Exponential Model

Goal: given \(\frac{dy}{dx}\propto y\), \(y(0)=y_0\), \(y(T)=y_T\) and \(T\ne0\), find \(y(x)\) in terms of \(y_0\), \(y_T\) and \(T\).

dydxydydx=ky1ydy=kdxlny=kx+Cy=exp(kx+C)y=exp(C)exp(kx)y(x)=Cexp(kx)\begin{aligned} \frac{dy}{dx}&\propto y\\ \frac{dy}{dx}&=ky\\ \int \frac1y\,dy&=\int k\,dx\\ \ln|y|&=kx+C\\ y&=\exp(kx+C)\\ y&=\exp(C)\exp(kx)\\ y(x)&=C\exp(kx)\\ \end{aligned}

y(x)=Cexp(kx)y(0)=Cexp(k(0))y0=Cexp(0)y0=C(1)y0=C\begin{aligned} y(x)&=C\exp(kx)\\ y(0)&=C\exp(k(0))\\ y_0&=C\exp(0)\\ y_0&=C(1)\\ y_0&=C\\ \end{aligned}

y(x)=Cexp(kx)y(x)=y0exp(kx)y(T)=y0exp(kT)yTy0=exp(kT)ln(yTy0)=kT1Tln(yTy0)=k\begin{aligned} y(x)&=C\exp(kx)\\ y(x)&=y_0\exp(kx)\\ y(T)&=y_0\exp(kT)\\ \frac{y_T}{y_0}&=\exp(kT)\\ \ln\left(\frac{y_T}{y_0}\right)&=kT\\ \frac1T\ln\left(\frac{y_T}{y_0}\right)&=k\\ \end{aligned}

y(x)=Cexp(kx)y(x)=y0exp(1Tln(yTy0)x)y(x)=y0(exp(ln(yTy0)))1Tx\begin{aligned} y(x)&=C\exp(kx)\\ y(x)&=y_0\exp\left(\frac1T\ln\left(\frac{y_T}{y_0}\right)x\right)\\ y(x)&=y_0\left(\exp\left(\ln\left(\frac{y_T}{y_0}\right)\right)\right)^{\frac1Tx}\\ \end{aligned}

y(x)=y0(yTy0)xT\therefore\boxed{y(x)=y_0\left(\frac{y_T}{y_0}\right)^{\frac xT}}

#Calculus

Note by Gandoff Tan
1 year, 4 months ago

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