### 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:

\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)