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Thread: Solving for Original Equation

  1. #1
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    Solving for Original Equation

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

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

    Is my working correct?

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

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

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

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

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

    $\displaystyle 8y - 14x + 4x^2 - 2C = 0$
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  2. #2
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    e^(i*pi)'s Avatar
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    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
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  3. #3
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    Quote Originally Posted by sparky View Post
    Question: Solve for the following differential equation for the original equation:

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

    Is my working correct?

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

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

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

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

    $\displaystyle 8y - 14x + 4x^2 - 2C = 0$
    Yes your working is correct. But your answer could be further simplified by dividing the whole equation by 2.

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

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

    $\displaystyle y=\frac{1}{4}\left(7x-2x^2+C\right)$
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  4. #4
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    Thanks for your reply e^(i*pi)

    Yes, you are right, thanks a lot:

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

    $\displaystyle 4y - 7x + 2x^2 - C$
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  5. #5
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    Thank you Sudharaka
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