Thread: Solve Differential equation for the original equation

1. Solve Differential equation for the original equation

Hi,

Is my working correct for the following question?

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

Here is my working:

$\displaystyle $\frac{dy}{dx} = \frac{-x}{2y}$$

$\displaystyle \int2ydy = \int-xdx$

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

Multiply both sides by 2:

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

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

2. 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.

3. Hi Ackbeet,

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

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

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

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

4. 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.

5. Ok thanks Ackbeet!

6. You're welcome. Have a good one!

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