# Equation

• Feb 22nd 2010, 03:33 PM
lehder
Equation
Hi everybody,

We consider the following differential equation:

$(E) : y''+2y'+2y=-(x+1)e^{-x}+2$

$g(x)=\int_0^{x} te^{-t} dt$ is a solution to this equation E

1)-I must deduct the solutions of the equation (E)

I know that the solutions are: $e^{-2x}(\alpha cos(x)+ \beta sin(x))$

But how to determine $\alpha$ and $\beta$??
• Feb 22nd 2010, 04:39 PM
Jester
Quote:

Originally Posted by lehder
Hi everybody,

We consider the following differential equation:

$(E) : y''+2y'+2y=-(x+1)e^{-x}+2$

$g(x)=\int_0^{x} te^{-t} dt$ is a solution to this equation E

1)-I must deduct the solutions of the equation (E)

I know that the solutions are: $e^{-2x}(\alpha cos(x)+ \beta sin(x))$

But how to determine $\alpha$ and $\beta$??

I'll need initial or boundary conditions to determine those.
• Feb 23rd 2010, 12:44 AM
chisigma
The 'incomplete' DE is...

$y^{''} + 2\cdot y^{'} + 2\cdot y=0$ (1)

... and its 'characteristic equation' is...

$x^{2} + 2\cdot x + 2 =0$ (2)

... the solution of which are $x= -1 \pm i$, so that the general integral of the (1) is...

$y= e^{-x}\cdot (\alpha\cdot \cos x + \beta\cdot \sin x)$ (3)

... that is different from the solution of the starting post.

Now a particular solution of the 'complete' DE...

$y^{''} + 2\cdot y^{'} + 2\cdot y= -(x+1)\cdot e^{-x} + 2$ (4)

... You can verify is ...

$g(x)= \int_{0}^{x} t\cdot e^{-t}\cdot dt$ (5)

... so that the general integral of (4) is...

$y= e^{-x}\cdot (\alpha\cdot \cos x + \beta\cdot \sin x) + g(x)$ (6)

Kind regards

$\chi$ $\sigma$