Solve Differential equation for the original equation

**sparky** · February 21st 2011, 11:17 AM

Hi,

Is my working correct for the following question?

Question: Solve for the following differential equation for the original equation: $<br /> \frac{dy}{dx} = \frac{-x}{2y}<br />$

Here is my working:

$<br /> \frac{dy}{dx} = \frac{-x}{2y}<br />$

$\int2ydy = \int-xdx$

$\frac{2y^2}{2} = \frac{-x^2}{2} + C$

Multiply both sides by 2:

$2y^2 = -x^2 + 2C$

What do I do next in order to solve the equation for the original equation?

**Ackbeet** · February 21st 2011, 11:20 AM

Well, you're pretty much done, aren't you? The only thing you could do is absorb the 2 into the C. If you want to solve for y, you could, I suppose. Make sure that when you take your square root, and you get two roots, that the ones you present as your solution actually satisfy the original DE.

**sparky** · February 21st 2011, 01:05 PM

Hi Ackbeet,

On the teacher's answer sheet, I noticed he had the answer as

$X^2 + 2Y^2 = 0$

My question is, what happened to the C (or 2C)?

If I bring everything except the C over, I get $2Y^2 + X^2 = 2C$ . How did he get 0?

**Ackbeet** · February 21st 2011, 01:17 PM

There's only one way to get C=0: you have an initial condition (which would have to be y(0)=0). Otherwise, you must have the C there.

**sparky** · February 21st 2011, 01:22 PM

Ok thanks Ackbeet!

**Ackbeet** · February 21st 2011, 01:24 PM

You're welcome. Have a good one!

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