# Thread: Bizzare diffeq calculation problems

1. ## Bizzare diffeq calculation problems

I'm learning very basic diffeqs in another math class, and the type of problems were doing are first order separable differential equations, very basic stuff.

I'm trying to use Mathematica/Wolfram Alpha to check my answers, but everytime, wolfram will output the incorrect answer and I will always have the correct answer and I can't figure out why

For example:

$\displaystyle 7\sqrt{xy}\frac{dy}{dx}=4$

I calculated y to be:
$\displaystyle y=(\frac{3}{14}8\sqrt{x}+C)^\frac{2}{3}$

When simplified, this matches the back of my book. Great. When I type the same problem for Wolfram Alpha to solve, I get back this:

Input:
Solve $\displaystyle \frac{dy}{dx}=\frac{4}{7\sqrt{xy}}$

Output:
$\displaystyle y = \frac{3}{14}^\frac{2}{3} (8 \sqrt{x} + 7*C)^\frac{2}{3}$

Which is completely wrong. Could this be a bug in the CAS?

2. Originally Posted by Cluuia09320
I'm learning very basic diffeqs in another math class, and the type of problems were doing are first order separable differential equations, very basic stuff.

I'm trying to use Mathematica/Wolfram Alpha to check my answers, but everytime, wolfram will output the incorrect answer and I will always have the correct answer and I can't figure out why

For example:

$\displaystyle 7\sqrt{xy}\frac{dy}{dx}=4$

I calculated y to be:
$\displaystyle y=(\frac{3}{14}8\sqrt{x}+C)^\frac{2}{3}$

When simplified, this matches the back of my book. Great. When I type the same problem for Wolfram Alpha to solve, I get back this:

Input:
Solve $\displaystyle \frac{dy}{dx}=\frac{4}{7\sqrt{xy}}$

Output:
$\displaystyle y = \frac{3}{14}^\frac{2}{3} (8 \sqrt{x} + 7*C)^\frac{2}{3}$

Which is completely wrong. Could this be a bug in the CAS?

$\displaystyle = \left(\frac{3}{14} \cdot 8 \sqrt{x} + \frac{3}{14} \cdot 7 \cdot C\right)^\frac{2}{3} = \left(\frac{3}{14} \cdot 8 \sqrt{x} + B \right)^\frac{2}{3}$

where the B here is just as arbitrary as your C.

3. Originally Posted by mr fantastic

$\displaystyle = \left(\frac{3}{14} \cdot 8 \sqrt{x} + \frac{3}{14} \cdot 7 \cdot C\right)^\frac{2}{3} = \left(\frac{3}{14} \cdot 8 \sqrt{x} + B \right)^\frac{2}{3}$

where the B here is just as arbitrary as your C.
But the answers can't be equivalent...

If we let the constant = 4.5 in both equations and we let x = 7 in both equations and solve for y

$\displaystyle 4.3381=(\frac{3}{14}8\sqrt{7}+4.5)^\frac{2}{3}$
$\displaystyle 5.0313= \frac{3}{14}^\frac{2}{3} (8 \sqrt{7} + 7*4.5)^\frac{2}{3}$

Both equations are not equal

4. Originally Posted by Cluuia09320
But the answers can't be equivalent...

If we let the constant = 4.5 in both equations and we let x = 7 in both equations and solve for y

$\displaystyle 4.3381=(\frac{3}{14}8\sqrt{7}+4.5)^\frac{2}{3}$
$\displaystyle 5.0313= \frac{3}{14}^\frac{2}{3} (8 \sqrt{7} + 7*4.5)^\frac{2}{3}$

Both equations are not equal
They aren't literally equal but they are both solutions. Differentiate both.

5. Originally Posted by Cluuia09320
But the answers can't be equivalent...

If we let the constant = 4.5 in both equations and we let x = 7 in both equations and solve for y

$\displaystyle 4.3381=(\frac{3}{14}8\sqrt{7}+4.5)^\frac{2}{3}$
$\displaystyle 5.0313= \frac{3}{14}^\frac{2}{3} (8 \sqrt{7} + 7*4.5)^\frac{2}{3}$

Both equations are not equal
I say again: "One of your mistakes is in assuming that the C in your answer is the same as the C in Wolfram's answer."

You continue to make that mistake by letting "the constant = 4.5 in both equations."

In my previous reply I showed you why the answers were equivalent (equivalent does not mean equal, by the way).

Calculate the arbitrary constant in each solution for the initial condition y = 1, x = 0 and you will discover that each gives the same rule for y (provided you do this correctly, of course).