# Differential equation - general and parictular solution

• May 20th 2009, 04:37 AM
jokke22
Differential equation - general and parictular solution
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?
• May 20th 2009, 04:48 AM
TheAbstractionist
Hi jokke22.

Quote:

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. (Happy)
• May 20th 2009, 08:15 AM
jokke22
Quote:

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. (Happy)

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... (Thinking)
• May 20th 2009, 08:40 AM
TheAbstractionist
Quote:

Originally Posted by jokke22
I end up with...

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

$\frac a4\ =\ 1$

$\frac{b-4a}4\ =\ -4$
• May 20th 2009, 03:21 PM
HallsofIvy
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.
• May 22nd 2009, 01:34 AM
jokke22
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,

$y_n= y^g_n + y^p_n$

$= Ce^\frac{x}{2}Dxe^\frac{x}{2} +$ (Wondering)
• May 22nd 2009, 03:08 AM
jokke22
Maybe I get it wrong... Since $\frac{a}{4}=1$ and $\frac{b-4a}{4}=-4$

The particular solution is:

$Ce^\frac{1}{2} + Dxe^\frac{-4}{2}$

(Wondering)