# ODE using variation of parameters

1. May 5, 2010

### cheddacheeze

1. The problem statement, all variables and given/known data
You are given that two solutions to the homogeneous Euler-Cauchy equation
$x^2 \frac{d^2}{dx^2}y(x) - 5x \frac{d}{dx} y(x) + 5y(x) = 0$
$y1=x, y2=x^5$

$y''-\frac{5}{x}y'+\frac{5}{x^2}y=-\frac{49}{x^4}$
changing the equation to standard form
use variation of parameters to find a particular solution to the inhomogenous Euler-Cauchy equation

2. Relevant equations
Wronskian
$W=4x^5$

yp (y particular)
$$yp=uy1+vy2$$

$$u= \frac{-49}{12x^3}$$

3. The attempt at a solution
$v' = \frac{y1r}{w}$
$v' = \frac{(x) (-49/x^4)}{4x^5}$
$v' = -\frac{49}{4x^8}$
$v = -\frac{49}{4} \int \frac{1}{x^8}$
$v = (-\frac{49}{4}) (-\frac{1}{7x^7})$
$v = \frac{49}{28x^7}$
$yp = \frac{-49}{12x^3}*x + \frac{49}{28x^7}*x^5$
$yp = \frac{-49}{12x^2} + \frac{49}{28x^2}$

Last edited: May 5, 2010
2. May 5, 2010

### cheddacheeze

have tried and checked and still have not got the right answer

3. May 5, 2010

### cheddacheeze

$yp = \frac{-7}{3x^2}$

tried plugging into the equation and didnt work, something must have gone wrong either in
yp=uy1+vy2
or finding what v' was
can anybody see whats wrong

4. May 6, 2010

### LCKurtz

Remember, when you use the formula for variation of parameters, you must have the leading coefficient = 1, so you are working with this last equation.

There is nothing wrong with your yp. I'm guessing you put it back in the first equation instead of the third one, and you got -49/x2. That is what you should get if you plug into the first equation because it has been multiplied on both sides by x2 relative to the third equation. If you plug your yp into the third equation, it will work exactly.

5. May 6, 2010

### cheddacheeze

i differentiated yp twice
plugged in yp'' yp' and yp into the differential
and i got $\frac{7}{3x^4}$

Last edited: May 6, 2010
6. May 6, 2010

### LCKurtz

Your particular solution yp= -7/(3x2) is correct. If you plug it in for the y in

$$y''-\frac{5}{x}y'+\frac{5}{x^2}y=-\frac{49}{x^4}$$

it works. Or you can multiply that equation through by x2:

$$x^2y''-5xy'+5y=-\frac{49}{x^2}$$

and it works in that equation too.

7. May 7, 2010

### cheddacheeze

turns out computer didnt like the answer in equation form
thanks