# Calculus Help

• Apr 3rd 2006, 03:41 PM
frozenflames
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.
• Apr 3rd 2006, 06:26 PM
ThePerfectHacker
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,
$y=\frac{1}{2}\ln (2x^3+C)$
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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,
$y=\frac{1}{2}\ln (2x^3+e)$
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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
$x>-\sqrt[3]{1/2\cdot e}=-\frac{1}{2}\sqrt[3]{4e}$.
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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.