Differential equation - general and parictular solution

**jokke22** · May 20th 2009, 03:37 AM

Hi!

I have been given an equation as following:

$y^{//}-y^{/}+\frac{1}{4}y = x-4$

Which I am to find the general $y^g_n$ and a particular solution $y^p_n$

I have found that this gives a general solution:

by
$y^g_n = r^2-r+\frac{1}{4}$

One solution = $\frac{1}{2}$

Which gives:
$y^g_n = Ce^\frac{x}{2} + Dxe^\frac{x}{2}$

I am then to find the particular solution $y^p_n$

How to proceed is then my question...

By setting:

$y^p_n = an+b$

and calculate from there?

How would that look?

**TheAbstractionist** · May 20th 2009, 03:48 AM

Hi jokke22.

Originally Posted by jokke22

I am then to find the particular solution $y^p_n$

How to proceed is then my question...

By setting:

$y^p_n = an+b$

and calculate from there?

How would that look?

$y\ =\ ax+b$

$y'\ =\ a$

$y''\ =\ 0$

So $y''-y'+\frac14y\ =\ -a+\frac{ax+b}4\ =\ x-4.$ Then just equate coefficients.

**jokke22** · May 20th 2009, 07:15 AM

Originally Posted by TheAbstractionist

Hi jokke22.

$y\ =\ ax+b$

$y'\ =\ a$

$y''\ =\ 0$

So $y''-y'+\frac14y\ =\ -a+\frac{ax+b}4\ =\ x-4.$ Then just equate coefficients.

Hmmm... I think I misinterpret what I am to do...

I end up with...

$\frac{-4a+ax+b}{4} = x-4$

Grateful if you could show me...

**TheAbstractionist** · May 20th 2009, 07:40 AM

Originally Posted by jokke22

I end up with...

$\frac{-4a+ax+b}{4} = x-4$

$\frac a4\ =\ 1$

$\frac{b-4a}4\ =\ -4$

**HallsofIvy** · May 20th 2009, 02:21 PM

If two polynomials are the same for all x, you can equate individual coefficients. That is what The Abstractionist did.

You can also set x to any two values and then solve the resulting two equations for a and b. For example, since $\frac{-4a+ax+b}{4} = x-4$ , taking x= 0 gives $\frac{-4a+ b}{4}= -4$ or -4a+ b= -16. Taking x= 1 gives $\frac{-4a+ a+ b}{4}= 1- 4$ or $\frac{-3a+ b}{4}= -3$ so $-3a+ b= -12$ . Those are easy to solve.

**jokke22** · May 22nd 2009, 12:34 AM

Okey! I think I get it know... Thank you for the help!

However, how would this then look if you were to put a solution where,

$y_n= y^g_n + y^p_n$

$= Ce^\frac{x}{2}Dxe^\frac{x}{2} +$