Mixing Tank Separable Differential Equations Examples
Yet again, this equation is clearly separable, since there is no t variable on the right hand side. We thus solve in the standard way: Z dS −0.1S +0.1 = Z dt −10ln|− 0.1S +0.1| = t+C −0.1S +0.1 = Ce−0.1t S = 1+Ce−0.1t. Here C may be positive or negative, depending on the initial conditions. We see that as t
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