# Diffrential equations!

• December 7th 2008, 01:18 PM
AshleyT
Diffrential equations!
Find differential equation that satisfy all values of the arbitrary constants :
$y = Ax^2 + e^x$
$x\frac{dy}{dx} = 2y + (x - 2)e^x$

Right, so i start off by eliminating e^x
$Equation one : y = Ax^2 + e^x$
$Equation two : \frac{dy}{dx} = 2Ax + e^x$
$Equation three: \frac{d^2y}{dx^2} = 2A + e^x$

Eliminate $e^x$
Minus both equations 1 & 2 and i get
$y - \frac{dy}{dx} = A(x^2 - 2x)$
Minus both equations 2 & 3 and i get
$\frac{d^2y}{dx^2} - \frac{dy}{dx} = A(2 - 2x)$

Now eliminate A
$\frac{y - \frac{dy}{dx}}{(x^2 - 2x)} - \frac{\frac{d^2y}{dx^2} - \frac{dy}{dx}}{(2 - 2x)} = 0$

SO then i cross multiply and get the wrong answer...am i doing something wrong above?

Thank-you
• December 7th 2008, 04:00 PM
mr fantastic
Quote:

Originally Posted by AshleyT
Find differential equation that satisfy all values of the arbitrary constants :
$y = Ax^2 + e^x$
$x\frac{dy}{dx} = 2y + (x - 2)e^x$

Right, so i start off by eliminating e^x
$Equation one : y = Ax^2 + e^x$
$Equation two : \frac{dy}{dx} = 2Ax + e^x$
$Equation three: \frac{d^2y}{dx^2} = 2A + e^x$

Eliminate $e^x$
Minus both equations 1 & 2 and i get
$y - \frac{dy}{dx} = A(x^2 - 2x)$
Minus both equations 2 & 3 and i get
$\frac{d^2y}{dx^2} - \frac{dy}{dx} = A(2 - 2x)$

Now eliminate A
$\frac{y - \frac{dy}{dx}}{(x^2 - 2x)} - \frac{\frac{d^2y}{dx^2} - \frac{dy}{dx}}{(2 - 2x)} = 0$

SO then i cross multiply and get the wrong answer...am i doing something wrong above?

Thank-you

$y = Ax^2 + e^x$ .... (1)

$\Rightarrow \frac{dy}{dx} = 2Ax + e^x \Rightarrow A = \left( \frac{dy}{dx} - e^x\right) \cdot \frac{1}{2x}$ .... (2)

Substitute (2) into (1) to eliminate A and the solution falls out easily.
• December 9th 2008, 10:35 AM
AshleyT
Quote:

Originally Posted by mr fantastic
$y = Ax^2 + e^x$ .... (1)

$\Rightarrow \frac{dy}{dx = 2Ax + e^x \Rightarrow A = \left( \frac{dy}{dx} - e^x\right) \cdot \frac{1}{2x}$ .... (2)

Substitute (2) into (1) to eliminate A and the solution falls out easily.

Uh, there's an error in your post so i cannot read it?
Thank-you very much for replying though :).

Is the second equation in your post meant to be dy/dx ?
• December 9th 2008, 01:33 PM
mr fantastic
Quote:

Originally Posted by AshleyT
Uh, there's an error in your post so i cannot read it?
Thank-you very much for replying though :).

Is the second equation in your post meant to be dy/dx ?

Sorry about that. I was posting in a hurry. Fixed now.
• December 10th 2008, 08:50 AM
AshleyT
Thank-you very much!