Question on Variation of Parameters

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yungman
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I have a question on the integration part of the Variation of Parameters. Given .[tex]y''+P(x)y'+Q(x)y=f(x)[/tex]

The associate homogeneous solution .[tex]y_c=c_1y_1 + c_2y_2[/tex].

The particular solution .[tex]y_p=u_1y_1 + c_2y_2[/tex].

[tex]u'_1 = -\frac{W_1}{W} = -\frac{y_2f(x)}{W}[/tex]

This is where I have question. Some books use indefinite integral with the integration constant equal 0.

[tex]u_1= -\int \frac{y_2f(x)}{W}dx[/tex]

But other books gave:

[tex]u_1= -\int_{x_0}^x \frac{y_2f(s)}{W}ds[/tex]

Where [tex]x_0[/tex] is any number in I.

None of the books explain this. Can anyone explain to me about this?

Thanks
 
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If yp = u1y1 + u2y2 is a particular solution you found, then compare this to:

Yp = (u1+A)y1 + (u2+B)y2

=u1y1 + u2y2 + Ay1+ By2

When you plug the second expression into the equation the last two terms give 0 because they satisfy the homogeneous equation. So the constants don't give anything extra.
 
LCKurtz said:
If yp = u1y1 + u2y2 is a particular solution you found, then compare this to:

Yp = (u1+A)y1 + (u2+B)y2

=u1y1 + u2y2 + Ay1+ By2

When you plug the second expression into the equation the last two terms give 0 because they satisfy the homogeneous equation. So the constants don't give anything extra.

Thanks for the respond.

So you mean even using definite integral, substituding in x0 only produce a constant as in your example of A and B. These will be absorbed into the associate homogeneous part ( into c1 and c2).
 
yungman said:
This is where I have question. Some books use indefinite integral with the integration constant equal 0.

[tex]u_1= -\int \frac{y_2f(x)}{W}dx[/tex]

But other books gave:

[tex]u_1= -\int_{x_0}^x \frac{y_2f(s)}{W}ds[/tex]

Where [tex]x_0[/tex] is any number in I.

None of the books explain this. Can anyone explain to me about this?

Thanks

Does it matter? We only want a particular solution.