# Thread: Euler Differential Equation help.

1. ## Euler Differential Equation help.

Hey everyone!

I have a question:

$\displaystyle 2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0$

So I let:
$\displaystyle y = x^k$
$\displaystyle y' = kx^{k-1}$
$\displaystyle y'' = k(k-1)x^{k-2}$

Therefore,

$\displaystyle 2x^2k(k-1)x^{k-2} + 3xkx^{k-1} - 15x^k = 0$

So would I be right in saying:

$\displaystyle x^k(2k(k-1) + 3k - 15) = 0$

$\displaystyle x^k \ne 0$

$\displaystyle 2k(k-1) + 3k - 15 = 0$

Therefore, k = 2.5, k = -3

$\displaystyle y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}$

Would this solution be right?

2. Originally Posted by falconed
Hey everyone!

I have a question:

$\displaystyle 2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0$

So I let:
$\displaystyle y = x^k$
$\displaystyle y' = kx^{k-1}$
$\displaystyle y'' = k^2x^{k-2}$

Therefore,

$\displaystyle 2x^2k^2x^{k-2} + 3xkx^{k-1} - 15x^k = 0$

So would I be right in saying:

$\displaystyle x^k(2k(k-1) + 3k - 15) = 0$

$\displaystyle x^k \ne 0$

$\displaystyle 2k(k-1) + 3k - 15 = 0$

Therefore, k = 2.5, k = -3

$\displaystyle y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}$

Would this solution be right?

I think your signs are backwards

$\displaystyle 2k^2-k-15=(2k+5)(k-3)$

So $\displaystyle k=3$ and $\displaystyle k=-\frac{5}{2}$

3. Originally Posted by falconed
Hey everyone!

I have a question:

$\displaystyle 2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0$

So I let:
$\displaystyle y = x^k$
$\displaystyle y' = kx^{k-1}$
$\displaystyle y'' = k(k-1)x^{k-2}$

Therefore,

$\displaystyle 2x^2k(k-1)x^{k-2} + 3xkx^{k-1} - 15x^k = 0$

So would I be right in saying:

$\displaystyle x^k(2k(k-1) + 3k - 15) = 0$

$\displaystyle x^k \ne 0$

$\displaystyle 2k(k-1) + 3k - 15 = 0$

Therefore, k = 2.5, k = -3

$\displaystyle y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}$

Would this solution be right?

I fixed a mistake in your second derivative.

Yes, that solution you have would be correct.

4. Originally Posted by TheEmptySet
I think your signs are backwards

$\displaystyle 2k^2-k-15=(2k+5)(k-3)$

So $\displaystyle k=3$ and $\displaystyle k=-\frac{5}{2}$
It should be $\displaystyle 2k^2\,{\color{red}+}\,k-15=(2k-5)(k+3)$...right?

5. Originally Posted by Chris L T521
It should be $\displaystyle 2k^2\,{\color{red}+}\,k-15=(2k-5)(k+3)$...right?