2.2.4.3 Solved using first_order_ode_dAlembert

Entering first order ode dAlembert solver

\[\begin {aligned} x^{\prime }&=k \left (A -n x\right ) \left (M -m x\right ) \end {aligned}\]

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

\begin{align*} p = k \left (-x n +A \right ) \left (-m x +M \right ) \end{align*}

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

\begin{align*} x &= \frac {A k m +M k n +\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\tag {1A}\\ x &= \frac {A k m +M k n -\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\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 &= \frac {A k m +M k n +\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n p \right ) k}}{2 k m n} \end{align*}

Hence (2) becomes

\begin{align*} p = \frac {p^{\prime }\left (t \right )}{\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}\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&=\frac {A k m +M k n +\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n} \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 ) = p \left (t \right ) \sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p \left (t \right )}\tag {4} \end{align*}

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

Integration could not be done or too complicated. Since initial conditions \(\left (t_0,p_0\right ) \) are given, then the result can be written as

\begin{align*} \int _{}^{p \left (t \right )}-\frac {1}{\tau \sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \tau \right ) k}}d \tau +t +c_1 = 0 \end{align*}

Singular solutions are found by solving

\begin{align*} p \sqrt {\left (\left (A m -M n \right )^{2} k +4 m n p \right ) k}&= 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 )&=0\\ p \left (t \right )&=-\frac {\left (A^{2} m^{2}-2 A M m n +M^{2} n^{2}\right ) k}{4 m n} \end{align*}

Solving for \(p\) from above gives

\[ p = \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}-\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_1 \right ) \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n +\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n +\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}-\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_1 \right )\right ) k}}{2 k m n} \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n +\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n +\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n} \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n +\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n +1}{2 k m n} \]

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 &= \frac {A k m +M k n -\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n p \right ) k}}{2 k m n} \end{align*}

Hence (2) becomes

\begin{align*} p = -\frac {p^{\prime }\left (t \right )}{\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}\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&=\frac {A k m +M k n -\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n} \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 ) = -p \left (t \right ) \sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p \left (t \right )}\tag {4} \end{align*}

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

Integration could not be done or too complicated. Since initial conditions \(\left (t_0,p_0\right ) \) are given, then the result can be written as

\begin{align*} \int _{}^{p \left (t \right )}\frac {1}{\tau \sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \tau \right ) k}}d \tau +t +c_2 = 0 \end{align*}

Singular solutions are found by solving

\begin{align*} -p \sqrt {\left (\left (A m -M n \right )^{2} k +4 m n p \right ) k}&= 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 )&=0\\ p \left (t \right )&=-\frac {\left (A^{2} m^{2}-2 A M m n +M^{2} n^{2}\right ) k}{4 m n} \end{align*}

Solving for \(p\) from above gives

\[ p = \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_2 \right ) \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n -\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n -\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_2 \right )\right ) k}}{2 k m n} \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n -\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n -\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n} \]

Substituing the above solution for \(p\) in \(x = \frac {A k m +M k n -\sqrt {A^{2} k^{2} m^{2}-2 A M \,k^{2} m n +M^{2} k^{2} n^{2}+4 k m n p}}{2 k m n}\) gives

\[ x = \frac {A k m +M k n -1}{2 k m n} \]

Summary of solutions found

\begin{align*} x&=\frac {A k m +M k n +\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n}\\ x&=\frac {A k m +M k n +\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}-\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_1 \right )\right ) k}}{2 k m n}\\ x&=\frac {A k m +M k n +1}{2 k m n}\\ x&=\frac {A k m +M k n -\sqrt {\left (A m -M n \right )^{2} k^{2}}}{2 k m n}\\ x&=\frac {A k m +M k n -\sqrt {\left (\left (A m -M n \right )^{2} k +4 m n \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\tau \sqrt {\left (A^{2} k \,m^{2}-2 A M k m n +M^{2} k \,n^{2}+4 m n \tau \right ) k}}d \tau +t +c_2 \right )\right ) k}}{2 k m n}\\ x&=\frac {A k m +M k n -1}{2 k m n} \end{align*}