2.1.10.4 Solved using first_order_ode_dAlembert

Entering first order ode dAlembert solver

\[\begin {aligned} x^{\prime }&=\sqrt {x^{2}-1}\\ x \left (0\right ) &= 1\\ \end {aligned}\]

Let \(p=x^{\prime }\) the ode becomes

\begin{align*} p = \sqrt {x^{2}-1} \end{align*}

Solving for \(x\) from the above results in

\begin{align*} x &= \sqrt {p^{2}+1}\tag {1A}\\ x &= -\sqrt {p^{2}+1}\tag {2A} \end{align*}

This has the form

\begin{align*} x&=t f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=x'(t)\). Each of the above ode’s is dAlembert ode which are now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(t\) gives

\begin{align*} p &= f+(t f'+g') \frac {dp}{dt}\\ p-f &= (t f'+g') \frac {dp}{dt}\tag {2} \end{align*}

Comparing the form \(x=t f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= \sqrt {p^{2}+1} \end{align*}

Hence (2) becomes

\begin{align*} p = \frac {p p^{\prime }\left (t \right )}{\sqrt {p^{2}+1}}\tag {3} \end{align*}

The singular solution is found by setting \(\frac {dp}{dt}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} x&=1 \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}t}}\neq 0\) from eq. (3). This results in

\begin{align*} p^{\prime }\left (t \right ) = \sqrt {p \left (t \right )^{2}+1}\tag {4} \end{align*}

This ODE is now solved for \(p \left (t \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {p^{2}+1}}d p &= dt\\ \operatorname {arcsinh}\left (p \right )&= t +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {p^{2}+1}&= 0 \end{align*}

for \(p \left (t \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (t \right )&=-i\\ p \left (t \right )&=i \end{align*}

Solving for \(p\) from above gives

\[ p = \sqrt {\sinh \left (t +c_1 \right )^{2}+1} \]

Substituing the above solution for \(p\) in \(x = \sqrt {p^{2}+1}\) gives

\[ x = \sqrt {\sinh \left (t +c_1 \right )^{2}+1} \]

Initial condition \(x \left (0\right ) = 1\) is now applied. Applying the initial condition \(x \left (0\right ) = 1\), the solution becomes

\begin{align*} x &= \sqrt {\sinh \left (t \right )^{2}+1} \\ \end{align*}

Substituing the above solution for \(p\) in \(x = \sqrt {p^{2}+1}\) gives

\[ x = 0 \]

Substituing the above solution for \(p\) in \(x = \sqrt {p^{2}+1}\) gives

\[ x = 0 \]

Solving ode 2A

Taking derivative of (*) w.r.t. \(t\) gives

\begin{align*} p &= f+(t f'+g') \frac {dp}{dt}\\ p-f &= (t f'+g') \frac {dp}{dt}\tag {2} \end{align*}

Comparing the form \(x=t f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= -\sqrt {p^{2}+1} \end{align*}

Hence (2) becomes

\begin{align*} p = -\frac {p p^{\prime }\left (t \right )}{\sqrt {p^{2}+1}}\tag {3} \end{align*}

The singular solution is found by setting \(\frac {dp}{dt}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}t}}\neq 0\) from eq. (3). This results in

\begin{align*} p^{\prime }\left (t \right ) = -\sqrt {p \left (t \right )^{2}+1}\tag {4} \end{align*}

This ODE is now solved for \(p \left (t \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {p^{2}+1}}d p &= dt\\ -\operatorname {arcsinh}\left (p \right )&= t +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {p^{2}+1}&= 0 \end{align*}

for \(p \left (t \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (t \right )&=-i\\ p \left (t \right )&=i \end{align*}

Solving for \(p\) from above gives

\[ p = -\sqrt {\sinh \left (t +c_2 \right )^{2}+1} \]

Substituing the above solution for \(p\) in \(x = -\sqrt {p^{2}+1}\) gives

\[ x = -\sqrt {\sinh \left (t +c_2 \right )^{2}+1} \]

Initial condition \(x \left (0\right ) = 1\) is now applied. Unable to resolve initial conditions. Substituing the above solution for \(p\) in \(x = -\sqrt {p^{2}+1}\) gives

\[ x = 0 \]

Substituing the above solution for \(p\) in \(x = -\sqrt {p^{2}+1}\) gives

\[ x = 0 \]

Direction field with Isoclines	Direction field and Solutions plot

Summary of solutions found

\begin{align*} x&=1\\ x&=\sqrt {\sinh \left (t \right )^{2}+1}\\ x&=0 \end{align*}