Problem:
Solve the following differential equation:$$ \frac{ d^2 s }{dt^2} = g(1 -s^2) $$with the following initial condition:\begin{align*}\dfrac{ ds}{dt} \left( 0 \right) &= 0 \\s(0) &= \dfrac{1}{4} \\\end{align*}and $g$ is a positive constant.
Note: This differential equation is from a physics problem I am working on.
Answer:
\begin{align*}\frac{ d^2 s }{dt^2} &= g(1 -s^2) \\\frac{ d^2 s} {dt^2} + gs^2 &= g \\\frac{ d^2 s} {dt^2} &= g - gs^2 \\d^2s &= g (1- s^2) \,\, dt^2 \\\dfrac{ d^2 s}{1-s^2} &= g \,\, dt^2 \\\end{align*}Now we integrate both sides. To integrate the left side, we apply the technique of partial fractions.\begin{align*}\dfrac{ 1}{1-s^2} &= \dfrac{A}{1-s} + \dfrac{B}{1+s} \\1 &= A(1+s) + B(1-s) \\A + B &= 1 \\A - B &= 0 \\A &= B \\A &= \dfrac{1}{2} \\B &= \dfrac{1}{2} \\\end{align*}Hence we have:$$ \int \dfrac{ 1 }{1-s^2} \,\, ds= \dfrac{1}{2} \ln{|1-s|} + \dfrac{1}{2} \ln{|1+s|} + C_1 $$Going back to the differential equation we have:\begin{align*}\left( \dfrac{1}{2} \ln{|1-s|} + \dfrac{1}{2} \ln{|1+s|} \right) \,\, ds&= gt \,\, dt + C \\%\dfrac{ds}{dt} &=\dfrac{ gt }{ \dfrac{1}{2} \ln{|1-s|} + \dfrac{1}{2} \ln{|1+s| } } + C \\\end{align*}We know that $\dfrac{ ds }{dt}\left( 0 \right) = 0$. We also know that when $t = 0$that $s = 1/4$. Hence we eliminate the constant.\begin{align*}0 &=\dfrac{ g(0) }{ \dfrac{1}{2} \ln{|1-\dfrac{1}{4}|}+ \dfrac{1}{2} \ln{|1+\dfrac{1}{4}| } } + C \\0 &= 0 + C \\C &= 0 \\\dfrac{ds}{dt} &= \dfrac{ gt }{ \dfrac{1}{2} \ln{|1-s|} + \dfrac{1}{2} \ln{|1+s| } } \\\end{align*}Now we have the following first order differential equation.$$ \ln{|1-s|} + \ln{| 1+s | } \,\,\, ds = 2gt \,\, dt $$Now we need to integrate both sides. However, there is a problem when$s = 1$ because $\ln{0}$ is not defined. What did I do wrong?
