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Math Help - Solve a Differential Equation

  1. #1
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    Solve a Differential Equation

    The equation is dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't seem to get the equation from the 1st form to the 2nd for some reason. I haven't taken calc in over a year so I'm a bit rusty.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfgiants13 View Post
    The equation is dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't seem to get the equation from the 1st form to the 2nd for some reason. I haven't taken calc in over a year so I'm a bit rusty.
    \frac{dy}{dt}=-ay+b=-a\left(\frac{-ay}{-a}+\frac{b}{-a}\right)=-a\left(y-\frac{b}{a}\right) cross multiplying gives \frac{dy}{y-\frac{b}{a}}=-a\text{ }dt. Integrate both sides.
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  3. #3
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    dy/dt=-ay+b where a and b are positve numbers. I have the solutions guide and it says that equation also equals dy/dt=-a(y-(b/a)) and then dy/(y-b/a)=-a. I can't


     \frac{dy}{dt} = -ay+b

    Can be written as

     \frac{dy}{dt} = -ay+\frac{-a}{-a}\times b

    taking out -a as a factor

     \frac{dy}{dt} = -a\left(y+\frac{b}{-a}\right)

    Then separating

     \frac{dy}{\left(y+\frac{b}{-a}\right)} = -a~dt

    Now integrate both sides
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    Thanks guys I get that now. Next part asks to sketch the solution for several different initial conditions. Is that just a matter of plugging in real numbers for a and b and sketching them?
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    Quote Originally Posted by sfgiants13 View Post
    Thanks guys I get that now. Next part asks to sketch the solution for several different initial conditions. Is that just a matter of plugging in real numbers for a and b and sketching them?
    No. As far as I can see, a and b have nothing to do with the initial conditions.

    I hope you realise that when you integrate, an arbitrary constant of integration is introduced. The value of that constant can be determined if intitial conditions are given eg. y = 1 when t = 0.

    By the way, if you're studying differential equations you better make it your urgent business to rust rid your calculus.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Note for dy/dt =- ay + b

    y = b/a is an equiilbrium solution.

    your initial conditions do depend on a and b in the sense that if y(0) < b/a

    then the solution curve approaches the line y= b/a from below.

    If y(0) > b/a then the solution curve decays to the line y =b/a from above.
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