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    Hi please can someone help me out here?

    The question is: -

    Show that the function y = 2x^2 - x satisfies the equation dy/dx + 4y = 8x^2 - 1

    I'm not sure because it's dy/dx plus 4y instead of multiplied by it which is what I have dealt with in the past.

    Thanks very much in advance!
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    Substitute your expressions for dy/dx and y into the left hand side of the equation, then simplify.
    Do you get the right hand side?
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    Just do the left hand side = d/dx[2x^2 - x] +4[2x^2 - x] = 4x - 1 + 8x^2 -4x = 8x^2 -1 = RHS

    So it satisfies the equation
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    Rearange to get f(x)dx=g(y)dy and INTEGRATE.
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    (Original post by limetang)
    Rearange to get f(x)dx=g(y)dy and INTEGRATE.
    The differential equation cannot be written in that form because it is not separable. Plus the question isn't asking him to solve the differential equation but just show that the solution they've given actually works, which is a simple case of plugging it in and showing LHS=RHS.
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    If y=2x^2 - x, work out dy/dx. After doing this, work out 4y by multiplying the equation 'y=2x^2 - x' by 4. Then substitute these values in place of 'dy/dx' and '4y' and you should get the LHS=RHS.
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    Thanks very much. I was looking too much into it I think. I'm worried that I couldn't figure that out by myself lol!!
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    (Original post by chaz1992)
    If y=2x^2 - x, work out dy/dx. After doing this, work out 4y by multiplying the equation 'y=2x^2 - x' by 4. Then substitute these values in place of 'dy/dx' and '4y' and you should get the LHS=RHS.
    This. You beat me to it
 
 
 
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Updated: April 15, 2011
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