DV equation.

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April 30th 2006, 11:48 AM
Bert

DV equation.

Hello,

Given is $xy^{'}- (x+1)y=x^2-x^3$

Before I start I multiply by $\frac{1}{x}$

Then I get $y'-\frac{x+1}{x}y=\frac{x^2-x^3}{x}$

First I will find the solution of the homogeneous DV $y'-\frac{x+1}{x}y=0$

I can solove this easely $\int dy=\int\frac{x+1}{x}$ and I get $y=x+\ln x+k(x)$

I deveritate this and I get $y'=1+\frac{1}{x}+k'(x)$ and now I need to put in but then I miss ???

Who can help me? Greets.
April 30th 2006, 11:59 AM
CaptainBlack

Quote:

Originally Posted by Bert

First I will find the solution of the homogeneous DV $y'-\frac{x+1}{x}y=0$

I can solove this easely $\int dy=\int\frac{x+1}{x}$ and I get $y=x+\ln x+k(x)$

except you have:

$<br /> \int \frac{1}{y}dy=\int\frac{x+1}{x}dx<br />$

RonL
April 30th 2006, 12:05 PM
Bert

do I need to integrate $y'-\frac{x+1}{x}y$ and not $y'-\frac{x+1}{x}$ ???

Greets.
April 30th 2006, 02:02 PM
ThePerfectHacker

It seems that, $x\geq 0$
$y=x^2+Cxe^x$ are all solutions to this differential equation.
May 1st 2006, 02:34 PM
ThePerfectHacker

That one I solved by using a theorem about linear differencial equations. I have another way.

"The general solution to a non-homogenous differencial equation is the sum of the general solution to the homogenous equation and a specific solution to the non-homogenous equation"

With that theorem,
We will solve,
$y'-(1+1/x)y=x-x^2$
Begin by solving,
$\frac{dy}{dx}-(1+1/x)y=0$
Dis is seperable which gives,
$\frac{1}{y}dy=(1+1/x)dx$
Integrating,
$\ln y=x+\ln x+C$
Thus,
$y=\exp (x+\ln x+C)=Cxe^x$

Now find a particular solution to,
$y'-(1+1/x)y=x-x^2$
It is reasonable to search for solutions of the form,
$y=ax^2+bx+c$
Substitute,
$2ax+b-(1+1/x)(ax^2+bx+c)=x-x^2$
Thus,
$2ax+b-ax^2-bx-c-ax-b-c/x=x-x^2$
We immediately see that $c=0$
Thus, you are left with,
$ax-ax^2-bx=x-x^2$
We also see that $a=1$
From here we see that $b=0$
Thus, $y=1x^2+0x+0=x^2$ is a specific solution.
Thus, all solutions are,
$y=x^2+Cxe^x$