1. ## DV equation.

Hello,

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

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

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

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

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

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

Who can help me? Greets.

2. Originally Posted by Bert

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

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

$\displaystyle \int \frac{1}{y}dy=\int\frac{x+1}{x}dx$

RonL

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

Greets.

4. It seems that, $\displaystyle x\geq 0$
$\displaystyle y=x^2+Cxe^x$ are all solutions to this differential equation.

5. 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,
$\displaystyle y'-(1+1/x)y=x-x^2$
Begin by solving,
$\displaystyle \frac{dy}{dx}-(1+1/x)y=0$
Dis is seperable which gives,
$\displaystyle \frac{1}{y}dy=(1+1/x)dx$
Integrating,
$\displaystyle \ln y=x+\ln x+C$
Thus,
$\displaystyle y=\exp (x+\ln x+C)=Cxe^x$

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