# Thread: Solving for Original Equation

1. ## Solving for Original Equation

Is my working and answer correct for the following question?

Question: Solve for the following differential equation for the original equation:

$\displaystyle y' = \frac{2x-3y}{3x+8y}$

Here is my working:

$\displaystyle \frac{dy}{dx} = \frac{2x-3y}{3x+8y}$

$\displaystyle (3x+8y)dy = (2x-3y)dx$

$\displaystyle \int 3xdy + \int 8ydy = \int 2xdx - \int 3ydx$ (why does the intergral of 3ydx = 0?)

$\displaystyle \frac{3x^2}{2} + \frac{8y^2}{2} = \frac{2x^2}{2} - 0 + C$

$\displaystyle 3x^2 + 8y^2 - 2x^2 - C = 0$

$\displaystyle x^2 + 8y^2 - C = 0$

1. This method doesn't work, since you don't know what y is a priori. If you did, the DE would be solved!
2. The DE is homogeneous. Use the appropriate technique.

3. Originally Posted by sparky
Is my working and answer correct for the following question?

Question: Solve for the following differential equation for the original equation:

$\displaystyle y' = \frac{2x-3y}{3x+8y}$

Here is my working:

$\displaystyle \frac{dy}{dx} = \frac{2x-3y}{3x+8y}$

$\displaystyle (3x+8y)dy = (2x-3y)dx$

$\displaystyle \int 3xdy + \int 8ydy = \int 2xdx - \int 3ydx$ (why does the intergral of 3ydx = 0?)

$\displaystyle \frac{3x^2}{2} + \frac{8y^2}{2} = \frac{2x^2}{2} - 0 + C$

$\displaystyle 3x^2 + 8y^2 - 2x^2 - C = 0$

$\displaystyle x^2 + 8y^2 - C = 0$
Just like in your other problems, it's a matter of grouping.
$\displaystyle (3x + 8y) dy = (2x - 3y) dx$

$\displaystyle 3x dy + 8y dy = 2x dx - 3y dx$

Regroup:
$\displaystyle (3x dy + 3y dx) + 8y dy = 2x dx$

Factor the 3:
$\displaystyle 3 (x dy + y dx) + 8y dy = 2x dx$

Now I showed you how to handle x dy + y dx in a different thread. Do you remember what I did?

-Dan