# urgent, impossible differential equation

• Jul 19th 2007, 08:43 AM
james30
urgent, impossible differential equation
ive been given a differential equation:

(d2y/dx2)-3(dy/dx)+2y=xe^2x

and cannot work out the particular solution. i have tried
$y=Ax^{2} e^{x}$

getting my y' as $2Cxe^{x} + Cx^{2}e^{x}$

and y'' as $Cx^{2}e^{x} + 4Cxe^{x} + 2Ce^{x}$
but then end up getting 2C=0

is this question impossible or have i used the wrong particular solution and if so how would i go about it?

HELP!!
• Jul 19th 2007, 09:06 AM
topsquark
Quote:

Originally Posted by james30
ive been given a differential equation:

(d2y/dx2)-3(dy/dx)+2y=xe^2x

and cannot work out the particular solution. i have tried
$y=Ax^{2} e^{x}$

getting my y' as $2Cxe^{x} + Cx^{2}e^{x}$

and y'' as $Cx^{2}e^{x} + 4Cxe^{x} + 2Ce^{x}$
but then end up getting 2C=0

is this question impossible or have i used the wrong particular solution and if so how would i go about it?

HELP!!

$\frac{d^2y}{dx^2} -3 \frac{dy}{dx} + 2y = xe^{2x}$

The problem is that if we try only one term we get multiple conditions on it. So typically we'd try $y_p(x) = Cxe^{2x} + De^{2x}$, but the last term is already a part of the homogeneous solution. So try instead:
$y_p(x) = Cx^2e^{2x} + Dxe^{2x}$

$y_p^{\prime}(x) = 2Cx^2e^{2x} + (2C + 2D)xe^{2x} + De^{2x}$

$y_p^{\prime \prime}(x) = 4Cx^2e^{2x} + (8C + 4D)xe^{2x} + (2C + 4D)e^{2x}$

So
$(4Cx^2e^{2x} + (8C + 4D)xe^{2x} + (2C + 4D)e^{2x}) -3(2Cx^2e^{2x} + (2C + 2D)xe^{2x}$ $+ De^{2x}) + 2(Cx^2e^{2x} + Dxe^{2x}) = xe^{2x}$

Simplifying:
$2Cxe^{2x} + (2C + D)e^{2x} = xe^{2x}$

So
$2C = 1$
and
$2C + D = 0$

Thus
$C = \frac{1}{2}$ and $D = -1$.

Thus the particular solution is
$y_p(x) = \frac{1}{2}x^2e^{2x} - xe^{2x}$

-Dan
• Jul 23rd 2007, 12:26 AM
james30
So
$2C = 1$

how does this work? i dont see how you get the 2C=1
• Jul 23rd 2007, 12:40 AM
Jhevon
I'd try $y_p = (Ax + B)e^{2x}$
• Jul 23rd 2007, 12:51 AM
james30
• Jul 23rd 2007, 07:00 AM
topsquark
Quote:

Originally Posted by james30

I'm setting the coefficients of like terms on each side of the equation to be equal to each other.

There is an $xe^{2x}$ on each side so the coefficients must be the same. Thus 2C = 1.

Similarly, there is no $e^{2x}$ on the RHS, so the coefficient 2C + D = 0.

-Dan
• Jul 23rd 2007, 07:01 AM
topsquark
Quote:

Originally Posted by Jhevon
I'd try $y_p = (Ax + B)e^{2x}$

This won't work as $Be^{2x}$ is a solution to the homogeneous equation. We simply won't have enough independent conditions left at the end to solve the system.

-Dan