=3+e^{\frac{x}{2}}
and asked to solve it using undetermined coefficients. Using the auxilary equation
y=e^{\lambda t}
I got y_{1}=e^{\frac{x}{2}}, y_{1}=xe^{\frac{x}{2}}

Now to solve the particular solution, I chose to guess:

y_{p}=axe^{\frac{x}{2}}+b and y_{p}=cx+d

From what I understand in this case, you're supposed to guess a polynomial of one order higher than what you have. However I basically get a giant mess:

y=axe^{\frac{x}{2}}+b+cx+d
y'=ae^{\frac{x}{2}}+a\frac{x}{2}e^{\frac{x}{2}}+c
y''=a\frac{x}{4}e^{\frac{x}{2}}+ae^{\frac{x}{2}}
Gives:
a\frac{x}{4}e^{\frac{x}{2}}+ae^{\frac{x}{2}}-ae^{\frac{x}{2}}+a\frac{x}{2}e^{\frac{x}{2}}+c+ \frac{1}{4}(axe^{\frac{x}{2}}+b+cx+d)=3+e^{x}{2}

I can't compare coefficients for all of the terms. Some of the exponentials are multiplied by x, which doesn't appear on the other side. What am I supposed to do?

 

[LCKurtz]

Try:

y_p = A + Bx^2e^{\frac x 2}

You don't need lower powers of x times the exponential because they are solutions of the homogeneous equation.

 

[Lancelot59]

Try:

y_p = A + Bx^2e^{\frac x 2}

You don't need lower powers of x times the exponential because they are solutions of the homogeneous equation.

So you need higher powers? That's the only place I can see the x² coming from.

 

[LCKurtz]

Hopefully your text discusses this. To really understand it you need to look at the "Annihilator Method". See:

http://faculty.swosu.edu/michael.dougherty/DiffEqI/lecture10.pdf

If you don't want to wade through all the theory, skip down to about page 5 for some examples.

 

[Lancelot59]

That's actually not being covered...I'll make a note to look into it once the semester is over and I have some spare time.