# Solution by variation of parameters

1. Nov 13, 2005

### asdf1

1) when proving that method works, why do you have to make u and v satisfy the 2nd condition uy1+vy2=0

2) when you're integrating to find yp, why do you leave out the constant that results from the integration?

2. Nov 13, 2005

### saltydog

When you let:

$$y=A(x)y_1(x)+B(x)y_2(x)$$

where $y_1(x)$ and $y_2(x)$ are solutions to the homogeneous equation, and you substitute that into:

$$y^{''}+p(x)y^{'}+q(x)y=R(x)$$

Letting:

$$A^{'}y_1+B^{'}y_2=0$$

reduces the complexity of the resulting equation to the simplified form:

$$A^{'}y_1^{'}+B^{'}y_2^{'}=R(x)$$

Technically you could let it be any non-zero function but zero makes the math easier.

As far as ignoring the constants of integration, note you're looking only for a particular solution to the original DE right: homogeneous part+particular part=solution. And a partcular solution does not contain any arbitrary constants.

3. Nov 14, 2005

### asdf1

thank you very much for clearing those things up!!! :)

4. Nov 14, 2005

### HallsofIvy

Remember that there are, in general, an infinite number of functions
A(x), B(x) such that $$y=A(x)y_1(x)+B(x)y_2(x)$$
Requiring that $$A'y_1+ B'y_2= 0$$
just "narrows the field" and guaranties that there will not be any second derivatives of A or B in the final equations.

Last edited by a moderator: Nov 14, 2005
5. Nov 14, 2005

### asdf1

thank you very much!!! :)