Solving for Original Equation

**sparky** · February 21st 2011, 03:48 PM

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

$y' = \frac{7-4x}{4}$

Is my working correct?

$\frac{dy}{dx} = \frac{7-4x}{4}$

$4dy = (7-4x)dx$

$\int 4dy = \int 7dx - \int 4xdx$

$4y = 7x - \frac{4x^2}{2} + C$

$8y = 14x - 4x^2 + 2C$ (multiply both sides by 2 to get rid of the half)

$8y - 14x + 4x^2 - 2C = 0$

**e^(i*pi)** · February 21st 2011, 03:49 PM

Yes, although how you didn't see 4/2 = 2 is a mystery. I'd have cancelled in the third line rather than multiplying through by 2.

2C and -2C are both constants and the convention is to either define a new constant or absorb the 2

**Sudharaka** · February 21st 2011, 03:55 PM

Originally Posted by sparky

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

$y' = \frac{7-4x}{4}$

Is my working correct?

$\frac{dy}{dx} = \frac{7-4x}{4}$

$\int 4dy = \int 7dx - \int 4xdx$

$4y = 7x - \frac{4x^2}{2} + C$

$8y = 14x - 4x^2 + 2C$

$8y - 14x + 4x^2 - 2C = 0$

Yes your working is correct. But your answer could be further simplified by dividing the whole equation by 2.

$4y=7x-\frac{4x^2}{2}+C$

$4y=7x-2x^2+C$

$y=\frac{1}{4}\left(7x-2x^2+C\right)$

**sparky** · February 21st 2011, 04:01 PM

Thanks for your reply e^(i*pi)

Yes, you are right, thanks a lot:

$4y = 7x - \frac{4x^2}{2} + C$

$4y - 7x + 2x^2 - C$

**sparky** · February 21st 2011, 04:02 PM

Thank you Sudharaka

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