# Solving diffrential equations

• December 5th 2008, 05:25 AM
AshleyT
Solving diffrential equations
Well i couldn't go to school today for various reasons.
So i'm trying to teach myself some stuff and am having trouble understanding the method.

Quote:

Find a differential equation satisfied by y = Ax^2 + Bx^0.5 for ALL values of A and B
The method goes like this :
if
$y = Ax^2 + Bx^0.5$
then
$\frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^0.5$
and $\frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^0.5$

Ok, i get that :) and i also get you can use elimination to then solve this...but i don't get what they've done...

By eliminating B from the first and second equation...they get
$y - 2x\frac{dy}{dx} = A(x^2 - 4x^2) + B(x^0.5 - x^0.5)$
This is where im confused...where the hell do they even get $2x\frac{dy}{dx}$
from?

Note: Its meant to be x to the power of a half...but latex doesnt like it lol.

Thank-you
• December 5th 2008, 09:06 AM
running-gag
Quote:

Originally Posted by AshleyT
Well i couldn't go to school today for various reasons.
So i'm trying to teach myself some stuff and am having trouble understanding the method.

The method goes like this :
if
$y = Ax^2 + Bx^0.5$
then
$\frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^0.5$
and $\frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^0.5$

Ok, i get that :) and i also get you can use elimination to then solve this...but i don't get what they've done...

By eliminating B from the first and second equation...they get
$y - 2x\frac{dy}{dx} = A(x^2 - 4x^2) + B(x^0.5 - x^0.5)$
This is where im confused...where the hell do they even get $2x\frac{dy}{dx}$
from?

Note: Its meant to be x to the power of a half...but latex doesnt like it lol.

Thank-you

Hi,
$y = Ax^2 + Bx^{\frac{1}{2}}$
$\frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^{-\frac{1}{2}}$
and $\frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^{-\frac{3}{2}}$
$2x \frac{dy}{dx} = 4Ax^2 + Bx^{\frac{1}{2}}$