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Thread: Calculus Help

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