Undetermined Coefficients Problem

  • Thread starter Lancelot59
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  • #1
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I'm given:
[tex]y''-y'+\frac{y}{4}=3+e^{\frac{x}{2}}[/tex]
and asked to solve it using undetermined coefficients. Using the auxilary equation
[tex]y=e^{\lambda t}[/tex]
I got [tex]y_{1}=e^{\frac{x}{2}}, y_{1}=xe^{\frac{x}{2}}[/tex]

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

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

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:

[tex]y=axe^{\frac{x}{2}}+b+cx+d[/tex]
[tex]y'=ae^{\frac{x}{2}}+a\frac{x}{2}e^{\frac{x}{2}}+c[/tex]
[tex]y''=a\frac{x}{4}e^{\frac{x}{2}}+ae^{\frac{x}{2}}[/tex]
Gives:
[tex]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}[/tex]

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?
 

Answers and Replies

  • #2
LCKurtz
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Try:

[tex]y_p = A + Bx^2e^{\frac x 2}[/tex]

You don't need lower powers of x times the exponential because they are solutions of the homogeneous equation.
 
  • #3
643
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Try:

[tex]y_p = A + Bx^2e^{\frac x 2}[/tex]

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.
 
  • #5
643
1
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.
 

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