Solve d^2y/dx^2 + 4y = 0?

**ZaZu** · October 1st 2009, 11:04 AM

Hi there,

$(d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
What I know is that if its $dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

Thank you

**Amer** · October 1st 2009, 12:29 PM

Originally Posted by ZaZu

Hi there,

$(d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
What I know is that if its $dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

Thank you

you can see this http://www.mathhelpforum.com/math-help/differential-equations/38182-differential-equations-tutorial.html

in the second order equation let $y=e^{rx}$ then find r

$r^2e^{rx} +4e^{rx} = 0$

$e^{rx}(r^2 + 4)=0\Rightarrow r=\mp 2i$ two complex root

see the link for more information ...

**Matt Westwood** · October 1st 2009, 01:26 PM

Originally Posted by ZaZu

Hi there,

$(d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
What I know is that if its $dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

Thank you

One way of doing it is by putting $p = \frac {dy}{dx}$ .

This leads to $\frac {d^2y}{dx^2} = \frac {dp}{dx} = \frac {dp}{dy} \frac {dy}{dx}$ (by the chain rule).

So $\frac {d^2y}{dx^2} = p \frac {dp}{dy}$

and so we can rewrite it:
$p \frac {dp}{dy} + 4 y = 0$
which is clearly separable.

So you can solve for $p$ , and this will then be a 1st order equation which (when you've done the algebra) you will find also separates.

**ZaZu** · October 4th 2009, 11:46 AM

Thank you very much guys

Greatly appreciated

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