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Math Help - DE IVP seperable variable

  1. #1
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    DE IVP seperable variable

    Hi again
    subject:first order linear DE
    Using variable separable method,
    I have no idea on how to start on these 3 problems, please guide me.
    a)dy/dx=(2-y)(2+y),y(0)=-1
    b)dy/dx=y(y-1),y(0)=2
    c)dy/dx=(y-2)(y-4),y(0)=3

    are they solve using the same method? so even if y move to other place,it does not change the method of solving?Thanks in advance
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  2. #2
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    oops it first order DE not linear DE
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  3. #3
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    a) \displaystyle \frac{1}{(2-y)(2+y)}\,\frac{dy}{dx} = 1. Integrate both sides with respect to \displaystyle x.


    b) \displaystyle \frac{1}{y(y-1)}\,\frac{dy}{dx} = 1. Integrate both sides with respect to \displaystyle x.


    c) \displaystyle \frac{1}{(y-2)(y - 4)}\,\frac{dy}{dx} = 1. Integrate both sides with respect to \displaystyle x.
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    i know it to do that but how to integrate the LHS, lets take a for example, it becomes ln|(2-y)(2+y)| or ln(4+y^2)?
    so all 3 uses the same method to solve?if so could you show me 1 example, then i will use it as base for the other 2, thanks in advance
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  5. #5
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    E.g. in the first one you should get to

    \displaystyle \int{\frac{1}{(2-y)(2+y)}\,dy} = \int{1\,dx}.

    You need to use Partial Fractions to evaluate the LHS.
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  6. #6
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    for case a) i manage to get A= 1/4 and B= 1/4 then i stuck again,how to continue from there?
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  7. #7
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    Quote Originally Posted by salohcinseah View Post
    for case a) i manage to get A= 1/4 and B= 1/4 then i stuck again,how to continue from there?
    I guess you mean you have \frac{1}{(2- y)(2+ y)}= \frac{\frac{1}{4}}{2- y}+ \frac{\frac{1}{4}}{2+ y}
    Now, you need to integrate
    \frac{1}{4}\int \frac{dy}{2- y}+ \frac{1}{4}\int\frac{dy}{2+ y}
    If you cannot see how to do that immediately, use substitutions- in the first let u= 2- y and in the second, u= 2+ y.

    May I ask what level course this is? Integration by partial fractions and substitution usually come fairly early in a Calculus course- certainly earlier than solving differential equations!
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  8. #8
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    it basic calculus for university it for engineering degree the thing is ,it sometimes very confusing,these questions are from DE so when i intergrate it it become 1/4 ln|2-y|+1/4 ln |2+y| = x + c then how i can do so ithat i can make c the subject to find it value?
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  9. #9
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    Substitute the values given in your initial condition...
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  10. #10
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    y(0)=-1 but there 2 ln and 2 y and 2 1/4 in the LHS that y i have no idea on how to my it into y= something something +c could you help me by solving it step by step so i can use it as a base to solve the other 2?
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  11. #11
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    You know that when \displaystyle x = 0, y = -1. So wherever you see an \displaystyle x, substitute \displaystyle 0, and wherever you see a \displaystyle y, substitute \displaystyle -1.
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  12. #12
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    arh stupid me, i was thinking of bring all y to RHS so if i put y as -1, all i need to do is bring x to LHS thus i get c, am i right?
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  13. #13
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    Yes. You can probably also use logarithm laws to simplify...
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  14. #14
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    thanks to all whom have helped me
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