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Math Help - Right or wrong -- Help me !!!

  1. #1
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    Right or wrong -- Help me !!!

    * A function y(t) satisfies the differential equation \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}

    a. What are the constant solutions of the equation ?
    My answer : Solution :  y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}, is not depend t , therefore y are the constant solutions of the equation.


    Right or wrong ?

    b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?
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  2. #2
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    Quote Originally Posted by butbi9x View Post
    * A function y(t) satisfies the differential equation \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}

    a. What are the constant solutions of the equation ?
    My answer : Solution :  y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}, is not depend t , therefore y are the constant solutions of the equation.


    Right or wrong ?

    b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?
    You'll notice that your equation can be written as

    \frac{dy}{dt}=y^2(y-1)(y-5)

    Now the question asks, for what constants (i.e. y = c) is this equation satisfied? Now between these constants either y' > 0 increasing or y' < 0 decreasing.
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  3. #3
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    Quote Originally Posted by butbi9x View Post
    * A function y(t) satisfies the differential equation \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}

    a. What are the constant solutions of the equation ?
    My answer : Solution :  y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}, is not depend t , therefore y are the constant solutions of the equation.


    Right or wrong ?
    Sorry, but that makes no sense at all. You cannot simply write an equation with only y and deduce from the equation that y does not depend on t!
    Perhaps more importantly, you integrated the left side with respect to t, \int (\frac{dy}{dt})dt= y but integrated the right side with respect to y. You can't do that! If y is a constant then dy/dx= 0. Solve y^4- 6y^3+ 5y^2= 0. As danny arrigo pointed out, that can be easily factored.

    b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?
    No, you are not. y will be increasing when dy/dx> 0 and, since you can factor the right hand side, that will happen when an even number of those factors are negative.

    y^2(y-1)(y-5) can change sign only at y= 0, y= 1, y= 5. If y< 0, all 4 factors will be negative so the product is positive. If y is between 0 and 1, two factors will be negative so the product is still positive. If y is between 1 and 5, only y- 5 is negative so the product is negative. If y> 5, none of the factors is negative so the product is positive.
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  4. #4
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    However, if the equation were

    \frac{dy}{dt}=t^4-6t^3+5t^2

    then we would have

    y=\frac{t^5}{5}-\frac{6t^4}{4}+\frac{5t^3}{3}+C.

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  5. #5
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    True- but it wasn't!
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