If $\;\; \displaystyle {dy\over dt} = 2y(12-3y),\;$ and $y(0)=1$, what is the maximum value?
Find the equation for the orthogonal family $y= Ce^{5x}.$
Find the equation of the orthogonal family
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1 Answers
1) $\displaystyle \frac{dy}{dt} = 2y (12-3y) \\ \displaystyle \int \frac{dy}{y(4-y)} = 6\int dt \\ \displaystyle \frac 14 \int \left [ \frac 1y + \frac 1{4-y} \right] = 6 \int dt \\ \displaystyle \ln y + \ln (4-y) = 24t + C \\ \displaystyle y (4-y) = D e^{24t}$
It's implicit equation for $y$. You need to find $D$ by substituting $y(0) = 1$, so $D = 1$ and $y(4-y) = e^{24t}$. This equation doesn't define a function, so you cannot use concepts applicable for functions to $y$, such as extrema. This equation has two branches, and each of them is a function.
2) $\displaystyle y = f(x) = C e^{5x} \\ \frac {df}{dx} = 5C e^{5x}$
Let's use $g(x)$ for orthogonal family, so
$\displaystyle \frac{df}{dx} \frac{dg}{dx} = -1 \\ \displaystyle \frac {dg}{dx} = -\frac 1{5C} e^{-5x} \\ \displaystyle g(x) = \frac 1{25C} e^{-5x}$