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  1. #1
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    Calculus Help

    Consider the differential equation
    dy/dx= 3x^2/e^2y

    a) Find a solution y=f(x) to the differential equation satisfying f(0)=1/2

    b) Find the domain and range of the function f found in part a.
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  2. #2
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    Quote Originally Posted by frozenflames
    Consider the differential equation
    dy/dx= 3x^2/e^2y

    a) Find a solution y=f(x) to the differential equation satisfying f(0)=1/2

    b) Find the domain and range of the function f found in part a.
    You have,
    \frac{\bold{dy}}{\bold{dx}}=\frac{3x^2}{e^{2y}}
    This, is a 'seperable differencial equation' .
    Thus, express as,
    e^{2y}\bold{dy}=3x^2\bold{dx}
    Thus, integrating,
    \int e^{2y}\bold{dy}=\int 3x^2\bold{dx}.
    Thus,
    \frac{1}{2}e^{2y}=x^3+C
    Thus,
    e^{2y}=2x^3+C
    Thus,
    2y=\ln (2x^3+C)
    Thus,
    y=\frac{1}{2}\ln (2x^3+C)
    --------------
    Now, we have that f(0)=1/2
    Thus,
    \frac{1}{2}\ln (2\cdot 0^3+C)=1/2
    Thus,
    C=e
    Meaning, the function is,
    y=\frac{1}{2}\ln (2x^3+e)
    --------------
    The domain of y=\frac{1}{2}\ln (2x^3+e)
    is what values x can take. Notice the logarithm, which is only defined for positive real numbers. Thus,
    2x^3+e>0
    Solving this inequality, and rationalizing the root
    x>-\sqrt[3]{1/2\cdot e}=-\frac{1}{2}\sqrt[3]{4e}.
    --------------
    The range for y=\frac{1}{2}\ln (2x^3+e)
    is the values or y such as, this equation is a solution for x. Notice, that this equation has a solution for any y thus, the range is all real numbers.
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