\(\frac{dv}{dt}=ct\) where \(v(t=0)=55\)
\begin{eqnarray*} dv = ct\;dt\\ \int_{55}^v dv' = \int_0^t ct'\;dt'\\ v-55 = c\frac{t^2}{2}\\ v(t) = \frac{c}{2}t^2+55 \end{eqnarray*}
\(\frac{dp}{dm}=-cp\) where \(p(m=0)=p_0\)
\begin{eqnarray*} \frac{dp}{p} &=& -c\;dm\\ \int_{p_0}^p \frac{dp'}{p'} &=& -\int_0^m c\;m'\\ \ln{p'}\bigg|_{p_0}^{p} &=& -cm\\ \ln{p}-\ln{p_0} &=& -cm \\ \ln{\left(\frac{p}{p_0}\right)} &=& -cm \\ p(m) &=& p_0\;e^{-cm} \end{eqnarray*}
\(\frac{dz}{dk}=-b-cz\) where \(z(k=0)=z_0\)
\begin{eqnarray*} \frac{dz}{b+cz} &=& -dk\\ \int_{z_0}^z \frac{dz'}{b+cz'} &=& -\int_0^k dk'\\ \mbox{let }u &=& b+cz\\ du &=& cdz\\ \int_{b+cz_0}^{b+cz} \frac{du}{u} &=& -\int_0^k dk'\\ \ln{u}\bigg|_{b+cz_0}^{b+cz} &=& -k\\ \ln{b+cz}-\ln{b+cz_0} &=& -k \\ \ln{\left(\frac{b+cz}{b+cz_0}\right)} &=& -k \\ z(k) &=&\left(\frac{b}{c}+z_0\right)\;e^{-k}-\frac{b}{c} \end{eqnarray*}
\(\frac{df}{dg}=-b-cf^2\) where \(f(g=0)= 0\)
This one is way harder and you might not be able to do it for a couple of weeks.
This one is harder because it needs a trigonometric substitution: \begin{eqnarray*} \frac{df}{b+cf^2} &=& -dg\\ \int_{0}^f \frac{df'}{b(1+\frac{c}{b}f'^2)} &=& -\int_0^g dg'\\ \mbox{let }(\sqrt{c/b})f &=& \tan u\\ (\sqrt{c/b})df &=& \sec^2 u\;du\\ \int_{0}^{\arctan(f\sqrt{c/b})} \left(\sqrt{\frac{b}{c}}\right)\frac{1}{b}\frac{\sec^2 u\;du}{1+\tan^2 u} &=& -\int_0^g dg'\\ \int_{0}^{\arctan(f\sqrt{c/b})} \frac{1}{\sqrt{bc}}\frac{\sec^2 u\;du}{\sec^2 u} &=& -\int_0^g dg'\\ \int_{0}^{\arctan(f\sqrt{c/b})} \frac{1}{\sqrt{bc}}du &=& -\int_0^g dg'\\ \frac{1}{\sqrt{bc}} u \bigg|_0^{\arctan(f\sqrt{c/b})} &=& -g \\ \arctan(f\sqrt{c/b}) &=& -\sqrt{bc}\;g \\ f(g) = -\sqrt{\frac{b}{c}}\;\tan{\left(\sqrt{bc}\;g\right)} \end{eqnarray*}
Calculate: Treat Newton's 2nd law as a separable differential equation and solve for the velocity and position as a function of time of an object
I'll start with Newton's Second Law:
\begin{eqnarray*} \Sigma \vec{F} = m\vec{a} = m \frac{d\vec{v}}{dt} \end{eqnarray*} Since there is only one, constant force (\(C\)), set the mass times acceleration equal to a constant. \begin{eqnarray*} \Sigma \vec{F} = C\\ \therefore C = m\frac{dv}{dt} \end{eqnarray*} This equation is separable, so I'll separate of time and velocity and integrate each side. \begin{eqnarray*} C dt = mdv\\ \int_{0}^{t} C dt' = \int_{v_0}^{v}mdv' \\ Ct'\Bigr|_{0}^{t} = mv' \Bigr|_{v_0}^{v}\\ C(t-0) = m(v-v_0)\\ v(t)=\frac{C}{m}t+v_0 \end{eqnarray*} To get the position, I'll recognize that \(v\) is \(\frac{dx}{dt}\). \begin{eqnarray*} \frac{dx}{dt}=\frac{C}{m}t+v_0 \end{eqnarray*} Again, this equation is separable and I'll separate of position and tine and integrate over each side. \begin{eqnarray*} dx=\Bigr(\frac{C}{m}t+v_0\Bigr)dt\\ \int_{x_0}^{x}dx'=\int_{0}^{t}\Bigr(\frac{C}{m}t'+v_0\Bigr)dt'\\ x'\Biggr|_{x_0}^{x}=\Bigr(\frac{C}{m}\frac{t'^2}{2}+v_0t'\Bigr)\Biggr|_{0}^{t}\\ x-x_0=\frac{1}{2}\frac{C}{m}t^2+v_0t\\ x(t)=\frac{1}{2}\frac{C}{m}t^2+v_0t+x_0 \end{eqnarray*}
Reflect: Do your answers look familiar? If yes, from where? If not, how would you have to modify these equations to be similar to equations you know?
\(C/m\) is a constant acceleration. Substituting \(C/m=a\), I get: \begin{eqnarray*} v(t)=at+v_0 \\ x(t) = \frac{1}{2}at^2+v_0t+x_0 \end{eqnarray*} Which are the kinematic equations I learned in introductory physics.