Solution 1
This solution involves trigonometric substitution and integrals, imaginary numbers, and hyperbolic identities (derivation made by Rigel Melaan).
\(\frac{1}{a v^2-g} d v=d t\)
Integrate both sides of the equation:
\(\int \frac{1}{a v^2-g} d v=\int d t\)
Factor out \(-g\) from the denominator and then the constants:
\(\int-\frac{1}{g\left(1-\frac{a v^2}{g}\right)} d v=\int d t\)
\(-\frac{1}{g} \int \frac{1}{1-\frac{a v^2}{g}} d v=\int d t\)
Apply substitution of variables by defining variables \(u\) and \(du\). The \(\Longrightarrow\) indicates that we can rearrange the expression into another expression.
\(u=i v \sqrt{\frac{a}{g}} \Longrightarrow u^2=-\frac{a v^2}{g}\)
\(\frac{d u}{d v}=i \sqrt{\frac{a}{g}} \Longrightarrow d v=\frac{1}{i \sqrt{\frac{a}{g}}} d u\)
Substitute it back to the previous expression:
\(-\frac{1}{g} \int \frac{1}{1-\frac{a v^2}{g}} d v=\int d t\)
\(-\frac{1}{g} \int \frac{1}{u^2+1} \times \frac{1}{i \sqrt{\frac{a}{g}}} d u=\int d t\)
Factor out constants and use the following identities. Note that \(\wedge\) means "implies." It indicates an identity.
\(-\frac{1}{g} \times \frac{1}{i \sqrt{\frac{a}{g}}} \int \frac{1}{u^2+1} d u=\int d t \wedge \frac{1}{i}=-i\)
\(\frac{i}{a g} \int \frac{1}{u^2+1} d u=\int d t \wedge \int \frac{1}{u^2+1} d u=\arctan u\)
\(\frac{i \arctan u}{\sqrt{a g}}+c=t\)
Substitute \(u\) back and use the \(\operatorname{arctanh}\) identity:
\(\frac{i \arctan i v \sqrt{\frac{a}{g}}}{\sqrt{a g}}+c=t \wedge i \arctan (i u)=-\operatorname{arctanh}(u)\)
\(\frac{-\operatorname{arctanh}\left(v \sqrt{\frac{a}{g}}\right)}{\sqrt{a g}}+c=t\)
Substitute \(a\) back:
\(-\sqrt{\frac{m}{k g}} \operatorname{arctanh}\left(v \sqrt{\frac{k}{m g}}\right)+c=t\)
Rearrange the equation to get the general solution to the ODE:
\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(c \sqrt{\frac{k g}{m}}-t \sqrt{\frac{k g}{m}}\right)\)
Solution 2
This solution involves trigonometric substitution, natural logarithms, and hyperbolic identities.
\(\frac{1}{a v^2-g} d v=d t\)
Integrate both sides of the equation:
\(\int \frac{1}{a v^2-g} d v=\int d t\)
Apply integration by substitution; Let \(v\) and \(dv\) be expressed in terms of \(x\):
\(v=\sqrt{\frac{g}{a}} x\)
\(d v=\sqrt{\frac{g}{a}} d x\)
\(\int \frac{1}{a\left(\frac{g x^2}{a}\right)-g} \times \sqrt{\frac{g}{a}} d x=\int d t\)
Simplify. Eliminate \(a\) in denominator and factor out \(g\) and other constants:
\(\int \frac{1}{g x^2-g} \times \sqrt{\frac{g}{a}} d x=\int d t\)
\(\int \frac{1}{g\left(x^2-1\right)} \times \sqrt{\frac{g}{a}} d x=\int d t\)
\(\frac{1}{\sqrt{a g}} \int \frac{1}{x^2-1} d x=\int d t\)
Resolve the left side into partial fractions:
\(-\frac{1}{2 \sqrt{a g}} \int\left[\frac{1}{1+x}+\frac{1}{1-x}\right] d x=\int d t\)
Integrate both sides of the equation:
\(-\frac{1}{2 \sqrt{a g}}[\ln (1+x)-\ln (1-x)]+c=t\)
Use properties of logarithms to simplify and apply identity:
\(-\frac{1}{\sqrt{a g}}\left[\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\right]+c=t\wedge\frac{1}{2} \ln \left(\frac{1+u}{1-u}\right)=\operatorname{arctanh} u)\)
\(-\frac{\operatorname{arctanh} x}{\sqrt{a g}}+c=t\)
Substitute \(v\) back to equation:
\(x=\sqrt{\frac{a}{g}} v\)
\(-\frac{1}{\sqrt{a g}} \operatorname{arctanh} \sqrt{\frac{a}{g}} v+c=t\)
\(a=\frac{k}{m}\)
\(-\sqrt{\frac{m}{k g}} \operatorname{arctanh}\left(v \sqrt{\frac{k}{m g}}\right)+c=t\)
Rearrange the equation to get the general solution:
\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(c \sqrt{\frac{k g}{m}}-t \sqrt{\frac{k g}{m}}\right)\)
Particular Solution
To find its particular solution, we let \(v=v_o\) when \(t=0\) to simulate free fall body:
\(v(0)=\sqrt{\frac{m g}{k}} \tanh \left(c \sqrt{\frac{k g}{m}}\right)\)
\(c=\sqrt{\frac{m}{k g}} \operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)\)
Substitute \(c\) back to the general solution:
\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
