Eliminating the arbitrary constant

**^_^Engineer_Adam^_^** · November 19th 2007, 06:35 AM

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

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

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

i just do not know what to equate can you please help me w/ this

answer:
$x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$

**kalagota** · November 19th 2007, 06:41 AM

Originally Posted by ^_^Engineer_Adam^_^

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

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

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

i just do not know what to equate can you please help me w/ this

answer:
$x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$

kuya, what is the equation?

**^_^Engineer_Adam^_^** · November 19th 2007, 06:52 AM

Originally Posted by kalagota

kuya, what is the equation?

$y = c_{1}x^2 + c_{2}e^{2x}$

**topsquark** · November 19th 2007, 06:57 AM

Originally Posted by ^_^Engineer_Adam^_^

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

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

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

i just do not know what to equate can you please help me w/ this

answer:
$x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0$

You have your two derivatives.

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

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

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

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

-Dan

**^_^Engineer_Adam^_^** · November 19th 2007, 06:58 AM

thank you

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