Posted by **manny** on Friday, September 7, 2007 at 7:23pm.

This is what im given

(1 + x)dy/dx - xy = x + x2

____________________________________________________________________________________________________________________________________________________________

From my understanding I should move everything to one side like:

(1 + x)dy/dx -xy - x -x2 = 0

then dy/ dx = xy + x + x2/(1 + x)

and from here i am lost

Can you guys help!

Thanks

- math- differential equations -
**retada**, Friday, September 7, 2007 at 7:30pm
hello i am retada and i love u

- please answer! -
**manny**, Friday, September 7, 2007 at 8:05pm
retada i believe that was unecessary

- math- differential equations -
**Count Iblis**, Friday, September 7, 2007 at 8:23pm
(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!]

- math- differential equations -
**manny**, Friday, September 7, 2007 at 8:35pm
ok thanks

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