Posted by manny on Friday, September 7, 2007 at 7:23pm.
hello i am retada and i love u
retada i believe that was unecessary
(1 + x)dy/dx - xy = x + x^2 --->
y' - x/(1+x) y = x
First solve the homogeneous part of the equation:
y_h' - x/(1+x) y_h =0 --->
y_h = A exp(x)/(1+x)
Then you promote A from a constant to a function of x and put:
y = A(x) Exp(x)/(1+x)
If you subsitute this in the differential equation, then only the term involving the derivative of A will survive. That's because the terms that do not involve the derivative of A are exactly what you get when you take y to be y_h. y_h satisfies the homogeneous equation and therefore they add up to zero.
So, we get:
A' Exp(x)/(1+x) = x --->
A' = x(1+x)Exp(-x)
Integral of Exp(ax) = 1/a Exp(ax)
Differentiate both sides w.r.t. a:
Integral of x Exp(ax) =
(x/a -1/a^2) Exp(ax)
Differentiate again:
Integral of x^2 Exp(ax) =
( x^2/a -2x/a^2 +2/a^3) Exp(ax)
We thus see that:
A(x) = (-3x -3 -x^2)Exp(-x) + const.
[Note: I didn't check if I made any errors in the calculations!]
ok thanks
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