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Math Help - Domain of Solution to Differential Equation

  1. #1
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    Domain of Solution to Differential Equation

    \frac{dy}{dx} = \frac{1 + y}{x}, where x does not equal zero

    I was to find the particular solution of this equation with the initial condition f(-1) = 1. I did so and found y = -1 (hopefully this is correct!)

    However, I'm also asked to give the domain of this solution. I'm really confused about when you should and shouldn't restrict domain (when to have absolute value signs on ln, for example), so I don't know how to answer this.

    Help, please?
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  2. #2
    Master Of Puppets
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    Quote Originally Posted by seuzy13 View Post
    \frac{dy}{dx} = \frac{1 + y}{x}, where x does not equal zero

    I was to find the particular solution of this equation with the initial condition f(-1) = 1. I did so and found y = -1 (hopefully this is correct!)

    However, I'm also asked to give the domain of this solution. I'm really confused about when you should and shouldn't restrict domain (when to have absolute value signs on ln, for example), so I don't know how to answer this.

    Help, please?
    What did you find for the solution to y =f(x) ?
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  3. #3
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    Never mind what I said earlier. The solution should be y = -2x -1, but I don't know what the domain should be.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The DE can be written as...

    x\cdot y^{'} - y = 1 (1)

    ... so that it is linear. In similar situation it is requested that the coefficient of the term y^{'} doesn't vanish so that it must be x \ne 0 and that determines the domain of the solutions. The DE can be resolved with the 'standard' method of sepation of variables and we obtain...

    \frac{dy}{1+y} = \frac{dx}{x} \rightarrow \ln (1+y) = \ln x + \ln c \rightarrow y= c\cdot x -1 (2)

    With the 'initial condition' y(-1)=1 the soltution is y= -2\cdot x -1...

    Kind regards

    \chi \sigma
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  5. #5
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    I have looked the problem up elsewhere and found that the domain should be x < 0, but I do not understand why.
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  6. #6
    MHF Contributor chisigma's Avatar
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    The fact is that the general solution is...

    y= c\cdot x -1 (1)

    ... being c an 'arbitrary constant'. That means symply that You are not able to impose any 'initial conditions' in x=0 different from y(0)=-1, but also in this case You able to derive the value of c because all solutions 'pass' from the point [0,-1] ...

    Kind regards

    \chi \sigma
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