Eliminating the arbitrary constant

• Nov 19th 2007, 06:35 AM
Eliminating the arbitrary constant
im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
$\displaystyle y = c_{1}x^2 + c_{2}e^{2x}$

derivating
$\displaystyle y' = 2c_{1}x + 2c_{2}e^{2x}$

$\displaystyle y'' = 2c_{1} +4c_{2}e^{2x}$

$\displaystyle x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$
• Nov 19th 2007, 06:41 AM
kalagota
Quote:

im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
$\displaystyle y = c_{1}x^2 + c_{2}e^{2x}$

derivating
$\displaystyle y' = 2c_{1}x + 2c_{2}e^{2x}$

$\displaystyle y'' = 2c_{1} +4c_{2}e^{2x}$

$\displaystyle x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$

kuya, what is the equation?
• Nov 19th 2007, 06:52 AM
Quote:

Originally Posted by kalagota
kuya, what is the equation?

$\displaystyle y = c_{1}x^2 + c_{2}e^{2x}$
:)
• Nov 19th 2007, 06:57 AM
topsquark
Quote:

im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
$\displaystyle y = c_{1}x^2 + c_{2}e^{2x}$

derivating
$\displaystyle y' = 2c_{1}x + 2c_{2}e^{2x}$

$\displaystyle y'' = 2c_{1} +4c_{2}e^{2x}$

$\displaystyle x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$

Solve the first derivative for $\displaystyle c_1$:
$\displaystyle c_{1} = \frac{y' - 2c_{2}e^{2x}}{2x}$

Plug this value for $\displaystyle c_1$ into your solution and second derivative:
$\displaystyle y = \left ( \frac{y' - 2c_{2}e^{2x}}{2x} \right ) x^2 + c_{2}e^{2x}$

$\displaystyle y'' = 2 \left ( \frac{y' - 2c_{2}e^{2x}}{2x} \right ) + 4c_{2}e^{2x}$

Now solve your solution for $\displaystyle c_2$ and plug that into your second derivative. Now do a lot of algebra. :)

-Dan
• Nov 19th 2007, 06:58 AM